Structural Analysis

STRUCTURAL ANALYSIS

For CIVIL ENGINEERING

STRUCTURAL ANALYSIS SYLLABUS Structural Analysis: Statically determinate and indeterminate structures by force/ energy methods; Method of superposition; Analysis of trusses, arches, beams, cables and frames; Displacement methods: Slope deflection and moment distribution methods; Influence lines; Stiffness and flexibility methods of structural analysis. ANALYSIS OF GATE PAPERS Exam Year 1 Mark Ques. 2 Mark Ques. Total 2003 2 - 12 2004 3 - 9 2005 2 - 10 2006 - 3 6 2007 1 4 9 2008 - 3 6 2009 - 1 2 2010 1 1 3 2011 - - - 2012 - - - 2013 1 3 7 2014 Set-1 2 2 6 2014 Set-2 1 1 3 2015 Set-1 1 1 3 2015 Set-2 1 1 3 2016 Set-1 1 1 3 2016 Set-2 1 1 3 2017 Set-1 1 3 7 2017 Set-2 1 1 3 2018 Set-1 2018 Set-2 1 1 3 7 1 3

Topics Page No.

1. STATIC INDETERMINACY & KINEMATIC INDETERMINACY

1.1 Structure Analysis 1.2 Structure 1 1.3 Scope of Structural Analysis 011 1.4 Idealization of Structure 01 1.5 Structural Stability 01 1.6 Support Reactions 02 1.7 Equation of Static Equilibrium 0 1.8 Degree of Static Indeterminacy 0 1.9 Kinematic Indeterminacy 02 1.10 Degree of Freedom of Various Joints 03 Gate Questions 04 04 06 Assignment 10 2. ANALYSIS OF TRUSS

2.1 Introduction 1 2.2 Assumption for Design of Truss Members and Connection 2.3 Method of Analysis 15 2.4 Zero Force Members 151 2.5 Truss with Internal Hinge 15 Gate Questions 16 7 8 Assignment 28 3. ARCH

3.1 Introduction 3.2 Types of Arches 3.3 Advantages of Arch over Beam 35 35 35

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 3.4 Three Hinged Arches 3.5 Two Hinged Arches Gate Questions 35 Assignment 361 39 4. INFLUENCE LINE DIAGRAM 4

4.1 Introduction 4.2 Sign Convention 4.3 Maximum S.F and B.M values due to Moving Loads 49 4.4 Beam Subjected to a Single Point Load W 49 4.5 Uniformly Distributed Load Longer Than the Span 49 4.6 Uniformly Distributed Load Smaller Than the Span 49 4.7 A Train of Concentrated Load 501 Gate Questions 50 Assignment 52 55 5. MOMENT DISTRIBUTION METHOD 6

5.1 Introduction 5.2 Basic Concept 5.3 Sign Convention 69 5.4 Stiffness 69 5.5 Distribution Factor 69 5.6 Carry Over Factor 69 5.7 Distribution Factor for Fixed and Pin Ends 69 5.8 Causes of Sway of Frames 70 Gate Questions 702 Assignment 71 7 6. SLOPE DEFLECTION METHOD 79

6.1 Introduction 6.2 Fundamental of Slope Deflection 6.3 Sign Convention 84 6.4 Equilibrium Equation 85 6.5 Steps for Analysis in Slope Deflection Method 85 6.6 Examples of Fixed End Moments 85 6.7 Shear Equation in Frame 86 Assignment 86 87 88

7.1 Introduction 7.2 Force Method 7.3 Examples of Force Method 90 7.4 Displacement Method 90 7.5 Examples of Displacement Methods 91 7.6 Various Force Methods 912 Gate Questions 91 Assignment 9 96 8. MATRIX METHOD OF ANALYSIS 98

8.1 Definition and Notations 1 8.2 Types of Matrix 1 8.3 Definition and Properties of Determinants 1012 8.4 Properties of Determinant 1012 8.5 Basic Matrix Operations 102 8.6 Development of Flexibility Matrix 10 8.7 Development of Stiffness Matrix 10 Gate Questions 104 04 06

1.1 Structure Analysis: It is the application of solid mechanics to predict the response (in terms of force and displacements) of given structure subjected to specified load.

1.2 Structure: It may be defined as an assemblage of load bearing elements in a construction.

1.3 Scope of structural Analysis: Determination of response quantities in terms of support reactions, internal forces (SF, BM, torsion and axial force) and displacements (deflections and rotations).

1.4 Idealization of structure:

(A) Line Elements (Skeletal): When one dimension of member is very large compare to other two dimensions, it is idealized as line element. Ex. Truss elements, beams, columns. (B) Surface Elements (Area element): When two dimensions of member are very large compare to other two dimensions, it id idealized as line element. Ex. Slab, Shear wall, Shell. (C) Solid Elements: When all three dimensions of member are large, it is known as Solid elements. Ex. Massive foundation.

1.5 Structural stability: If structure deforms elastically and immediate elastic restrained is developed under the action of external load, structure is called stable.

NOTE: Structure is unstable externally if all unknown reactions are PARALLEL or CONCURRENT.

Determinate and unstable Indeterminate and unstable (Reactions are parallel) (Reactions are parallel)

Determinate and unstable Indeterminate and unstable

(Reactions are concurrent) (Reactions are concurrent)

Types of support y x Reactions z Plane structure Space structure Fixed support Fx, Fy, Mz Fx, Fy, Fz, Mx, My, Mz

Hinged or Pinned support Fx, Fy Fx, Fy, Fz

Roller support Fy Fy

Vertical Guided fixed support Fx, Mz Fx , Mx, My, Mz

Horizontal guided fixed support Fy, Mz Fy , Mx, My, Mz OR Horizontal shear release support

1.7 Equation of Static Equilibrium

• In a 2-D structure or planer structure (in which all members and forces are in plane only) the equations of equilibrium are ∑=F0 X  ∑=FY 0 3 nos.  ∑=M0 • In a. 3-D structures or space structures (in which members and forces are in 3rD). The equations of equilibrium are

∑=FXX 0 ∑ M = 0

∑=FYY 0 ∑ M = 0} 6 nos.

∑=FZZ 0 ∑ M = 0

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 1.8 Degree of Static Indeterminacy (DS ) : If all reactions of structure cannot be determine by using static equilibrium equations alone, it is called statically indeterminate structure.

Note: In this case additional equations needed which are obtained by relating the applied loads & reactions to the displacements or slopes known at different points on the structure. These equations are called Compatibility equations. D s=Dse + Dsi

Where, Dse =External indeterminacy Dsi =internal indeterminacy

Rigid jointed plane Frame Rigid jointed space Frame

Dse = R – r – r’ Dse = R – r Dsi = 3C Dsi = 6C Where, Where, R = No of unknown reactions R = No of unknown reactions r = No of equilibrium equations = 3 r = No of equilibrium equations = 6 C = No of closed loops C = No of closed loops r’ = No of extra equilibrium equations because of internal hinge or link OR OR Ds = 3m + R – 3j Ds = 6m + R – 6j Where, Where, m = No of members m = No of members j = No of joints j = No of joints

Pinned jointed plane Frame (Plane Truss) Pinned jointed space Frame (Space Truss) Dse = R – r Dsi = m + r – 2j Dse = R – r Where, Dsi = m + r – 3j R = No of unknown reactions Where, r = No of equilibrium equations = 3 R = No of unknown reactions r = No of equilibrium equations = 6 Ds = m + R – 2j Ds = m + R – 3j

NOTE: Internal hinge: If internal hinge is provided anywhere in the structure, one part of structure cannot transmit moment from one part to other part which provides extra

r’ = no of member connecting at internal hinge – 1 conditional equation, i.e, ∑M = 0.

structure cannot transmit moment & horizontal force from one part to other part which

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Rigid jointed Rigid jointed Pinned Pinned Stability & plane Frame space Frame jointed jointed space Indeterminancy plane Frame Frame R – r < R – r < m + R < 2j m + R < 2j Unstable & determinate R – r = R – r = m + R = 2j m + R = 2j Stable & determinate R – r > 0 R – r > 0 m + R > 2j m + R > 2j Stable & indeterminate 0 0 1.9 0 Kinematic Indeterminacy0 (Degree of freedom): The degree of kinematic indeterminacy may be defined as the total number of degree of freedom at various joints in skeletal structure.

Note : To determine no of unknown DOF, one need same no of compatibility equations.

Rigid jointed plane Frame Rigid jointed space Frame

Dk = 3j – R + r’ Dk = 6j – R

Where, Where, R = No of unknown reactions R = No of unknown reactions m = No of members m = No of members j = No of joints j = No of joints r’ = No of DOF because of internal hinge r’ = No of DOF because of internal hinge Pinned jointed plane Frame (Plane Pinned jointed space Frame (Space Truss) Truss)

Dk = 2j – R Dk = 3j – R

Dknad = KI neglecting axial deformation

= Dk - m

1.10 Degree of freedom of various joints:

Types of support y x DOF Description z

Fixed support

0 0

Hinged or Pinned support 1 θ1

Vertical Guided fixed support 1 δy

Horizontal guided fixed support 1 δx OR Horizontal shear release support

4 δy1, δy2, δx, θ1

4 δx1, δx2, δy, θ1

NOTE: Internal hinge: If internal hinge is provided anywhere in the structure,

DOF = No of member connected at internal hinge + 2

Q.1 The following two statements are and the degree of kinematic made with reference to the planar indeterminacy, k, are truss shown below:

a) I. The truss is statically determinate b)d = 3 and k = 13 II. The truss is kinematically c)d = 3 and k = 10 determinate d)d = 9 and k =13 With reference to the above d = 9 and k = 10 [GATE – 2004] statements, which of the following Considering beam as axially rigid, applies? Q.4 the degree of freedom of a plane a)Both statements are true frame shown below is b)Both statements are false c)II is true but I false d)I is true but II is false [GATE- 2000] a) 9 b) 8 Q.2 The degree of static c) 7 d) 6 indeterminacy, Ns and the degree [GATE – 2005] of kinematic indeterminacy, Nk for the plane frame shown below, Q.5 The degree of static indeterminacy of assuming axial deformations to the rigid frame having two internal be negligible, are given by: hinges as shown in the figure below, is

a) Ns = 6 and Nk = 11 b) Ns = 6 and Nk = 6 c) Ns = 4 and Nk = 6 a) 8 b)7 d) Ns = 4 and Nk = 4 c) 6 d) 5 [GATE-2001] [GATE-2008] Q.3 For the plane frame with an Q.6 The degree of static indeterminacy overhang as shown below, assuming of a rigidly jointed frame in a negligible axial deformation, the horizontal plane and subjected to degree of static indeterminacy, d,

b c d a) ∞, 0, ∞ )[GATE ∞, ∞, -∞2015] ) 0, ∞, ∞ ) ∞, ∞, 0 a) 6 b)4 Q.10 The kinematic indeterminacy of the c) 3 d) 1 plane truss shown in the figure is [GATE-2009]

Q.7 The degree of static indeterminacy of a rigid jointed frame PQR supported as shown in the figure is

a) 11 b) 8 c) 3 d [GATE-2016] Q.11 A planar tower shown in) figure 0 .

a) Zero b) One c) Two d) Unstable [GATE-2014] Consider the following statements for Q.8 The static indeterminacy of the two- external and internal determinacies of truss. span continuous beam with an (P) Externally determinate internal hinge, shown below, is _____ (Q) External SI = 1 (R) External SI = 2 (S) Internally determinate. (T) Internal SI = 1 [GATE-2014] (U) Internal SI = 2 Which one of following options is correct? Q.9 A guided support as shown in the P Q R S T U figure below is represented by three A False True False False False True springs (horizontal, vertical and B False True False False True False rotational) with stiffness kx, ky and C False False True False False True k respectively. The limiting values D True True False True False True

ofθ kx, ky and k are θ Q.12 Consider frame shown in figure. If axial and shear deformation in different members are assumed to

a) 5 b) 6 c) 7 d) 8

ANSWER KEY:

1 2 3 4 5 6 7 8 9 (d) (c) (d) (b) (d) (a) (a) (a)

10 11 11 12 (a) (a) (a) (d) 0

Q.1 (d) Dsi = 3C = Ds = m + R – 2j Ds = 6 = 12 + 6 – 2(9) 0 = Dk = 2j – R = 2(9)0 – 6 Q.7 (a) =12 Dse = R – r – r’ = 4 – 3 - Hence the given truss is statically Dsi = determinate and kinematically Ds = 1= 0 indeterminate. 3C = 0 Q.2 (c) Q.8. Dse = 0R – r – r’ = 4 – 3 - Ns = Ds = Dse + Dsi Dsi = Dse = 7 – 3 = 4 Ds = 1= 0 Dsi 3C = 0 Ns = 4 Q.9 (a) 0 As rotation and horizontal Dk == 3 0j – R =3(6) – 7 = 11 deflection in zero as per given figure. Therefore its stiffness is '∞' Nk = Dknad = Dk – m = 11 – 5 = 6

Q.3 (d) Force Stiffness = Dse = R – r = 6 – 3 = 3 as deflectiondeflection = 0. Dsi = 3C = 3(2) = 6 and stiffness is zero in y direction Ds = 3 + 6 = 9 Dk = 3j – R Q.10 (a) = – 6 = 24 Dk = 2j – R Nk = Dknad = Dk – m = 24 – 11 = 13 = 2(7) – 3 3(10) =11 Q.4 (b) Q.11 (a) Dse = R – r = 4 – 3 = 1 Dk = 3j – R =3(4) – 3 = 9 Dsi = m + r – 2j = 15 + 3 – 2(8) =2 Dknad = Dk – m = 9 – 1 = 8 (beam is only axially rigid ) Q.12 (d) Dk = 3j – R Q.5 (d) =3(6) – 4 = 14 Dknad = Dk – m = 14 – 6 = 8 Dse = R – r – r’ = 4 – 3 - 2= -1 Dsi = 3C = 3(2) = 6 Ds = -1 + 6 = 5

Q.6 (a) Space frame: Dse = R – r = 12 – 6 = 6

Q.7 The degree of static indeterminacy of the pin-joined plane frame shown Q.1 Total degree of indeterminacy (both in figure is internal and external) of the plane frame shown in figure is

b) 12 c) 12 d) 15 a) 10 Q.2 If there are m unknown member forces, r unknown reaction a) 1 b) 2 components and j number of joins, c) 3 d) 4 then the degree of static indeterminacy of a pin-joined plane Q.8 The portal frame as shown in the frame is given by given frame is statically a) m + r + 2j b) m – r + 2j indeterminate to the c) m + r – 2j d) m + r – 3j

Q.3 Degree of kinematic indeterminacy of a pin joined plane frame is given by a) 2j - r b) j – 2r c) 3j – r d) None of these a) first degree b) second degree Q.4 A pin-joined plane frame is unstable c) third degree d) None of these if Q.9 A perfect plane frame having n a) (m + r) < 2j b) m + r = 2j number of members and j number c) (m + r) > 2j d) None of these of joints should satisfy the relation a) n < 2j - 3 b) n = 2j – 3 Q.5 The degree of static indeterminacy c) n > 2j - 3 d) n = 3 – 2j of a pin-joined space frame is given by Q.10 Neglecting axial deformation, the a) m + r – 2j b) m + r – 3j kinematic indeterminacy of the c) 3m + r – 3j d) m + r + 3j structure in the figure below is

Q.6 The degree of kinematic indeterminacy of a pin-joined space frame is given by a) 2j – r b) 3j - r c) j – 3r d) j – 3r

Q.11 c)Consider 20 the following statements: 1) The displacement method is more useful when degree of kinematic indeterminacy is greater than the degree of static a) 8 b) 7 indeterminacy greater than the c) 6 d) 5 degree of static indeterminacy Q.14 What is the degree of static 2) The displacement method is indeterminacy of the plane structure more useful when degree of as shown in the figure below? kinematic indeterminacy is less than the degree of static indeterminacy. Indeterminacy is less than the degree of static indeterminacy 3) The force method is more useful when degree of static indeterminacy is greater than a) 3 b) 4 the degree of kinematic c) 5 d) 6 indeterminacy. Q.15 The prismatic is show in the figure 4) The force method is more useful below. when degree of static indeterminacy is less than the degree of kinematic indeterminacy. Which of the statements are correct? Consider the following statements: a) 1 and 3 b) 2 and 3 1) The structure is unstable. c) 1 and 4 d) 2 and 4 2) The bending moment is zero at supports and internal hinge. Q.12 The pin-joined frame shown in the 3) It is a mechanism. figure is 4) It is statically indeterminate. Which of these statements are correct? a) 1, 2, 3 and 4 b) 1, 2 and 3 c) 1 and 2 d) 3 and 4 Q.16 What is the statical indeterminacy for the frame shown below? a) a perfect image b) a redundant frame c) a deficient frame d) None of the above

Q.13 The degree of static indeterminacy of the rigid frame having two a) 12 b) 15 internal hinges as shown in the c) 11 d 14 figure is Q.17 The statical indeterminacy for the given 3D frame is

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Q.19 Consider the following pin-joined a) 8 b) 6 plane frames c) 9 d) 12 Q.18 The statical indeterminacy for the given 2D frame is

a) 8 b) 6 c) 9 d) 11 Which of these frames are stable? a) 1, 2 and 3 b) both 3 and 4 c) 2, 3 and 4 d) both 2 and 4

Q.20 The kinetic indeterminacy of the beam as shown in figure is

a) 5 b) 9 c) 14 d) 15

ANSWER KEY:

1 2 3 4 5 6 7 8 9 (c) (c) (a) (a) (b) (b) (d) (c) (b) 10 11 12 13 14 15 16 17 18 (b) (d) (c) (d) (b) (b) (c) (c) (a) 19 20 (c) (b)

Q.1 (c) Force method is useful when se D = R – r – r’ = 12 – 3 - 3= 6 DsK< D Dsi = 3C = 3(2) = 6 Displacement method is useful Ds = 6 + 6 = 12 when D< D (Note: r’=No of member connected ks at internal hinge – 1= 4-1 = 3) Q.12 (c) This is pinned jointed space frame Q.2 (c) Degree of indeterminacy No of members, m =9 Q.3 (a) Number of joints, j = 6 D=+− m R 3j Q.4 (a) se

Dse =+− 9 4 3*6 Q.5 (b) D5se = − Since the degree of indeterminacy is

Q.6 (b) Negative so it is a deficient frame. Q.7 (d) Q.13 (d) Dse = R – r – r’ = 4 – 3 - 2= -1 No of members, m = 21 Dsi = 3C = 3(2) = 6 Number of joints, j = 11 Ds = -1+ 6 = 5 =+− Dse m R 2j (Note: r’=No of member connected

Dse = 21 +− 5 2*11 at internal hinge – 1= 2-1 = 1 There are 2 similar internal hinge) D4se = Q.14 (b) Q.8 (c) For plane truss degree of indeterminacy Dse = R – r – r’ = 6 – 3 - D= m + r – 2j Dsi = se

Ds = 3 + = 3 0= 3 re = 4; m = 10 ; j = 5 3C = 3(0) =0 D=+ 4 10 – 2 ×= 5 4 0 s Q.9 (b) Q.15 (b) A perfect plane frame means a The degree of indeterminacy determinate structure, so =3–1 −=− 1 1 n –() 2j−= 3 0 So structure is deficient and ∴ n= 2j – 3 unstable. It is a mechanism.

Q.10 (b) Q.16 (c) Dk = 3j – R Dse = R – r – r’ = 12 – 3 - 1= 8 =3(11) – 8 = 25 Dsi = 3C = 3(1) = 3 Dknad = Dk – m = 25 – 11 =14 Ds = 8+ 3 = 11 (Note: r’=No of member connected Q.11 (d) at internal hinge – 1= 2-1 = 1)

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Q.17 (c) Dse = R – r – r’ = 18 – 6 - 9= 3 Dsi = 6C = 6(1) = 6 Ds = 3 + 6 = 9 (Note:r’=3(No of member connected at internal hinge – 1) r’1= 3(2-1) = 3 r’2= 3(3-1) = 6)

Q.18 (a) Dse = R – r – r’ = 11 –3 - 3= 5 Dsi = 3C = 3(1) = 3 Ds = 5 + 3 = 8 (Note:r’=(No of member connected at internal hinge – 1= (2-1) =1 There are 3 similar internal hinge)

Q.19 (c)

Q.20 (b) The kinematic indeterminacy of the beam is given by

Dkr= 3j – R + r Now, j = 5 R7=

r1r = ∴ Dk=× 3 5 – 7 += 1 9

2.1 Introduction: A truss is a structure composed of 2.3.1 Method of Joint: slender members joined together at end points by bolting/ Riveting or In a planer-truss, at every joint there welding. Ends of the members are are two conditions of equilibrium joined to a common plate called = ∑F0x gusset plate. = ∑F0y Since all the members at a joint are assumed to pass through a single point, moment about the joint will always be zero. Hence ∑=M0 will not be of any consequence. Sign Convention: Tension (+) ve, Compression (-) ve. Analysis should start at joint having at least one known force and at most two unknown forces. For example in A roof truss the following figure, if we take joint and its A then free body diagram of joint A component can be drawn as in Figure (ii). s A typical Roof Truss

2.2 Assumptions for Design of Truss Members and Connection :

1. The members are joined together by smooth pins. 2. All loadings are applied to joints. 3. Self-weight of the members is negligible. 4. All members are straight. Now from the equilibrium of forces we have 2.3 Method of Analysis: (Statically =⇒+ = ∑Px 0 F AB F AF cosθ0 Determinate and Stable Trusses) 3P =⇒+= ∑FY 0 F AF sin θ0 There are two methods of analysis 2 for statically determinate and stable Thus two unknowns can be found trusses. They are: out from two equilibrium equations. 1. Method of Joint Likewise we can proceed to other 2. Method of Section joints and find out member forces 3. Tension coefficient method by using equilibrium equation if no.

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission of unknown forces at the joint is at through point ‘C’, there moment about most two in number. ‘O will be zero. Hence only one

unknown i.e. FGF remains, which can be 2.3.2 Method of Section found out.

2. Similarly, F is found by taking A section is cut through the truss BC such that its cuts at the maximum of moments about G or from equilibrium three members in which forces are equations like = i.e., + += unknown. ∑F0X FBC F GC cosθF GF 0 Thus, if truss is cut into two parts at = i.e., = ∑F0Y FGC sin θP section (a)-(a), three equilibrium

equations can be written for each 2.4 Zero Force Members: part. If max no. of unknown forces in

cut members are three, they can be 1. If only two non-collinear found out. members exist at a truss joint Hence max no. of member (in which Note: and no external force or support forces are unknown) cut should be at most reaction is applied to the joint, three. the members must be zero force members. For example in the figure shown BC and CD are non-collinear members and no force exist at joint C, hence BC and CD are zero force members. Similarly, AB and AE are zero force

members since A is not carrying any force and AB and AE are non- collinear.

1. Fgf = Found out by taking moment about C of all forces in the cut part of

truss. As FBC and FGC are passing

Note: These members (CB and CD) i.e. zero force members, are provided for stability of truss during construction and to provide support in case the applied loading changes.

2. If three members join at a point and out of them, two are collinear and also no external load acts at joint, the third member is a zero force member.

F0DF = On similar lines as discussed above,

Since CD and DE are collinear and no load FBH= 0, F DC = F DE = 0 acts at D] Also, since, FBH = 0 and from equilibrium of

F0FC = joint E.

Since GF and FE are collinear, FED= F EF , F EF = 0 Force in DF = 0 and no load acts at joint F] FAG = 0 because at G no vertical force exist to balance force in GA.

2.5 Truss with Internal Hinge:

4 = no. of unknown support reactions RHR,1, 1 2 H 2 3 = no. of eq. of static equilibrium External indeterminacy =−=431 Internal indeterminacy =−−[m () 2J 3 ] =−18() 2 x 11 −=− 3 1 [m = no. of members, J = no. of joints] Overall indeterminacy = internal indeterminacy + external indeterminacy = -1 + 1 = 0 The truss is overall determinate.

Note: In such cases, Hinge ‘A’ will provide additional condition for calculating external reactions i.e. BM at A = 0 i.e. moment of all forces either to the left of the section or to the right of the section is zero.

Q.1 A truss, as shown in the figure, is Q.3 For the plane truss shown in the carrying 180 kN load at node L2. figure, the number of zero force The force in the diagonal member members for the given loading is M2U4 will be

a) 4 b) 8 a) 100 kN tension c) 11 d) 13 b)100kN compression [GATE–2004] c)80 kN tension d)80 kN compression Common Data for Q.4 and Q.5: [Gate–2003] A truss is shown in the figure. Members are of equal cross section A and same modulus Q.2 In a redundant joint model, three of elasticity E. A vertical force P is applied at bar members are pin connected at point C. Q as shown in the figure. Under some load placed at Q, the elongation of the members MQ and OQ are found to be 48 mm and 35 mm respectively. Then the horizontal displacement `u’ and the vertical displacement 'v' of the node Q, in mm, will be respectively, Q.4 Force in the member AB of the truss is P P a) b) 2 3 P c) d) P 2 [GATE-2005]

Q.5 Deflection of the point C is 2 2+ 1 PL PL a)-6.64and56.14 b) 6.64and56.14 a)  b) 2 2 EA EA c) 0.0 and 59.14 d) 59.41 and 0.0  [GATE-2003] PL PL c) ()22+ 1 d) ()21+ EA EA

Q.6 The members EJ and IJ of a steel truss shown in the figure below are subjected to a temperature rise of 30°C. The coefficient of thermal expansion of steel is 0.000012 per °C per unit length. The displacement (mm) of joint E relative to joint H along [GATE-2013] the direction HE of the truss, is Q.9 Mathematical idealization of a crane has three bars with their vertices arranged as shown in the figure with a load of 80 kN hanging vertically. The coordinates of the vertices are given in parentheses. The force in the member QR, FQR a) 0.255 b)0.58 will be c) 0.764 d) 1.026 [GATE-2008] Q.7 For the truss shown in the figure, the force in the member QR is

p a) 30 kN Compressive a) Zero b) b) 30 kN Tensile 2 c) 50 kN Compressive c) p d) 2 d) 50 kN Tensile [GATE–2010] [GATE-2014] Q.8 The pin-jointed 2-D truss is For the truss shown below, the loaded with a horizontal force of Q.10 member PQ is short by 3 mm. The 15 kN at joint S and another 15 magnitude of vertical displacement kN vertical force at joint U, as of joint R (in mm) is ______shown. Find the force in member RS (in kN) and report your answer taking tension as positive & compression as negative.

[GATE-2014]

Q.11 For the 2D truss with the applied loads shown below, the strain energy in the member XY is_____ kN-m. For member XY, assume AE = 30 kN, where A is cross-section [GATE-2016] area and E is the modulus of elasticity. Q.13 A plane truss with applied loads is shown in the figure.

The members which do not carry any force are a) FT, TG, HU, MP, PL b) ET, GS, UR, VR, QL c) FT, GS, HU, MP, QL d) MP, PL, HU, FT, UR [GATE-2016] Q-14 Consider the deformable pinned [GATE-2015] jointed truss as shown in figure.

Q.12 Consider the plane truss with load P as shown in the figure. Let the horizontal and vertical reactions at the joint B be HB and reaction at the joint C. Which one of the following sets gives the correct values of VB, HB and VC? a) VB = 0; HB = 0; Vc = P b) VB = P/2; HB = 0; Vc = P/2 c) VB= P/2; HB = P (sin 60°); Vc = P/2

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Given that E = 2 * 1011 N/m2 , A = 10 mm2 , L=1 m, P = 1 kN. The horizontal displacement at joint C (in mm up to one decimal) is______[GATE-2018] Q-15 Consider the member of planer truss having same cross sectional area (A) and Modulus of elasticity (E) . a) No reactions develop at supports A and D. b) The bar BD will be subject to a compressive force. c) The bar AC will be subject to a compressive force. d) The bar BC will be subject to a tensile force. [GATE-2001] For load shown on truss, the statement that represent the nature of forces in the Q-17 The right triangular truss is made member of truss is: of members having equal cross sectional area of 1550mm2 and (A) There are 3 members are in Young's of 2x105 MPa. The tension and 2 members are in horizontal deflection of the joint Q is compression (B) There are 2 members are in tension and 2 members are in compression, 1 zero force member (C) There are 2 members are in tension and 1 member is in compression, 2 zero force member (D) There are 2 members are in tension and, 3 zero force members

Q-16 Identify the FALSE statement from a) 2.47 mm b) 10.25 m the following, pertaining to the c) 14.31 mm d) 15.68 mm effects due to a temperature rise T [GATE-2007] of the bar BD alone in the plane truss shown below: ∆ ANSWER KEY:

1 2 3 4 5 6 7 8 A B B C A C C 7.5 9 10 11 12 13 14 15 16 A 2 5 A A 2.7 D B 17 D

Q.1 (a) 11 sinθ,= cosθ = The force in member 22√√22 M2L4 = 0. Let the horizontal displacement of Hence remove this member for the joint Q is ‘u' and Vertical sake of analysis. displacement is 'v' By resolving the Reactions, displacement components in, RL0= 120 kN Along member 'OQ' RL6= 60 kN U sin + V cos = 35 _ __ _ _ (1) Along member ‘MQ’ U sinθ2 + V cosθ2 = 48 _ _ _ _ (2) Solving equations 1&2, we get, u = 6.64θ1 mm & θv1 = 56.14 mm

Use method of sections. Q.3 (b) Apply a section 'XX' as shown. If at any joint, three forces act and Consider right part of section if two of them are in the same line, then the third force must be zero. Using this principle.

Apply V = 0 for right part of section. Σ = FU4 M 2 cosθ 60 Dsc = 3 – 3 = 0 60 Dsi = 9 – 12 = 3 = 0 FM= U4 2 cosθ Total 8 members having zero 60* 5 forces. Pl follow the order of = =100kN (Tension) members as shown in fig for 3 (Force Pulling the Joint is Tension) analysis.

Q.2 (b) Q.4 (c)

45 sinθ,= cosθ = 11√√41 41 50 Vertical reaction at support ‘B’ is ∟P ABC = 4 (upward) 2 Resolving vertically

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission P At B, sin450 = 2 PP F2= √= (comp) BD 2 √ 2 Resolving horizontally at B, 0 FFAB= BDcos45 P1P ×= √√222 1 FEJ == 0.707 Q.5 (a) 2 We know that if three forces act at F0= a joint, and if two of them are in JI the same line, then the third force Relative displacement is zero. According to this rule,

various members having zero force =0.707*3000*0.000012*30 are marked in the figure. =0.764= ∑FLαT mm Forces in other members are also shown in figure (F) Now remove Q.7 (c) Analyzing at joint ‘S’ member SR the external load place a unit and force ‘P’ are in the same line. vertical load at 'C’ Now forces in Hence the third force ‘TS’ = 0. various members say Further force in member RS = P. (K) Now simplified diagram is Now V at ‘C’ FKl = ∑ δAE P 1 L2 8 224 = + AE Apply V = 0 at ‘R’ P1L     P12L    4        FRT sin 45° = P 223     22   3  ∑ + FRT = P 2 ( comp) AE AE PL 2 PL PL 1 F∴QR = FRT √cos 45 = +=2 + Apply ∑H = 0 at R AE 2AE AE 2 =P2 √×1 == P (Tensile) √ 2 PL 2 2+ 1 =  AE 2 Q.8 (7.5)

Q.6 (C)

As we need to calculate relative displacement between E and H, remove member between them and apply unit load. Now as temperature is raised in EJ and JI, so calculate member force in these two members only,

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Taking moment about Q = 0 80 × 1 + RR × 2 = 0 RR = -40 kN ⟹ From (i), we get RQ = 120 kN Taking⟹ moment about P = 0 FQR × 4 + 120 × 1 = 0 FQR=-30kN (-) mean compressive ⟹

∑MP = 0 15*8 + 15*4 – 8*VT = 0 VT = 22.5 kN ∑V = 0 VP + VT = 15 VP = -7.5 kN FQR = 0 and FQV = 0 Joint V FRV = -VP =7.5 kN Joint R

Q.10 (2) Since deflection at R is required. So, let us apply a virtual unit load at point R in upwards direction. 1 v = 0 UPR 2 At pointH =0 P ΣFUPR ⟹ PQsin = 0 θ = -FRV - FRU sin 45= 0 UPQ = -UPR F∑VRV = 0- Σ F 1 ⟹ cosθ + U = - .cos ⟹2sinθ cosθ FRs + FRU7.5 sin √2 45= kN 0 1 1θ 14 F∑HRS = 7.50 kN = −× =×=−2 2 tan θ 23 3 Q.9 (a)

= UPQ PQ −2 =𝑅𝑅 ×−()3 𝛿𝛿 3 × λ short]=2mm [∴ PQ is 3mm So, deflection at R = 2mm (upwards) v = 0 RQ + RR = 80 ……..(i)

ΣF ⟹

R×+ 31090 ×= A Q.13 (a) RA=-30 kN Conditions for zero force members RG = 35 kN are Taking⟹ joint A (i) The member meets at a joint and they are non-collinear and no external force acts at that joint. Both the members will be the zero force members. (ii) When the members meet at joint and two are collinear and no external force act at the Joint G joint then third member will be zero force member. According to the above statements We can say that FT, TG, HU, MP and PL members are zero force members. Q.14 (2.7)

Joint B Fx = 10kN F2 × L 10 ×× 10 3 U== = 5kNm − 2AE 2× 30

Q.12 (a)

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission B = 0 –1*L + VA*L = 0 ∑M VA = 1 Now,∴ ∴ VA – VB = 0 ∑V V B= =1 0 B = 0 ∴ P*L – VA*L = 0 ∴ HB –1= 0 ∑M VA = P ∑H H B= = 0 1 Now,∴ ∴ ∴ Joint B∴ – VA + VB – 2P = 0 –VB + FBC = 0 ∑V V B= =3P 0 ∑V –1 =+ 0FBC = 0 ∴ ∴ ∴ FBC = 1 –HB + P = 0 ∴ ∑H H B= =0 P ∴ HB – FAB = 0 Assume∴ all members are in tension ∑H 1 –= F0AB = 0 ∴ ∴ FAB = 1 Joint B ∴ VB + FBC = 0 Joint A ∴ ∑V 3P =+ 0FBC = 0 VA +FAC sin45 = 0 ∴ FBC = -3P ∑V 1 += F0AC sin45= 0 ∴ ∴ FAC =– √2 ∴ –HB – FAB = 0 ∴ ∑H –P = – 0 FAB = 0 ∴ ∴ FAB = -P ∴ Joint A ∴ Part L P K AE PKL/AE –VA +FAC sin45 = 0 AB L –P 1 AE –PL/AE ∑V = 0 –P + FAC sin45= 0 AC – 2AE – ∴ FAC = P√2 BC L –3P 1 AE –3 PL/AE ∴ √2L P√2 √2 √2 PL/AE ∴ HC – 5.41PL/AE −5.42*1000*1000 Now apply unit load at a point where = δ = ∑PKL/AE = 5 deflection is required. 10*2*10 = –2.705 mm Q.15 (D)

Joint A Analyze the given truss for unit FAD = 0 horizontal load at ‘o’ ∑V = 0 Say, Force in members, ∴ compressive as +ve and tension as Joint C –ve. Say, Force due to unit horizontal load @ Q in a member = k FBC = 0 ∑V = 0 ∴

Member P k l PQ 225 1.67 7.5 QR - 180 - 1.32 6 PR - 135 - 1 4.5 Here force P is making couple of P*L which will be balanced by couple due to ∑Pkl 4862.025 Q δQh = = reactions HD and HB. It means inclined AE 1550× 200 member will also carry zero force. = 0.01568 m = 15.68 mm

Q.16 (B)

As temperature increases, the bar `BD' will tend to elongate. But the Joints B & D will offer resistance. Hence the bar 'BD' will be in compression.

Q.17 (D)

Analyze the given truss for external load.

c) 150 kN tension Q.1 A simply supported truss shown in d) 100 kN compression the given figure carries load as shown. The force in member BE is Q.4 In the cantilever truss shown in the figure the reaction at A is

a) 2 t (tensile) b) 2 t (compressive) a) 10 t b) 20 t c) 1t (tensile) c) 30 t d) 40 t d) 1t (compressive) Q.5 Force in the member BC of the Q.2 A cantilever pin-jointed truss truss shown in the given figure carries one load P at the point Z as shown in the given figure. The hinges in the vertical wall are at A and D. The truss has only three horizontal members AC, CG and GZ

a) 5 t, tensile b) zero c) 2.88 t, compressive d) 5 t, compressive The nature of force in member AB a) is tensile Q.6 A simply supported truss shown in b) is zero the given figure carries a load if 20 c) is compressive kN at F the forces in the members d) cannot be predicted EF and BE are respectively.

Q.3 The force in member FD in the given figure is

a) 50 kN compression b) 50 kN tension a) zero and 10 kN (compression) b) zero and 10 kN (tension)

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission c) 10kN (tension) and 10kN (compression) d) 10kN (compression) and 10kN (tension) Q.7 Axial force in the member BC of the truss shown in the given figure

is (where α = 30°) a) 1.5kN (tension) b) 4.5 kN (tension) c) 1.5 kN compression d) 4.5 kN (compression)

Q.10 For the truss shown in the figure, a) 15 kN b) 10 3 kN which one of the following members has zero forced induced c) 15 3 kN d) 30 kN in it?

Q.8 Axial force in the members 1 – 2 and 1 – 5 of the truss shown in the figure are respectively.

a) BC b) AD c) DE d) BD

Q.11 The force induced in the vertical member CD of the symmetrical a) 50 kN (compressive) and 25 kN plane truss shown in the figure (tensile) 50 b) 25 kN (tensile) and kN 2 (compressive) c) 50 kN (tensile) and 50 2 (compressive) 50 d) 25 kN (compressive) and 2 a) 50 kN (tension) kN (compressive) b) 100 kN (tension) c) 50 kN (compression) Q.9 A pin-joined tower truss is loaded d) Zero as shown in the below figure. The force induced in the member DF is Q.12 A loaded pin-joined truss is shown in the given figure. The force in member AC is

a) 10 2 kN (Tensile) a) zero b) 10 kN (Compressive) b) 50 kN (tensile) c) zero c) 50 kN (compressive) d) 10 kN (Tensile) 50 2 Q.13 Which of the following is the d) kN (tensile) correct statements regarding the 2 force and deflection at point B in Q.16 Due to horizontal pull 60 kN at C, trusses I and II shown in the what is the force induced in the figure? member AB?

a) I will have less member force & less deflection at B compared t a) 0 b) 40 kN b) I will have less member force c) 80 kN d) 100 kN and more deflection at B compared to II Q.17 The force in member FD in the c) I will have more member force given figure is & deflection at B compared to II

d) I will have more member force & less deflection at B compared to II

Q.14 In the plane truss shown below, how many members have zero a) 50 kN (compressive) force? b) 50 kN (tensile) c) 150 kN (tension) d) 100 kN (compression) Q.18 A truss ABC carries two horizontal and two vertical loads, as shown in a) 3 b) 5 the figure. The horizontal and c) 7 d) 9 vertical components of reactions at A will be Q.15 In the pin-joined plane truss shown below, what is the

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Q.20 Consider the following: Assumption in the analysis of a plane truss. 1. The individual members are straight. 2. The individual members are connected by frictionless hinges. a) HA = 1000 kN; VA = 1706.25 kN b) HA = 2000 kN; VA = 2550 kN 3. The loads and reactions acts c) HA = 1000 kN; VA = 2350 kN only at the joints. d) HA = 2000 kN; VA = 1706.25 kN Which of the following assumptions are valid? Q.19 What does the Williot-Mohr a) 1 and 2 b) 1 and 3 diagram yield? c) 2 and 3 d) 1, 2 and 3 a) Forces in members of a truss b) Moments in a fixed beam c) Reaction at the support d) Joint displacement of a pin jointed plane frame

ANSWER KEY:

1 2 3 4 5 6 7 8 9 (c) (c) (b) (c) (d) (b) (d) (c) c

10 11 12 13 14 15 16 17 18 (a) (a) (c) (a) (c) (a) (d) (b) d 19 20 (d) (d)

Q.1 (c) Force in EF is zero if joint At joint D there is no external force equilibrium at E is considered so force in FD and BD is zero. force in BE will be Tensile, and its Taking moment about A, the value will be 10 kN by considering reaction at B joint equilibrium at B. 2×2 R = = 1t B 4 Q.7 (d) Considering joint B, force in The given truss is symmetrical. So member BF force in member BO and EO will be same in magnitude and nature FBF = 2t (compressive) both. Joint equilibrium at O. Force in member BE

FBE =1t (tensile)

Q.2 (c)

Q.3 (b) Considering joint equilibrium at D. 15 F = = 30 kN The force equilibrium equation in bo sin300 vertical direction gives FD = 50 kN Joint equilibrium at B. tensile Q.4 (c) Only a horizontal reaction can exist at A. Taking moments about hinge.

HA × 5 = 10 ×15 H = 30t A ∴ FBC = 30 KN Q.5 (d) By considering joint equilibrium at Q.8 (d) B. Cutting a section through 1-2; 2-5; and 5-6. Taking moment of left part about 5. 50×6 F = = 100 kN (Tensile) 1-2 3 Cut a section through 1-2, 1-5 and 4-5 and balance vertical force for FF= BC BA left part only. 00+= FBC cos 60 F BA cos 60 5 F 1-5 = 50 FBC = 5t () compressive 2

FBA = 5t ()tensile ∴F1-5 = 50 2 kN (Compressive) Q.6 (b)

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Q.9 (c) Fl22 Fl U = U + U = × 2 = AB BC 2AE AE Q.10 (a) Pl2 Cutting a section through AC, CD U=1 2 and DF and taking moment of 4AE sin θ δV Pl upper portion About C. The forces ∆ ==1 in the members AC and CD meet at 1 δP 2AE sin2 θ C so they not produce Pl ∆1t = Any moment. The force in member 2 θ 2 AE sin  DF. 2 2×3 F = = 1.5kN (compression) Thus ∆∆1t > 1 DF 4 Q.14 (c) Q.11 (a) Six members having zero forces At join D, the equilibrium shows are obvious and seventh member the force in member CD is 50 kN can be from top chord member tensile. between the applied loads. Q.12 (c) Q.15 (a) Cutting a section through members At joint C there is no external force CA, CD, DE and EB and taking so force in member BC is zero. moment of The upper part about E. Q.16 (d) Cut a section through members AB, BD and CD, and take moment of upper part about D. 4 12 DE = AD sin θ = 3×= 55

∴ FCA × 6 + 10 × 6 - 20 × 3 = 0

∴ FCA = 0

Q.13 (a) Force in AB or AC [Consider joint DE × F = 60 × 4 B] AB P 60×4×5 Case-I: sin θ FAB = = 100 2 12 P θ Case-II: sin  Q.17 (b) 22 θ as sinθ > sin Q.18 (d) 2 So in case I force in members will Q.19 (d) be less than that in case II. From symmetry there will be only Q.20 (d) vertical deflection of joint B. For plane truss assumptions are: i. Weight of all members are

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission neglected. All members are straight. ii. All connections are smooth, frictionless pins. iii. External loads are applied directly to the pin joints. iv. All frames are perfect, hence statically determine. i

 A three hinged arch is a statically 3.1 Introduction: determinate structure, having a hinged at each abutment or springing, and also Arches are a structure which at the crown. eliminates tensile stresses in spanning  In three hinged arch three equations great amount of open space. All the forces are available from static equilibrium are resolved into compressive stresses. and one additional equation is available It also helps to reduce bending from the fact that B.M at the hinge at the moment in long span structure. crown is zero.

3.2 Types of Arches: Standard cases

Types of arches based on number of hinges Case 1 : A three hinged parabolic arch of 1. Two hinged arch span L rise ‘h’ carries a udl of w over the whole span

2. Third hinged arch

3. Fixed arch or hinge less arch a) The horizontal reaction at each support WL2 is H = 8h b) BMxx = 0  A three hinged arch is a statically SFxx =0 determinate structure where the rest two aches are statically indeterminate Case 2 : A three hinged semi-circular  In bridge construction, especially is arch of radius 'R' carries a udl of w on railroad bridges, the more frequently the whole span. used aches the two hinged and the fixed end ones. 3.3 Advantages of Arch over beam

 Section required for arch is small compare to beam for same span and loading as bending moment at any section is less in arch compare to a) The horizontal thrust at each end; beam. WR H = 2 3.4 Three Hinged Arches b) The maximum bending moment occurs at θ=30 °

Case3: A three hinged arch consisting of two quadrant parts AC and CB of radii

R1 and R 2 . The arch carries a concentrated load of W on the crown The horizontal thrust at each support is given by WI2 H = 2 2h+ h ( 12)

Case 6 : A three hinged parabolic arch of span 1 has

W its abutments A and B at depth h1 and h2 V = V == H AB 2 below the crown C. The arch carries a concentrated load W at the crown. The Case 4 : A symmetrical three-hinged horizontal thrust at each support is given by parabolic arch of span land rise carries a WI2 point load, w, which may be placed H = 2

anywhere on the span 2h12+ h ( ) 3.5 Two Hinged Arches:

Two hinged arch is an indeterminate

structure. Va and Vb can be determined by taking moment about either end. The horizontal thrust at each support may be

determined from the condition that the Wx H = horizontal displacement of the either hinge 2h with respect to other is Zero. The absolute maximum bending moment 1 occurs at a distance of on either side 23 of the crown.

Case 5 : A three hinged parabolic arch of span 1 has its abutments at depth h1 and h2 below the crown the arch carries an M.yd udl of w per unit length over the whole ∫ s span EI H = 2 y .d s ∫ EI Where, M is beam moment

Standard Cases

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Case 1: A two hinged semi-circular arch of A two hinged semi-circular arch of radius R radius R carries a concentrated load W at carries a distributed load uniformly varying the crown from zero at the left end to w per unit run at the right end.

The horizontal thrust on each support is given by The horizontal thrust at each support for w 2 WR H = this case is H = π 3 π Case 2 : A two hinged semi-circular arch of radius R Case 5 : carries a load W at a section, the radius A two hinged parabolic arch carries a udl of vector corresponding to which makes an W per unit run on entire span of the span of angle with the horizontal. the arch is L and its rise is h.

The horizontal thrust at each support is given by The horizontal thrust at each is given by w 2 H= sin α WL2 π H = Case 3 : 8h A two hinged semi-circular arch of radius R BM = 0 (everywhere) carries a udl W per unit length over the whole span. Case 6 : When half of the parabolic arch is loaded by udl. Then the horizontal reaction at WL2 support is given by H = 16h

Case 7 : When two hinged parabolic arch carries uniformly varying distributed load, for zero to w the horizontal thrust is given The horizontal thrust at each support is WL2 4 WR by H = given by H = 16h 3 π Case 8 : Case 4 :

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission A to hinged parabolic arch of span L and rise h carries a concentrated load W at the crown

The horizontal thrust at

The Linear Arch or Theoretical Arch:

a) The different members of the linear arch are subjected to axial compressive forces only. The joints of this linear arch are in equilibrium b) The shape of linear arch follows the shape of the free bending moment diagram for a beam of the same spam and subjected to the same loading c) The B.M at any section of an arch is proportional to the ordinate or the intercept between the given arch and the linear arch. This principle is called Eddy’s theorem d) The actual BM at the section X is proportional to the ordinate x1 x.

a) 9.01 KN and 56.31° b) 9.01 kN and 33.69° Q.1 A three hinged parabolic arch ABC c) 7.50 kN and 56.31° has a span of 20 m and a central rise d) 2.50 kW and 33.69° of 4 m. The arch has hinges at the [GATE-2010] ends and at the centre. A train of two point loads of 20 kN and 10 kN, Q.3 A figure shows two hinged parabolic 5 m apart, crosses this arch from left arch of span L subjected to UDL over to right, with 20 kN load leading_ entire span of intensity q per unit The maximum thrust induced at the length supports is a) 25.00 kN b) 28.13 kN c) 31.25 kN d) 32.81 kN [GATE – 2004] Q.2 A three hinged parabolic arch having a span of 20 m and a rise of 4 m carries a point load of 10 kN at The maximum bending moment in quarter span from the left end as the arch equal to shown in the figure. The resultant qL2 qL2 (a) (b) reaction at the left support and its 8 12 inclination with the horizontal are qL2 respectively (c)Zero (d) 10

ANSWER KEY:

1 2 3 (c) (a) (c)

Q.1 (c) VBB×= 10 H × 5 2.5× 10 ∴=H = 5kN ← B 5

= → HA 5kN

ILD for Horizontal thrust at Supports. Resultant Reaction, Central ordinate 22 l 20 RH=AA + V = = =1.25 4h 4× 4 =+=522 7.5 9.01 kN At quarter point, Inclination of Resultant with 1.25 Ordinate = = −1 7.5 0.625 Horizontal =tan =56.31 ° 2 5 According to ILD, Maximum thrust is =56.31 ° With horizontal. induced at supports if the leading load 20 kN is at the crown. Q.3 (c) Horizontal thrust at supports A two hinged parabolic arch carries H=20×1.25 + 10×0625 = 31.25 kN a udl of q per unit run on entire span of the span of the arch is L, BM is Q.2 (a) zero everywhere.

Apply ∑=M0A from right.

VB ×=× 20 10 5

∴=VB 2.5 kN ↑

VAB+= V 10 kN ∴=− ↑ VA 10 2.5 = 7.5 kN Applying moment at central hinge ‘C’,

∑=M0C from right,

Q.1 In a two hinged arch an increase in temperature induces Q.4 A symmetrical two-hinged parabolic a) no bending moment in the arch arch rib has a span of 32 m between rib abutment pins at the same level and b) Uniform bending moment in the a central rise of 5 m. When a rolling arch rib load of 100 kN crosses the span, the c) Maximum bending moment at maximum horizontal thrust at the the crown hinges will be d) Minimum bending moment at a) 100 kN b) 125 kN the crown c) 160 kN d) 240 kN

Q.2 Which one of the following is Q.5 A uniformly distributed load of 2 associated with the rib shortening in kN/m covers left half of the span of arches either due to change in a three-hinged parabolic arch, span temperature or lack of fit to cause 40 m and central rise 10 m. Which of stress in the arch members? the following statements related to a) Only two-hinged arches and not different functions at the loaded three-hinged arches quarter point are correct?

b) Two and three-hinged arches −1 1 1. The slope is tan  c) Two-hinged arches made of 2 reinforced concrete only. 2. The normal thrust is KN d) Only three-hinged arches but not 65 two-hinged arches. 3. The shear force is not zero 4. The bending moment is zero. Select the correct answer using the Q.3 A three-hinged parabolic arch rib of span L and crown rise ‘h’ carries a codes given below. uniformly distributed superimposed Codes: load of intensity ‘w’ per unit length. a) 1, 2 and 4 b) 2 and 3 The hinges are located on two c) 1 and 3 d) 3 and 4 abutments at the same level and the third hinge at a quarter span Q.6 A three hinged arch is shown in the location from left hand abutment. below figure. The influence line for The horizontal thrust on the the horizontal thrust is abutment is wL2 wL2 a) b) 4h 6h wL2 wL2 c) d) 8h 12h a) b)

Q.7 For the two hinged parabolic arch as shown in the figure below, which one of the following diagram represents the shape of the bending moment variation? 1. Line of action of reaction RA at A is inclined at 45° to the horizontal. 2. Va = VB = W/2 3. HB = 2HA 4. HA = HB = W/2 a) b) Which of the above is/are correct? a) 1 and 3 b) 2 and 3 c) 1, 2 and 4 d) 4 only c) d) Q.11 A parabolic arch, symmetrical with hinges at centre and ends, carries a point load p at a distance x from left support. The arch has a span of 20 m Q.8 A circular three-pinned arch of span and rise of 5 m. What is the value of 30 m and a rise of 8 m is hinged at ‘x’ if the left hinge reaction is the crown and springing. It carries a inclined with a slope of two vertical horizontal load of 100 kN per to one horizontal? vertical metre on the left side. The a) 8 m b) 5 m horizontal thrust at the right c) 4 m d) 2.5 m springing will be a) 200 kN b) 400 kN Q.12 A symmetrical circular arch of span c) 600 KN d) 800 kN 25 m with a central rise of 5 m is hinged at crown and springing. It Q.9 A symmetrical three-hinged carries a point load on the arch at a parabolic arch of span L and rise h is distance of 6.35 m from the left hinged at springing and crown. It is hinge. If the inclination of the thrust subjected to a UDL w throughout the span. What is the bending from horizontal, then what is the moment at a section L/4 from the at the right hinge is θ measured left support? a) 2.5 a) wL2 / 8 b) wL2 /16 b)value1.0 of tan θ? c) wL3 / 8h d) Zero c) 0.4 d) Not possible to calculate with the Q.10 A three-hinged circular arch ACB is given data formed by two quadrants of circles AC & BC of radii 2R & R respectively Q.13 The horizontal thrust at support A in with C as crown as shown in the a three hinged arch shown in the figure. Consider the following in given figure is respect of the reactive forces

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission a) 2 KN b) 4 KN 180 190 c) 8 KN d) 10 KN a) KN b) KN4t π π Q.14 For the semi-circular two-hinged 200 arch shown in the figure below a c) KN 2t d) None of these moment 50 t-cm applied at B π produces a displacement load of 10 t A three hinged arch is shown in the is applied at A, the rotation at B in Q.18 figure is quarter of a circle. If the the arch will be vertical and horizontal components of reaction at A are equal, the value

of θ is

a) 0.1 b) 0.001 c) 0.0001 d) 0.01

Q.15 A three hinged semi-circular arch of radius R carries a uniformly distributed load w per unit run over the whole span. The horizontal a) 60° b) 45° thrust is c)30° d) None of these wR a) wR b) A three hinged arch is subjected to 2 Q.19 point load at crown. The magnitude 4wR 2wR c) 2t d) 1t of horizontal reaction at A is 3π 3π

Q.16 The horizontal thrust due to rise in temperature in a semi-circular two hinged arch of radius R is proportional to a) R b) R 2 c) 1/R d) 1/R2 P P Q.17 A two hinged semi-circular arch of a) b) − 2 radius 20 meters carries a load 2() 21 system shown in figure. The P c) 2− 1 d) horizontal thrust at each support is 22 (Assume uniform flexural rigidity) Q.20 An arch is used in long span bridges because there is

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission a) considerable reduction in d) no change in shear and bending horizontal thrust moment b) considerable reduction in bending moment c) reduction in shear and bending moment

ANSWER KEY:

1 2 3 4 5 6 7 8 9 (c) (a) (c) (b) (c) (c) (b) (a) (d) 10 11 12 13 14 15 16 17 18 (c) (c) (c) (b) (a) (b) (d) (b) (b) 19 20 (a) (b)

L wL L Q.1 (c) V×−× Hh − ×= 0 Increase in temperature in a two A14 48 hinge arch will cause horizontal wL L 3h wL2 ⇒ ×−×H0 − = thrust only. 2 4 4 32 Moment due to horizontal thrust is – wL2 Py ∴= H So maximum bending moment will 8h be at crown as crown has highest Q.4 (b) value of y. For unit load maximum horizontal thrust Q.2 (a) 25L In the case of two hinged and fixed = 128h arches rib shortens by an amount For 100 KN load maximum HL horizontal thrust AE 25× 32 There is always horizontal = ×100 128× 5 compressive thrust due to 5 temperature increase or lack of fit in =×=100 125 kN the case of two hinged arches. While 4 in three hinged arches no such Q.5 (c) thrust exists and only the shape will wL2 change at crown. H = = 20KN 16h Q.3 (c) 3 VA = wL = 30KN WL 8 VV= = AB2 The equation of parabolic arch y 4h = ×−(L X) L2 12 sinθ= cos θ= 55 dy 4h ∴=(L − 2X) dx L2 L At X = 4 h h - h1 ∴ = 2h (L/2)22 (L/4) tanθ = L ⇒ 4 h–h= h -1 ()1 For given arch θ = tan() 1/2 = 3h 4h1 So statement (1) is correct B.M.D 3h 12 h1 = sinθ = cosθ = 4 55 Taking moment about hinge at

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission WX These expressions are same as that M = (L - 2X) for loaded half 8 for three hinged arch in the same L case. At X = 4 Q.8 (a) W L L wL2 M = × × = = 50kN-m 8 4 2 64 So bending moment is not zero and statement (4) is wrong

Normal thrust N = VA sin θ + Hcos θ =14 5 kN The vertical reaction At B is Radial shear VrA = V cos θ - Hsin θ VB × 40 = 100 ×× 8 4 = 8 5kN V= 80 kN B Q.6 (c) For horizontal thrust at B taking The horizontal reaction varies moment of right part about crown linearly from any end up to the (c) crown. VB ×=× 20 H 8 80×20 ∴H = = 200KN Q.7 (b) 8 The horizontal thrust influence line in two-hinged arches is a fourth Q.9 (d) order parabola. For a three hinged parabolic arch, However bending moment at any carrying UDL over the entire span, point bending moment is zero at all the = Beam B.M – Thrust B.M sections. Horizontal thrust due to udl through Q.10 (c) the span The moment at crown being zero. wL2 = Consider AC 8h HA .2R= V A . 2R ⇒=HV AA For half span loading shown in the For BCH= V figure, the symmetry shows that BB As = so = horizontal thrust HHABVVAB 2 Now wL VVWAB+= H = 16h so VAB= V = W/2 B.M for loaded half portion taking x positive from A Q.11 (c) 3 wx22 wl 4h Vertical at A H = wlx - - × × l - x 2 () P(20 - x) 8 2 16h l Va = wlx wx2 wx 20 H = - = ×() l-2x Horizontal thrust 848 V ×10 l B x ≤ H = = 2VB 2 5 2P wx ∴ X For right half M = - (l - 2x) 8 20 With x positive from B.

Q.14 (a) From Maxwell-Betti’s theorem

PA∆ AB = M BAθ BA Given that left hinge reaction is 10×0.5 ∴ θ= inclined with a slope of two vertical BA 30 on one horizontal. = 0.1 radians ∴VA = 2H P(20 - x) 4px = Q.15 (b) 20 20 (20 - x) = 4x Q.16 (d) X = 4m 4Elαt It is given by H = πR 2 Q.12 (c) Q.17 (b) 120 100 80 H = sin2o 30 + + sin 2o 60 π ππ 190 = kN π

Q.18 (b) Vertical reaction at A p×(25-6.25) V = =0.75P A 25

Vertical reaction at B VB = 0.25P Horizontal thrust 12.5V H = B = 0.625P 5 VB 0.25P Values of horizontal and vertical ∴tanθ = = = 0.4 components of reaction at A are H 0.625P ° since the vertical and horizontal components are Q.13 (b) equal at θ = 45 The vertical reaction at BV= B and ° the horizontal thrust at A and B be H thereforegiven by the sincomponents θ and are cos equalθ Taking moment of forces on right respectively, and sin θ = cos θ at 45 segment about C ×=× VB 10 H 5 Q.19 (a)

H= 2VB Vertical reaction at both support = P VB = 0.5H (by symmetry) For left segment taking moment 2 about C. Now taking moment about hinge at crown from LHS = 0 VA × 8 – H × 4 −×8 4 = 0

VA = 0.5H + 4

Q.20 (b)

4.1 Introduction this article, the following types of loads on a simply supported girder are considered: • An influence line represents the 1. Single point load variation of the reaction, shear, 2. Uniformly distributed load longer than moments or deflection at a specified the span point in a member as a concentrated 3. Uniformly distributed load shorter than unit force moves over the member. the span • Influence lines represent the effect of a 4. A train of point loads. moving load only at a specified point on a member, whereas shear and moment 4.4 Beam Subjected to a Single Point diagram represents the effect of fixed Load W loads at all points along the member. (a) Maximum S.F. & B.M. at a given point • Thus influence line helps in deciding, at • Let W be the moving load and the a glance, where should the moving values of maximum S.F. and B.M. loads be placed on the structure so that required be at C, which is at a distance z it creates greatest influence at the from A. ILD for S.F. and B.M. are as specified point. shown in Figure. 4.2 Sign Convention It is clear that maximum negative S.F. occurs when the load is just to the left • Influence line for reaction → (+) ve, if Wz reaction acts upward. of section C and its value is = . L • Influence line for shear and moment Similarly, maximum positive S.F. occurs when the load is just to the right of the Lz− section and its value is = W . L # Influence lines for statically From ILD for moment M , it is clear determinate structures consist of C that maximum bending moment will straight-line segment. occur when the load is on the section # Influence lines for statically Wz() L− z indeterminate structures will consist of itself and its value is = curved-line segments. L The following example illustrates the construction of influence line diagram. 4.3 Maximum S.F. and B.M. Values due to Moving Loads Simply supported beam with given point C A designer finds the position of moving; loads for which shear force and bending moment values are maximum. Influence line diagrams are used for this purpose. In

FC = ω×Area of ILD for FC in length CB 2 1  Lz−  wL()− z =−=w() Lz   2  L  2L • From Figure, it is clear that maximum moment at C will be, when the udl covers entire span, M,C.max = ω× Area of ILD for Mc 1 zL()−− z wzLz() =wL = Fig. (1) 2L 2 (b) Absolute maximum values anywhere in the beam (b) Absolute maximum values anywhere • For achieving this, ordinate of ILD in the beam should be maximum. Negative S.F. • Negative S.F. is maximum when z = L, ordinate is maximum when z = L and is i.e. at B when the load occupies entire equal to 1. span AB Absolute maximum negative S.F. • Therefore, absolute maximum negative 1 z() L - z wz() L - z S.F. = W and it occurs at B. = w× L = • For positive S.F., ILD ordinate has 2L 2 maximum value of 1, when z = 0; i.e., at A. • Similarly, maximum positive S.F. occurs Absolute maximum S.F. = W. when z = 0; i.e., at A when the load • Ordinate of ILD for moment is occupies entire span AB L Maximum positive S.F. maximum when z = . Hence, when a 1 wL 2 =w × ×× 1L = load is at mid-span, absolute maximum 22 WL • Maximum moment at any section moment occurs and its value is 4 11wz() L− z =× ×=×L wz() L − z 4.5 Uniformly Distributed Load Longer 2L 2 L Than the Span This is maximum, when z = i.e., at Let a uniformly distributed load of intensity 2 w move from left to right. mid-span Absolute maximum moment, (a) Maximum S.F. and B.M. at given 1 L wL2 =×××w() L − L/2 = at mid span Sections 22 8 • Load intensity and the area of ILD over loaded length give the value of stress 4.6 Uniformly Distributed Load Smaller resultant (SF/BM). Referring again to Than the Span Figure (1) • Negative S.F. is maximum, when the • Let the length of uniformly distributed load covers portion AC only. load ω / unit length is d. Let it move from left to right over beam AB of span • Maximum negative FC = x Area of ILD L. We are considering the case when, for FC in length AC < Now, position of this load for 2 d L. 1  z  wz maximum shear force and bending =wz   = 2  L  2L moment at section C are to be • Positive S.F. is maximum when the determined. uniformly distributed load occupies the

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission i.e., Lx = dz xz or = dL Simply supported beam with given point C i.e. Bending moment at a section is maximum when the load is so placed that the section divides the load in the same ratio as it divides the span. Once the position of moving load is identified for maximum values the required values can be easily found. Position of load for maximum +ve & –ve SF Position for absolute maximum moment Obviously for this yc should be maximum Now, zL()− z y = c L For yc to be maximum, Position of load for maximum moment dy c =0 = L − 2z dx • From ILD for shear force at C, it is clear L that maximum shear force will develop Or z = when the head of the load reaches the 2 section. i.e. Absolute maximum moment occurs at mid-span. The position of the load is to be • For maximum positive shear force the tail of the udl should reach the section. such that the section divides the load in the same ratio as it divides the span which • Maximum bending moment will means that for absolute maximum moment develop at C when the load is partly to C.G. of the load will be at the mid-span. the left of the section and partly to the right of section. Let the position of the section be as shown in Figure. Referring 4.7 A Train of Concentrated Load to this figure. A train of concentrated loads moving over a ++ xy()1c y() y c2 y simply supported beam from left to right is Mc =ωX +− ω() d x 22shown in Figure. It is required to find:

For MC to be maximum a) Maximum shear force at C b) Maximum bending moment at C dM ωy()1c++ y ωy() c2 y c =0 = − c) Absolute maximum shear force in the dx 2 2 beam i.e., d) Absolute maximum bending moment in Thus, moment at C will be maximum the beam. when the ordinates of ILD for Mc at head and tail of the udl are equal. (a) Maximum shear force at C Now, yy12= i.e., ()zx−()() Lz −−− dx yy= zCC Lz− ∴()()()zxLz − − = zLzdx −−+ Lz−− z22 Lx + xz = Lz −−+ z dz xz

R1 be at a distance x from C. Let ordinate of ILD for moment at C be

y1 under R1 and y2 under R2 and maximum ordinate at C be yC.

• Influence line diagram for shear force at C is shown in Figure. As soon as W1 enters the span negative shear force develops at C. It increases as the load moves on. Some more loads may enter the span and hence, the rate of increase in S.F. goes up. This will continue till the load W1 reaches the section C. • As soon as W crosses section C, it 1 Mc= Ry 11 + Ry 2 2 contributes to positive shear, thus, zx− ()()Lz−−− dx reducing the negative shear. Hence, =+×R1 yR c2 y c − there will be a drop in shear force value. z Lz For to be maximum • Further movement causes more Mc

increase in shear force till the second dMc R 1c y y c ==−−0Ry2c × load reaches C. There is a second peak dx z L− z value and a sudden drop, when the RR second load crosses. 12= z Lz− • Thus, shear force will have a peak value • i.e., the average load on the left-side whenever a load is on the section. portion of the beam is same as the Highest value among these peak values average load on the right side portion of is to be selected. the beam. • By two or three trial values, it is • But, seldom have we got exactly equal possible to get maximum negative shear average load on both sides of the force value. section. For example, when load W1 is • It is to be noted that for maximum to the left of the section, the average negative shear force, most of the loads load on left side may be heavier. When are to the left of the section. it just rolls over the section, the average • Similarly, for maximum positive shear load on right-hand side may become force, there are peak values whenever a heavier. Hence, the above condition for load comes on the section and the maximum bending moment can be maximum value is obtained when most interpreted as the bending moment is of the loads are to the right of the maximum when that load is on the section. section.

• Thus, due to a train of moving loads on (b) Maximum bending moment a simply supported beam, maximum moment at the given section develops • Let R be the resultant of the loads on 1 when the load W1 is on the section

the left of the section and R 2 be where the load W1 is such that as it rolls resultant of the loads on the right of the on the section and comes to the other section. side, heavier portion of the beam

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission becomes lighter and lighter portion additional load is more or decrease due becomes heavier. to the reduced contribution from • In case of some load entering and some leading loads is more. It needs a few leaving the span, the change of portion trials to arrive at conclusions. However, heavier becoming lighter and lighter it can be definitely said that one of the portion becoming heavier may happen loads should be on the support A to get under more than one particular load. All absolute maximum positive SF. such cases are to be considered to • Similarly, to get absolute maximum identify which position gives maximum negative shear force, one of the loads moment at the section. should be on support B (just to the left of the section) and a few trials may be (c) Absolute maximum shear force required to get absolute maximum negative shear force which occurs at At any section, influence line ordinate support B. z for negative shear is  and for L (d) Maximum moment under a load • Let a train of concentrated loads Lz− positive shear it is . Hence, W , W , W ... move on a simply L 123 supported beam AB from left to right as when z = 0, i.e., at support A, ILD shown in Figure. Now, the condition for ordinate for positive shear force is moment to be maximum under wheel maximum (= 1) and when z-L, i.e., at load is required. Let R be the support B, ILD ordinate for negative W2 shear force is maximum (= 1). For shear resultant of all loads. Let its distance force at support sections A and B are as from W2 be d and from support A be x. shown in Figure.

Simply supported beam subjected to a train of load Simply supported beam subjected to a train of load

RL()− x Now, R = A L

Therefore, moment under load W2 = +− • Obviously maximum shear force occurs M RA () x d Rd when one of the load is on support A. Lx− =R() x +− d Rd • When the load starts moving from left L to right, contribution of leading loads to R 2 shear force at A decreases but more =Lx −+ x Ld − xd − Rd L  number of loads may come on the beam For the moment to be maximum and they will contribute to additional dM R shear. ==0[] L −− 2x d • However, no general conclusions can be dx L drawn to say whether increase due to

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Ld or x = − 22 Distance of W from A 2 Ld Ld =+=xd −+ d= + 22 22 Thus, we can conclude, for moment to be maximum under any particular load, the load and the resultant should be equidistant from the mid-span.

(e) Absolute maximum bending moment

• Influence line diagram ordinate for bending moment is maximum at the center of span. Hence, bending moment will be maximum near the center of the span when heavier loads are near to the center. • Since, the maximum moment always occurs under a wheel load, it can be concluded that absolute maximum moment occur under one of the loads when the resultant of all the loads and the load under consideration are equidistant from the center of the beam. • The maximum moment under possible loads can be evaluated and the maximum of these selected as absolute maximum.

Q.1 A simply supported beam with an overhang is traversed by a unit concentrated moment from the left to the right as shown below: a) -80 b) -40

The influence line for reaction at B is c) 0 d) +40 given by [GATE - 2001] a)

Common Data for Q.3, Q.4 & 5: A beam PQRS is 18 m long and is simply supported at points. Q and R 10m apart. b) Overhangs PQ and RS are 3m and 5m respectively. a train of two point loads of 150 kN and 100 kN, 5 m apart, crosses this beam from left to right with 100 k1NT load leading. c) Q.3 The maximum sagging moment under the 150 kN load anywhere is a) 500 kNm b) 450 kNm d) Zero every where c) 400 kNm d) 375 kNm [GATE- 2000] [GATE – 2003]

Q.2 Identify, from the following, the Q.4 During the passage of the loads, the correct value of the bending maximum and the minimum moment MA (in kNm units) at the reactions at supports `R' in kN, are fixed end A in the statically respectively determinate beam shown below a) 300 and -30 b) 300 and -25 (with internal hinges at B and D), c) 225 and -30 d) 225 and -25 when a uniformly distributed load of [GATE – 2003] 10 kN/m is placed on the spans. (Hint: Sketching the influence line Q.5 The maximum hogging moment in for MA or applying the Principle of the beam anywhere is Virtual Displacements makes the a) 300 kNm b) 450 kNm solution easy). c) 500 kNm d) 750 kNm [GATE – 2003]

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Q.6 Muller Breslau Principle in structural analysis is used for Q.10 Beam PQRS has internal hinges in a) Drawing influence line diagram spans PQ and RS as shown. The for any force function beam may be subjected to a moving b) Writing virtual work equation distributed vertical load of c) Superposition of load effects maximum intensity 4 kN/m of any d) None of these length anywhere on the beam. The maximum absolute value of the [GATE-2003] shear force (in kN) that can occur Q.7 Consider the beam A_BCD and the influence line as shown below. The due to this loading just to the right influence line pertains to of support Q shall be:

a) 30 b) 40 c) 45 d) 55 [GATE-2013] a) Reaction at A, RA b) Shear force B, VB 11. In a beam of length L, four possible c) Shear force on the left of C, V-c influence line diagrams for shear d) Shear force on the right of C, V+c force at a section located at a [GATE - 2006] L distance of of from the left end Q.8 The influence line diagram (I L D) 4 shown is for the member support (marked as P, Q, R and S) are shown below. The correct influence line diagram is

a) PS b) RS c) PQ d) QS [GATE - 2007]

Q.9 The span(s) to be loaded uniformly for maximum positive (upward) reaction at support P, as shown in the figure below, is (are)

a) P b) Q c) R d) S [GATE-2014]

a) PQ only b) PQ and QR 12. A simply supported beam AB of c) QR. and RS d) PQ and RS span, L = 24 m is subjected to two [GATE - 2008] wheel loads acting at a distance, d =

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 5 m apart as shown in the figure below. Each wheel transmits a load, P = 3 kN and may occupy any position along the beam. If the beam is an I-section having section modulus, S = 16.2 cm3, the [GATE-2015] maximum bending stress (in GPa) due to the wheel loads is ______.

ANSWER KEY:

1 2 3 4 5 6 7 8 9 10 11 12 C C C A C A B A D C A 1.78

Q.1 (c)

B.M. at A = intensity of loading × area under the loading 11   =10 × ×× 4 2  + 10 − ×× 4 2 = 0 22   ‘Zero’ can be visualized by seeing Whatever may be the position of the diagram also. couple, reactions are same at supports of s.s. beam with overhang. Q.3 (c) ∑=M0A

RLB × =µ= 1 1 ∴=R B L Q.2 (c) Use Muller Breslau principle. To draw ILD for MA, release the moment at ‘A’ by placing a moment hinges at ‘A’. (1) ILD for maximum sagging Draw the deflected profile by moment under 150 kN C.G. of allowing unit rotation at ‘A’ Loading system

100×+ 5 150 × 0 y X = ordinate at ‘B’ =1 100+ 150 2 = 2 m from 150 KN load ∴ y = 2 For maximum sagging moment both Similar ordinate at ‘D’ = 2 C.G of load system and the chosen load 150 kN shall be placed at equal distances on either side of centre of span ‘QR’ as shown below.

Q.6 (a) Muller Breslau’s principle can be ILD for maximum sagging moment used for drawing ILD of any force 2.4 y function not only of statically From similar triangle = determinate structures, but also of 61 indeterminate structures. It uses y = 0.4 under 100 kN load. deflection profiles as the basis for Maximum sagging moment ILD. = 150 2.4 + 100 0.4 = 400 kNm Q.7 (b) Q.4 (a) ILD for maximum and minimum reaction

If the unit load is at ‘B’. Rb = 1. The reaction at a support is also equal to S.F at support. Hence considering ILD for Reaction at R free body diagram of AB, the ILD Maximum reaction shown is of S.F at B. = 150 1 + 100 1.5 = 300 kN Minimum reaction we will get if Q.8 (a) leading load is at point P. The ILD is changing, sign in the According to I.L.D. minimum region ‘R S’ which is focal length. reaction = - 100 0.3 = - 30kN The ILD changes sign either for diagonal members or vertical Q.5 (c) members. The focal length shown is Having maximum overhang of diagonal ‘PS’. Hence the ILD available near support R, Maximum shown is of Members ‘PS’. hogging moment occurs at support R. Apply unit moment at support R

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission To have the maximum S.F just right of ‘Q’, place u.d.l on span ‘PR’. Hence Max S.f = Intensity of u.d.l [area of I.L.D under U.D.L] 11 =4 × 10 × 0.25 +× 20 × 1 22 = 4 [1.25+=× 10] 4 11.25 Max S.F = 45 kN Q.9 (d) “According to Muller Breslau’s 11. (a) principle, the ILD for any force of a 1 = 2 given structure is given by the deflection profile by release of that 𝛉𝛉 𝛉𝛉 force to some scale”. For getting ILD for reaction at ‘P’ release vertical force and allow it to deflect to some scale.

ab So, = ⇒=b 3a L 3L ILD for reaction at ‘P’ : 4 4 The reaction is positive for loads Also a+=⇒= b 1 a1 = 0.25 between placed on spans ‘PQ’ and 4 ‘RS’. ⇒=b3 = 0.75 Load is to be placed on span PQ & 4 RS, in order to have maximum vertical reaction at ‘P’. Q.12 (1.78) For maximum bending moment both C.G of load system and the Q.10 (c) Apply Muller Breslau’s principles. chosen load 3 kN shall be placed at Release the S.F just right of ‘Q’ by equal distances on either side of making a cut. Give a unit centre of span ‘QR’ as shown below displacement to just the right of ‘Q’ by doing so the slop to the right of ‘Q’ and left of ‘Q’ should be same, as per Muller Breslau. The deflected profile; by giving nit displacement as discussed above is shown below.

In 20m the ordinate is 1 In 5m the ordinate is 0.25

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission X(L− X) Max ordinate at P = L 10.75(24− 10.75) = 24 = 5.935 5.935*8.25 = Ordinate at Q 13.25 = 3.695

BMmax =3*5.935 + 3*3.696 =28.89 kNm

MM σ= ⋅y = IZ 28.89× 106 = 16.2× 103 =1783.3MPa =1.783GPa

Q.1 The influence line diagram for bending moment at section X(Mx), at a distance of 4 m from the left support of a simply supported girder AB is shown in figure below. The shear force at section X due to a A uniformly distributed load of uniformly distributed dead load of intensity 2 t/m longer than the span intensity 2 t/m covering the entire crosses the girder from left to right. span will be a) 8t b) 4t c) 2t d) 1t

Q.4 A beam with cantilevered arm BC supporting a freely supported end span CD is shown in the figure. Which one of the figures represents The maximum bending moment at the influence line for shear force at A section X is equal to a) 12 tm b) 24 tm c) 48 tm d) 96 tm

Q.2 The Muller-Breslau principle can be a) used to 1. Determine the shape of the influence line 2. Indicate the parts of the structure to be loaded to obtain b) the maximum effect 3. Calculate the ordinates of the influence lines Which of these statements is/are c) correct? a) only 1 b) both 1 and 2 c) both 2 and 3 d) 1, 2 and 3

Q.3 The influence line for shear at d) section x(FX) at a distance of 4 m from the left supported girder AB is shown in figure

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Q.7 Consider the beam ABCD and the influence line as shown below. The influence line pertains to

Q.5 Which one of the following represents the correct influence line for bending moment at point C for the beam shown in the figure below. a) reaction at A1 Ra b) shear force at B1 VB c) shear force on the left of C1 Vc- d) shear force on the left of C1 Vc+ a) Q.8 The absolute maximum bending Moment in a simply supported beam of span 20 m due to a moving UDL of b) 4 t/m spanning over 5 m is a) 87.5 t-m at the support b) 87.5 t-m near the midpoint c) 3.5 t-m at the midpoint c) d) 87.5 t-m at the midpoint

Q.9 The influence line for horizontal d) thrust of a two-hinged parabolic arch of span ‘/ and rise ‘h’ will be shown in a)

Q.6 Three wheel loads 10t, 26t and 24t spaced 2 m apart roll on a girder from left to right with the 10t load leading. The girder has a span of 20 meter. For the condition of b) maximum bending moment at a section 8 meter from the left end a) The 10t load should be placed at the section. b) The 26t load should be placed at c) the section. c) The 24t load should be placed at the section. d) Either the 26t load or the 24t load d) None of the above should be placed at the section. Q.10 A simply supported beam with overhangs is shown in the figure.

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission The influence line diagram for shear d) 1 alone is correct in respect of a section just to the right of the support ‘A’ will be as Q.12 Which one of the following equations shown in represents influence line of fixed end moment at B of the fixed beam AB of length l with origin at A? 2 x2 () l - x x() l - x a) b) a) l2 l2 x() l - x x2 c) d) l l2

b) Q.13 Influence line diagram for a truss member is shown in the above figure.

Positive values indicate tension. Dead load of the truss is 20kN/m and the d) live load is 10kN/m. Live load is longer than the span. Maximum tensile force in the member is: a) 600kN b) 400kN Q.11 Consider the following statements: c) 300kN d) 200kN 1. An influence line for a function (example: moment shear force, Q.14 Which one of the following is the reaction, deflection) in a influence line for reaction at A of the structure is a curve which shows beam shown in the figure? its variation at a particular section of the structure for various positions of a moving unit load. 2. The influence line for bending a) moment/shear force must not be confused with bending moment diagram and shear force diagram for the structure, b) 3. The bending moment of diagram and shear force diagram shows the moment/shear values at all the sections of the structure. The c) influence line diagram for BM/SF is always drawn for moving unit price load and for a particular section only d) Of these statements: a) 1, 2 and 3 are correct b) 1 and 2 are correct c) 2 and 3 are correct

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Q.15 Select the correct influence line diagram for shear force at x of the following beam. c)

a) d)

Q.18 Consider the following statements: 1. Influence Line Diagram (ILD) for c) SF at the fixed end of a cantilever and SFD due to unit load at the free end are same. d) 2. ILD for BM at the fixed end of a cantilever and BMD due to unit load at the free end are same. Which of these statements is/are correct? Q.16 The maximum bending moment at a) Only 1 b)Only 2 the left quarter point of a simple c) Both 1 and 2 d)Neither 1 nor 2 beam due to crossing of UDL of Q.19 The influence line diagram (ILD) length shorter than the span in the shown is for the member direction left to right, would occur after the load has just crossed the section by a) one-fourth of its length b) half of its length c) three-fourth of its length d) its full length

Q.17 For the continuous beam shown in a) PS b) RS figure. The influence line diagram c) PQ d) QS for support reaction at D is best Q.20 A uniformly distributed line of 60 represents as kN per meter run of length 5 meters on a girder of span 16 meters. What is the maximum positive shear force at a section 6 meters from the left a) end? a) 140.625 kN b) 65.625 kN c) 90.625 kN d) 45.625 kN

1 2 3 4 5 6 7 8 9

10 11 12 13 14 15 16 17 18 (a) (a) (a) (b) (c) (b) (c) (c) (a) 19 20 (a) (a)

Q. 1 (c) R1 =+= 26 24 50t The maximum bending moment at section-X will occur when UDL R / L= 50 / 8 = 6.25 t / m > R / L occupies the whole span. 11 When 26t load crosses the section C ∴ Maximum bending = Area under R1 = 24t the influence line diagram R /L= 24/8 = 3t/m = R/L 11 11 = ×××+432 × 1232 ×× It means that maximum bending 22 moment is obtained when 26t load 24+ 72 = =48 t − m is on the section 2 Q. 7 (b) Q.2 (d) Q.8 (d) Q.3 (a) Absolute maximum bending SF will be equal to the area under moment will occur at the centre the influence line diagram when the loads is spread equally on 1 31 1 either side of the centre. SF= 2 × 12 × − ×× 4 2 42 4 wa a Mlmax = − =2[] 4.5 – 0.5 = 8 t 42 45×  5 Mmax = 20 − Q.4 (a) 42 =×=5 17.5 87.5 t-m centre Q.5 (d) Q.9 (a) Q.6 (b) Figure in (a) is the I.L.D of Maximum bending moment at a horizontal thrust for three hinged section occurs when a particular arch. load is on the section which changes the ratio R1/L1 > R/L to R1/L1 < R/L Q.10 (a) as the load passes over the section Q.11 (a)

Q.12 (a) For the influence line of fixed end

Where R1 → resultant of load on moment at B, release the moment at left side of section B and give unit rotation, the Resultant of all loads (R) deformed shape will represent the =++=10 26 24 60t influence line R / L= 60 / 20 = 3t / m When 10t load crosses section C

x2 (l - x) FEM = B l2 Due to unit load at x distance from A.

For maximum positive SF at D, the Q.13 (b) loading should be applied as shown Tensile force due to D.L in the figure. Maximum positive = =()20 −×= 10 20 200 kN load × area of ILD Maximum Tensile force due to L.L SF at D intensity covered by the load =20 ×= 10 200 kN 55 5 =×+60 kN Thus maximum tensile force 2 8 16 =200 + 200 1125 = 400 kN =kN = 140.625 kN 8

Q.14 (c)

Q.15 (b) Maximum shear force is at either of the support due to a point load

Q.16 (c) The load should be positioned such that section divides the span and load in the same ratio.

Q.17 (c) The ILD for support reaction at D can be obtained by giving unit displacement in the direction of reaction. The deflected shape of beam will represent ILD as in figure(c).

Q.18 (a) ILD for BM at fixed end will have maximum ordinate when the unit load is at free end. While the BMD due to unit load at free end will have zero ordinate at free end and maximize ordinate at fixed end.

Q.19 (a)

Q.20 (a) We must first draw the influence line

5.1 Introduction • Moment distribution method is the most suitable manual method for Ml θ= analysis of continuous beams and plane 4EI frames. The method was presented by M 4EI Prof. Hardy Cross of USA in 1929. Stiffness= K = • The method consists in solving θ l indirectly the equations of equilibrium as formulated in slope deflection method without finding the b) Far end Hinged displacements. This is an iterative procedure. This is also known as relaxation method. Ml 5.2 Basic Concept θ= 3EI In this method the analysis begins by M 3EI Stiffness= K = assuming each joint in the structure to be θ l fixed. Then, by unlocking and locking each joint in succession, The internal moments at the joints are distributed and balanced 5.5 Distribution Factor (D.F.):If moment until the joints have rotated to then final or is applied at a joint which distributes to nearly final positions). The following connecting members in the proportion of examples has been given to illustrate the their stiffness. basic concept. Hence, DF of a member is SIGN CONVENTION Stiffness of member 5.3 Sign Convention DF = Total stiffness of all members at the joint Clockwise moment is taken as (+) ve, OR anticlockwise moment is taken as (-) ve i.e. Relative stiffness of member DF = the sign convention is same as that in slope Total relative stiffness at the joint deflection method for member end moment calculation.

5.4 Stiffness

Stiffness of member is moment required to produce unit rotation at a joint. If a moment ‘M’ is applied at joint A’ due to a) Far end Fixed which the joint rotates by A then,

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Moment distributed in AD K DF for AB , = AB 4EI3Aθ K MAD = ∑ l3 K DF for AC , = AC Moment distributed in AC ∑ K 4EI θ M = 2A K AC l DF for AD , = AD 2 K Moment distribution in AB ∑ 3EI θ M = 1A AB l 1 5.6 Carry over Factor I II3 = 3 21 Carried over moment MAD : M AC : M AB : : Carry over Factor = l32 l 4l 1 Applied moment I 3 = Relative stiffness of AD ⇒ l3 I 2 = Relative stiffness of AC l2 3 I 1 = Relative stiffness of AB EI = constant throughout AB 4l1 M = applied Moment I ⇒=Relative stiffness when far end is fixed When moment ‘M’ is applied at pin, l M Moment carried over to fixed far end = 3I 2 ⇒=Relative stiffness when far end is hinged 4 l However, in the following case, carry over factor = 0. Note: Meaning of far end

5.7 Distribution Factor for Fixed and Pin Ends

Stiffness of member AD At end ‘A’ two members, Wall and AB joins. Hence, distribution factor at 4EI3 KAD = KAB l3 A0= = ∞+ Stiffness of member AC KAB

4EI2 DF of fixed end = 0 KAC = l2 Note: Stiffness of member AB Wall has infinite stiffness i.e. for all applied 3EI moments; rotation is zero as in case of fixed K = 1 AB MAB l1 ends = ∞ 0 = ++ ∑ KKAB K AC K AD At end ‘C’ there does only one member exist at joint C, i.e. CB

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission K ⇒ D.F at C =CB =1 KCB ⇒ D.F. of hinged end = 1 From these we conclude that wall or fixed ends does not rotate and absorbs all the moment carried over to it

5.8 Causes of sway of frames  Unsymmetrical loading  Different end condition od column  Non uniform section(change in EI)  Unsymmetrical outline  Horizontal loading  Settlement of supports

Q.1 A frame ABCD is supported by a Q.3 The plane frame below is analysed roller at A and is on a hinge at C as by neglecting axial deformations. shown below: The reaction at the Following statements are made with roller end A is given by respect to the analysis:

a) P b) 2P c) P/2 d) Zero 1. Column AB carries axial force [GATE-2000] only. 2. Vertical deflection at the center of beam BC is 1 mm. With Q.2 The frame below shows three beam reference to the above statements, elements OA, OB and OC, with which of the following applies? identical length L and flexural a) Both the statements are true rigidity EI, subject to an external b) Statement 1 is true but 2 is false moment M applied at the rigid joint c) Statement 2 is true but 1 is false O. The correct set of bending d) Both the statements are false

moments {}MOA , M OB , M OC [GATE - 2004]

Q.4 All members of the frame shown below have the same flexural rigidity EI and length E. if a moment M is applied at joint B, the rotation of the joint is

a) {3M/8, M/8, 4M/8} b) {3M/11, 4M/11, 4M/11} c) {M/3, M/3, M/3} d) {3M/7, 0, 4M/7} [GATE-2001]

Q.8 The bending moment at the middle support is a) wL2/4 b) wL2/8 c) 5wL/4 d) 5wL/8 [GATE - 2007] a)ML/12El b)ML/11El Q.9 All members in the rigid-jointed c)ML/8EI d)ML/7El frame shown are prismatic and have [GATE – 2005] the same flexural stiffness EL Find the magnitude of the bending Q.5 Vertical reaction developed at B in moment at Q (in kNm) due to the the frame below due to the applied given loading. load of 100 kW (with 150, 000 mm2 cross-sectional area and 3.125 x 109 mm4 moment of inertia for both members) is

[GATE-2013]

a) 5.9 kN b) 30.2 kN Q.10 Considering the symmetry of a rigid c) 66.3 kN d) 94.1 kN frame as shown below, the [GATE -2006] magnitude of the bending moment (in kNm) at P (preferably using the Q.6 Carry-over factor CAB for the beam moment distribution method) is shown in the figure below is

a) 1 / 4 b) 1 / 2 c) 3 / 4 d) 1 a) 170 b) 172 [GATE – 2006] c) 176 d) 178

Common .Data for Q.11 and Q.12: [GATE-2014] A two span continuous beam having equal spans each of length L is subjected to a Q.11 The portal frame shown in the uniformly distributed load w per unit figure is subjected to a uniformly length. The beam has constant flexural distributed vertical load w (per unit rigidity. length).

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission The bending moment in the beam at Q.14 The value of moment M in beam ABC the joint 'Q' is shown in figure such that joint B does not rotate.

The bending moment in the beam at the joint 'Q' is The value of support reaction (in kN) at B wL2 should be equal to______a) zero b) (hogging) 24 [GATE-2017-1] 2 2 wL wL c) (hogging) d) (hogging) Q.15 A prismatic beam PQR of flexural 12 8 rigidity EI= 1*104 kN.mm2 is subjected to [GATE-2016] moment of 180 kNm at Q as shown in Q.12 For beam shown below, the value of support moment is______kNm figure.

The rotation at Q (up to two decimal places) is ______[GATE-2015-1] [GATE-2018-2]

Q.13 consider portal frame shown in

4 8 figure. Assume E=2.5*10 MPa , I = 8*10 mm4 for all member of frames.

Total rotation (in degree up to one decimal place) at the rigid joint Q would be_____

[GATE-2017-2]

Q.1 (d) Vertical deflection at the centre of Apply ƩM = 0 the span Say horizontal reaction at A as 𝑐𝑐 H . 5WL42 5×× 10 5 A = = = 1 mm P.L P.L 384EI 384× 813 H . L =+ A 2 2 . HA = 0 Hence reaction at Roller end A = 0 Q.4 (b) Q.2 (d) Joint Part K DF= Member Joint Part K Moment 𝐾𝐾 BA 4EI ∑K =DF*M ∑K OA 3EI 3 ∑𝐾𝐾 3 l 11EI M B BC 3EI l 7EI 7 7 l O OB 0 0 0 l l BD OC 4EI 4 4 4EI M l 7 7 l M M.L rotation at B, θ == B K 11EL

Q.3 (a) Q.5 (a) The frame has no horizontal loads. At joint ‘A’ Hence no horizontal reaction at ‘A’. Deflection in Beam = Contraction in Column AB is subjected to axial Column ()100− R .L3 R.L force only. = 3EI AE ()100−× R (1000)2 R = 3×× 3.125 109 150,000 R = 5.88 kN (say) 5.9 kN

Q.6 (d) Carry over factor Moment developed at far end C = AB Moment applied at near end

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission MC = 0 5WL 5WL 5WL Total R ==+ Take momentB at884 C, Ʃ ∴ WL22 WL R B ×=L + let us apply moment ‘M’ at A 8 2 2 for RA ; take moment at C = 0 5WL ∴=R B Mc = 0 RA × L = M 8 The given continuous beam can be RA = M/L (upward) Ʃ M treated as two symmetrical propped & R = (downward) cantilevers. B L 5WL Mc = 0 from right side Each span R B = MB = RB× L 8 (It is a standard case, which can be Again Ʃ M ML= × remembered) B L MB =M Q.8 (b) As per above solution Carry over factor Moment at B M Q.9 (25kNm) = = =1 Moment at A M Joint Part K DF= Member Moment Q.7 (c) ∑K =DF*M TP 3EI 1𝐾𝐾 1 ∑𝐾𝐾 M 3 4EI 4 4 T TQ 4EI 2 2 M 2 4 4 TR 4EI 1 1 M 4 4 4 TS 0 0 0

22 =M = *100 Moment transfer at TQ 4 4 = 50kNm Carry over factor = 1/2 50 = Moment transfer at Q 2 = 2N5k m MA = 0 WL22 WL Q.10 (c) TRakeB ×= Lmoment +at A, Ʃ 8 2 5WL2 ∴=R B 8

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Moment at hinge is 10*1=10kNm Carry over moment at fixed end = 10/2 =5kNm Q.13 (1ο ) Moment at joint Q=2000*2-1650*2*1 =700 kNm 4EI 4EI ∑ K =+= 2EI 44 N EI= 2.5*104 *8*10 84 mm mm2 Axis of symmetry is passing through 4 3kN 8 −12 4 a column; hence it can be treated as EI= 2.5*10 *10 *8*10 *10 m m2 Member Stiffness D.F = 2 4EI 1 EI 20000kN.m BA M 700 350 350 64 θ= = = = 4EI() 4I 3 K 2EI EI 20000 BC 180 84 θ=0.0175* = 1ο FE11: π −WL2 −× 24 8×8 Q.14 (60 kN) M= = = − 128kNM BC 12 12 As joint B is not getting rotates means it has MCB = +128 kNM to treat as Fixed where rotation is always 0. Now, beam is treated as fixed beam.

RB =wl/2 =30*4/2 = 60 kN (Reaction at B will not come from BC span BM at C = 176 kNm as there is not any load on span)

Q.11 (a) Q.15 (0.01 rad) VP=wl/2 VS=wl/2 4EI 4EI So, HP=0 ∑ K=+= 1.8EI M =0 54 Q.12 (5kNm) EI= 10000kN.m2 ⇒ Q M 180 180 θ= = = =0.01rad K 1.8EI 1.8*10000

Q.1 The given figure shows a portal Q.4 The fixed end moment MA for the frame with one end fixed and beam shown in the figure is another end fixed and other hinged. The ration of the fixed end moments M BA due to side sway will be MCD a) +40 kN.m b) -40 kN.m c) +80 kN.m d) -80 kN.m

Q.5 A portal frame is shown in the given a) 1.0 b) 2.0 400 figure. If θ=θ=BB radian, then c) 2.5 d) 3.0 EI the value of moment at B will be Q.2 The distribution factors for members CB, CD and CG for the frame shown in the figure (EI constant) will be respectively

a) 120 kNm b) 240 kNm c) 360 kNM d) 480 kNm

Q.6 For the frame as shown in figure below, the final end moment MBC has been calculated as –40 kN-m.

a) 0.24, 0.28 and 0.48 What is the end moment MCD ? b) 0.24, 0.48 and 0.28 c) 0.48, 0.24 and 0.28 d) 0.28, 0.48 and 0.24

Q.3 A fixed beam AB is subjected to a triangular load varying from zero at end A to w per unit length at end B. The ration of fixed end moment at B a) +40kN-M b) -40kN-M to A will be c) +30kN-M d) -30kN-M 1 1 a) b) 2 3 Q.7 A horizontal fixed beam AB of span 2 3 6 m has uniform flexural rigidity of c) d) 4200 kN/ m2. During loading the 3 2 support B sinks downwards by 25

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission mm. The moment induced at the end Q.11 Four identical beams AE, BE, CE and A is DE have been rigidly joined at E. The a) 17.5 kN m(Anticlockwise) point C slip and routes along with b) 17.5 kN m(Clockwise) member firmly fixed at E c) 105 kN m(Anticlockwise) d) 105 kN m(Clockwise)

Q.8 The possible direction of sway of the rigid frame shown in figure

Which one among the following is correct? a) There is no moment on the members b) Except at C, there is no moment on the members of frame c) Except at C and E for member a) is towards left EC, no moment will be there on b) is towards right other members c) does not exist as there is no sway d) All the members are subjected to d) cannot be ascertained moment. Q.9 Members AB and BC in the figure Q.12 A uniform simply supported beam is shown are identical. Due to a subjected to a clockwise moment M moment 2M applied at B, what is the at the left end. What is the moment value of axial force in the member AB? required at the right end so that rotation of the right end is zero? a) 2M b) M c) M/2 d) M/3

a) M/l (compression) b) M/ l (tension) c) 1.5 M/ l (compression) d) 1.5 M/ l (tension)

Q.10 What are the distribution factors at joint B for the members BA and BC respectively in the figure?

a) 0.57 and 0.43 b) 0.43 and 0.57 c) 0.50 and 0.50 d) 0.36 and 0.64

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission ANSWER KEY: 1 2 3 4 5 6 7 8 (a) (b) (d) (b) (b) (d) (a) (c) 9 10 11 12 13 14 15 16 (d) (c) (d) (c)

Q.1 (a) Due to sway, the deflection of point B will be equal to that of point C. 6ElΔ M= BA L2 3E(0.5l)Δ 6elΔ M = = CD (L/2)22 L M ∴ BA = 1.0 MCD Q.5 (b) Q.2 (b) The deformed shape of structure will have translation at B and C 2EI M = 0 + (2θ + θ ) BC10 B C Member Relative Distribution 2EI 400 stiffness Factor = ×3× = 240 kN-m CB El/8 0.24 10 EI CD El/4 0.48 CG El/7 0.28 Q.6 (d) The shear force at end A should be Total 29El 1.00 equal and opposite to shear force at 56 D. Let -40 kN-m denotes the clockwise moment so moment at Q.3 (d) end B of column AB is 40 kN-m anticlockwise and the shear force at A is 10 kN towards left. Therefore shear force at D is 10 kN towards right. Thus end moment at M 3 C is 30 kN-m clockwise or -30 kN-m. B = M2A Q.7 (a) Q.4 (b) 6EIδ 6*4200*0.025 M AB = = For part B.D. L226

MAB =17.5( anticlockwise ) Q.8 (c)

Q.9 (d) The distribution factors for member AB and BC at B will be 0.5 for both.

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Thus member end moment at BC is M ∴ M' = M. 2 M Moment at C = 2 1.5M Reaction at B R= B l This reaction is away from B so axial force is member AB is tensile. This can be found from body diagram of AB. Similarly axial force in member 1.5M BC will also be but l compressive. Q.10 (c)

kBA Distribution Factors D=BA , kBA +k Bc

kBA D=BC kBA +k Bc For propped cantilever portion BA 4EI k = = EI BA 4 For simply supported portion BC 3EI k = = EI BC 3

Thus DBA = D BC = 0.5

Q.11 (d) Since the member are firmly fixed at E. It is a rigid joint at which joint at which all the members will have same rotation. So from slope deflection equations all the members will have moment.

Q.12 (c)

Rotation Left End Right End Clockwise ML ML 3EI 6EI Moment (M) (clockwise) (anticlockwise) at left end to keep rotation at right and zero, a moment should be applied in anticlockwise direction. Let the moment in ‘M’ ML ML = (for zero rotation) 3EI 6EI

6.1 Introduction

• Slope deflection method is useful to ωl MMAB+ BA analyses indeterminate structures like R A = − continuous beams and plane frames. 2l + • The unknowns in this method are ωl MMAB BA R B = + degree of freedom i.e. slope (0) and 2l deflection (A). • Combined these slopes and deflections are known as displacements. Thus slope-deflection method is a displacement method. • This is a classical method on which moment distribution method, Kani’s • Thus, in slope deflection method we method and stiffness matrix method are establish a relationship between the based. degree of freedom (0, A) and the Analysis of a beam and frame means member end moments. This determination of bending moment and relationship is called slope deflection shear force throughout the length of the relationship. member i.e. determination of BMD and • Finally by using equilibrium equation SFD for the member. BMB and SFD for a we find the slope deflection relationship member of structure can be drawn if we to obtain the member moments. know the internal end moments of a • To find out slope-deflection member. relationship, method of superposition is used. For example: if we have a fixed beam as shown below. Example: (Continuous Beam)

The BMD and SFD can be found if we know the internal end moments of member i.e. MAB and MBA ∆= deflection of joint B with respect to joint A

θA and θB = rotation of A and B

MAB = Internal member end moment at ‘A’

MBA = Internal member end moment at ‘B’ To find out the effect of external loading,

rotation θθAB, and displacement ''δ on

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission internal moments, we follow method of 2EIθ 4EIθ 6EIδ M = M +AB + - superposition. BA FBA l ll2 2EI 3δ 6.2 Fundamental of slope deflection MAB =M FAB + 2θ A +θ B - ll equation: .....(A) 2EI 3δ MBA =M FBA + 2θ B +θ A - 1. Consider all the joints to be fixed. The ll member end moment due to external This equation A is called slope deflection loads is MFAB and MFBA equation. 6.4 Equilibrium Equation • We know that in slope deflection Effect of external load 2. Allow the support B to settle with method; slope deflection equations are respect to A. The member end moments written which relates member end generated are as show below. moments to the slopes and deflections (i.e. degree of freedom of the structure). Hence, if values, of slopes and deflections are known, member end moments are calculated. 3. Allow end A to rotate. The member end • To find out the slope and deflections, moments are as shown below. equilibrium equations are to be written. No. of such equilibrium equations, required for complete solution is equal to the no. of degree of freedom of the structure. • The equilibrium equations are classified 4. Allow end B to rotate. The member end as: moments are as shown below. 1. Joint equilibrium equation. 2. Shear equations • Joint equilibrium equations are written to find out the values of 0 and shear equations are written to determine 5. Axial deformation is neglected. unknown displacement (δ ). 6. Shear deformations are neglected. The following example, illustrates how do we write down the joint equilibrium 6.3 Sign Convention: (For calculation equation and shear equation. of member end moment). • Clockwise moment is taken as (+) ve. • Clockwise rotation is taken as (+) ve. • If ''δ gives clockwise rotation, it is taken as positive ((+) ve). For example Joint equilibrium Equation in the above discussion if end support B settles by amount ''δ with respect to A, it is taken as (+) ve. Thus, from the principal of superposition, we have Equilibrium of Joint B MM0AB+= BC ....(i) 4EIθAB 2EIθ 6EIδ MAB =M FAB + + - 2 Equilibrium of Joint C M - P l = 0 ...... (ii) l ll CB 3 3

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Shear equations for each member. Form this, the support reactions can also be RA+ R B + R C = Wl 112 +++ PPP 3 calculated. Note: (f) Sketch shear force diagram (SFD) and a) If unknown displacement ()δ is bending moment diagram (BMD). horizontal, the shear equation is = ∑F0H 6.6 Examples of Fixed end moments b) If unknown displacement ()δ is = vertical, the shear equation is ∑F0v Where

EFH = Summation of all horizontal forces = 0

EFV = Summation of all vertical forces = 0.

6.5 Steps for Analysis in Slope Deflection Method

a) Computation of fixed end moments: The formulae for fixed end actions for various load cases are given in Table on the next page. The sign convention followed is that for moments, i.e. clockwise positive and anticlockwise - negative. b) Relate member end moments to joint displacement: The end moments are as follows : 2EI 3δ M2AB=θθ A +− B + M FAB LL 2EI 3δ MBA=θ A +−+ 2θ B M FBA LL These are also known as slope deflection equations. c) Formulate equilibrium equations: These are obtained by making algebraic sum of moments at each joint as zero. In case of frames with sway, additional equations are obtained considering shear condition. d) Solve the equations: This will give displacements (primary

unknowns), i.e. θAB , θ , δ etc. e) Back-substitution: In the expressions for end moments formed in step (b), substitute values of known displacements as obtained in step (d). This gives final end moments

6.7 Shear equation in frame

∑ H0=

PH−−=AB H 0

HHPAB+=

∑ M0B =

HA .h−−= M AB M BA 0 MM+ H = AB BA A h

∑ M0C =

HD .h−−= M CD M DC 0 MM+ H = CD DC D h

2EI 3∆ c) MBA =+− 30 2θ Q.1 The figure given above shows a rigid 44 frame. If D is a lateral translation of 2EI 3∆ the joints, slope deflection equation d) MBA =++ 30 2θ for the member BA be written as: 44 Q.2 Final moment values got through the analysis of a portal frame have been shown in the figure. What is the value of X?

2EI 3∆ a) MBA =−+ 30 2θ − 44 2EI 3∆ b) MBA =−− 30 2θ + 44

a) 96 kN b) 60 kN c) 30 kN d) 24 kN

ANSWER KEY: 1 2 3 4 5 6 7 8 c b

Q.1 (c) Taking: FEMBA = 30 kN-m Sign of moment due to sinking of support B is already considered in derivation of equation.

Q.2 (b) Free-body diagram of columns Applying horizontal force equilibrium condition for frame x 30+−=x 0 2

X = 60 N

7.1 Introduction Purposes of choosing statically • Differential settlement of supports, indeterminate structures as compared to temperature variation, change in statically determinate ones are as follows: length due to fabrication errors in indeterminate structures will introduce (a) For a given loading, the maximum internal stresses in structure. stress and deflection of an Note: indeterminate structure are generally In statically determinate structures smaller than that in statically internal stresses will not be introduced determinate structure. because of these factors. For example: In any statically indeterminate structures, it is necessary to satisfy: 1. Equilibrium equations 2. Compatibility equation 3. Force displacement requirement Pl Max BM = There are two different ways to satisfy 4 these requirements.

7.2 Force method

1. Also called Compatibility method or Pl Method of consistent deformation or Max BM = 8 Flexibility method 2. Unknown in this case are Forces (b) Indeterminate structures have a (reaction, BM, SF) tendency to redistribute its load to 3. Force-displacement equation are its redundant supports in case where written and solution for unknown faulty design or overloading occurs. forces is obtained from compatibility equations. Example :

As load P is increased, plastic hinge will form at supports first arid hence it will be treated as simply supported structures. Further load can also be resisted. However, in case of simple supports, hinge will form at centre hence collapse will be early. • Although there will be cost saving in material due to lesser stress in member, the cost of construction of supports and joint may sometime offset the saving in material.

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 4. Once unknown forces are known, the 4. Once unknown displacements are reactive forces like BM, SF etc. are known, internal forces are found using found using equilibrium equation. compatibility and load-displacement equations.

RL3 ∆= : Load displacement 2 3EI relationship Example : WL43 RL = : Compatibility equation 8EI 3EI Wx 2 BM= Rx − :Equilibrium equation X 2 5. Force method of analysis of θ = unknown displacement 2EI indeterminate structure is suitable M= M + ( −θ ): Load when degree of static indeterminacy is AB FABl B less then degree of kinematic displacement relation indeterminacy. 2EI M= M + ( −θ 2 ): Load BA FBAl B displacement relationship

MBA = 0: Equilibrium equation 4EIθ −Wl2 ⇒==B M lFBA 12 Wl3 No. of unknown reactions = 2 ⇒θ = No. of unknown joint displacement B 48EI

= 4()θABCD ,θ ,θ ,θ From θB, M AB can be found using load- displacement relationship. 7.3 Examples of force method 5. Displacement method is suitable when a) Castigliano’s theorem (method of degree of kinematic indeterminacy is least work) less than the degree of static b) Strain energy method indeterminacy. c) Virtual work method / Unit load method d) Claperon.s three moment equations [used in continous beam analysis]. e) Column analogy method (used in rigid frames with fixed supports). Unknown reaction = 4 f) Flexibility matrix method unknown displacement (θθ=BC , ) 2 ⇒ Displacement method is suitable 7.4 Displacement method 7. 5 Examples of displacement 1. Also called Stifness method methods 2. Unknowns in this case are a) Slope deflection method Displacement ( , 0) b) Moment distribution method 3. Force-displacement equations are c) Stiffness matrix method written and ∆solution for unknown d) Kani’s method displacement is obtained from equilibrium equations.

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission By compatibility condition, i.e. net 7.6 Various Force Methods: deflection of end B = 0 [as it is hinged] we have,

7.6.1 Method of Consistent Deformation ∆=BCR 2 δ BB

∆BC R 2 = δ BB # if we take M as redundant, then actual Example: beam can be thought of as being composed of primary structure & beam with redundant loading, i.e.

• The beam ABC is a redundant beam with degree of redundancy as one. • Any one of the three reactions can be taken as redundant. • Let us take R2 as redundant. At this stage we consider the beam to be composed of: a) Primary structure, which is obtained by removing the redundant and loading the resulting beam with external loading only.

θ+θ=M0 b) A beam with loading as redundant AC AA reaction. Where θ=AC rotation at A due to external loading in Primary Structure. θ= rotation at A due to unit couple at AC Let the deformations of beam due to A (unit couple). −θ these loading will be as under. M = AC θAA Example

Where, δBB = deflection at B due to unit load applied at B

fCB = Deflection at C due to unit load at B

∆= deflection at the location of P PQ2 2 due to Q-system of forces Hence, as per Betti’s theorem. PP∆∆+= Q ∆ + Q ∆ 1 PQ12 2 P Q 1 QP 1 2 Q 2 P

= Deflection at B due to unit load at C fBC Proof of Betti’s theorem

fCC = Deflection at C due to unit load at C From compatibility conditions net deflection at B & C is Zero.

⇒∆B +R B .f BB + R C .F BC = 0 ...(i)

⇒∆C +R B .f CB + R C .F CC = 0 ...(ii)

By solving (i) and (ii), R B and R C is Let is applied 1st on the portal calculated. Q12 ,Q

frame and then P1 and P2 are slowly 7.6.2 General Reciprocal Virtual Work applied Theorem (Betti’s Theorem) 11 1 Work done= Q∆ + Q ∆ + P ∆+ 221 QQ1 2 Q 21 Q 2 1 PP The virtual work done by a P-force 1 system in going through the PQQ∆+∆+∆ 2 PP22 1 PP 2 QP 2 deformation of Q-Force system is 2 Now if and are applied 1st on the equal to the virtual work done by the Q- P1 P2

force system in going through the portal frame and then Q1 and Q2 are deformation of P-force system. applied 11 1 Work done= P∆+ P ∆+ Q ∆ + 2212PP12 P P 2 1 QQ 1 1 ∆ +∆PQ +∆ P 2 2QQ22 1P1 Q 2 PQ Equating the two works done (as sequence of application will not have any effect on final work done). PP∆ +∆ = Q ∆ + Q ∆ 1 PQ12 2 P Q 1 QP 1 2 Q 2 P ∆=QP deflection at the location of Q1 1 The works are called virtual work because due to P-system of forces ∆ is not the deflection at P location due P1Q 1

to P1 itself.

7.6.3 A Special Case of Betti’s Law (Maxwell’s Reciprocal Theorem)

If only two forces P and Q are acting and magnitude of P and Q are unity.

∆PQ= ∆ QP Where ∂U ∆=PQ Deflection at the location of P due = 0 Taking M and R3 as redundant ∂M to unit load at the location of Q ∂U ∆= Deflection at the location of Q = 0 QP ∂R due to unit load at the location of P 3 Where U = strain energy in the beam AB i.e., deflection at the location of Q due and BC due to combined Action of to unit load at ‘P’ is equal to external loading and reactions. deflection at the location of P due to • Thus theorem of least work is a special unit load at Q. case of Castigliano’s 2nd theorem. Example • This theorem applies to beam, frame, 1. and truss everything.

7.6.5 Castigliano’s Theorem

∂U  = Pn δ=δAB BA  ∂∆n  1st theorem 2. ∂U = M  n  ∂θn  ∂U  = ∆ n  ∂Pn  2nd theorem ∂U ⇒δ =δ = θ  BA AB n  ∂Mn  i.e., Downward deflection at B due to clockwise unit couple at A is equal to clockwise slope at A due to downward Note: unit load at B. • Castigliano extended the principal of least work, later on, to the self-straining system. 7.6.4 Theorem of Least Work • If λ = Small displacement in the For any statically indeterminate direction of redundant force R then, structure, the redundant should be such ∂U =λ as to make the total internal energy ∂R within the structure a minimum. • Self-straining may be caused by settlement of support of a redundant structure by an amount λ or by the initial misfit of a member by an amount λ too short/long.

∂U 7.6.6 Strain energy = 0 Taking R2 as redundant ∂R 2

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Energy method is useful to calculate deformation of structure subjected to complicated load. Strain energy method is based on conservation of energy principle which states that work done by all external forces acting on structure Ue, is transformed in to internal work done Ui which develops when structure deforms.

Ue=Ui  Strain energy due to axial loading,

PL2 U = axial 2AE Where, AE= Axial rigidity  Strain energy due to bending(flexure),

2 L M U= (X) dx bending ∫0 2EI Where, EI= Flexural rigidity

 Strain energy due to Shear,

2 L F U= (X) dx shear ∫0 2GA Where, GA= Shear rigidity

 Strain energy due to Torsion,

2 L T U= (X) dx torsion ∫0 2GJ Where, GJ= Torsional rigidity

Codes: Q.1 For a linear elastic structural A B C D system, minimization of potential a) 1 2 1 2 energy yields b) 1 1 2 2 a) compatibility conditions c) 2 2 1 1 b) constitutive relations d) 2 1 2 1 c) equilibrium equations [GATE-2005] d) strain-displacement relation Q.4 Consider the stepped bar made [GATE – 2004] with linear elastic material and Q.2 The unit load method used subjected to an axial load of 1 kN as in Structural analysis is shown in figure. a) applicable only to statically indeterminate structures b) another name for stiffness method c) an extension of Maxwell's reciprocal theorem d) Derived from Castigliano's theorem [GATE – 2004] Q.3 Match List-I with List-II and select the correct answer using the codes given below the lists: List – I A. Slope deflection method B. Moment distribution method C. Method of three moments The strain energy in N.mm up to one D. Castigliano's second theorem decimal place in the bar due to an List – II axial load is ______1. Force Method [GATE-2017-1] 2. Displacement

EXPLANATIONS

Q.1 (a) Compatibility conditions deals with balancing of displacements and minimization of potential energy yields compatibility conditions

Q.2 (d) The Unit Load Method is derived from Castiglione’s theorem-1. Q.3 (c)

Q.4 (35Nmm)

PL2 U = axial 2AE 100022 *400 1000 *900 U = + axial 2*100*2*1055 2*60*3*10

Uaxial = 35Nmm

Q.1 The following methods are used for structural analysis 1.Macaulary method 2.Column analogy method 3.Kani’s method 4.Method of sections Those used for indeterminate a) b) structural analysis would include a) 1 and 2 b) 1 and 3 c) 2 and 3 d) 2, 3 and 4

Q.2 Match List-I (Names of persons with c) d) whom the methods of analysis are associated) with List – II (Method) and selected the correct answer using the codes given below the lists

List-I 1. Clapeyron Q.4 If the flexural rigidity of the beam 2. Hardy cross BC of the portal frame shown in the 3. Lame given figure is assumed to be zero, 4. Euler then the horizontal displacement of List – II the beam would be A. Moment distribution method B. Method for determining crippling load on a column C. Theorem of Three moment D. Thick cylinders Codes: 1 2 3 4 a) C A B D b) A C D B Ph3 Ph3 c) C A D B a) b) 3EI 24EI d) A B D C Ph3 Ph3 c) d) Q.3 A loaded portal frame is shown in 12EI 6EI figure. The profile of its bending moment diagram will be Q.5 Match List-I (Method) with List-II (Factor) and select the correct

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission answer using the code given below d) 3 1 5 2 the lists: List-I Q.6 Which one of the following A. Moment distribution method statements is correct? B. Slope deflection method The principle of superposition is C. Kani’s method applicable to a) nonlinear behaviour of material D. Force method and small displacement theory LIST-II b) nonlinear behaviour of material 1. Rotation factor and large displacement theory 2. Flexibility c) nonlinear elastic of material and 3. Hardy Cross small displacement theory 4. Displacements d) linear elastic behaviour of 5. Stiffness matrices material and large displacement Codes: theory. A B C D a) 3 4 1 2 b) 2 1 5 3 c) 2 4 1 3

ANSWER KEY: 1 2 3 4 5 6 (c) (c) (d) (d) (a) (c)

Q.1 (c) Macaaulay’s method is used for A – 3, B – 4, C – 1, D - 2 deflection and slope calculation due to point loads in prismatic beams. Method of sections is used for Q.6 (c) statically determine trusses. The principle of superposition states that the displacement resulting from Q.2 (c) each of a number of forces may be 1-C, 2-A, 3-D, 4-B added to obtain the displacements resulting from the sum of forces. Q.3 (d) This method depends upon the - The horizontal force will be linearity of the governing relations distributed to both the supports. between the load and deflection. - The bending moment at hinged The linearity depends upon the two end will be zero and fixed end factors. will have some B.M. - The B.M.D for columns will be (i) The linearity between bending linear. moment and curvature which - The B.M.D for beam will be depends upon the linear elastic parabolic. materials. If non-linear materials The choice now remain between (b) superposition of curvatures is and (d). not possible. The beam at left end in (b) shows zero B.M which is not possible (ii) The linearity between So only option D is valid. curvatures and deflection Q.4 (d) depends upon the As the beam BC has no rigidity. So assumption that the deflections end B of column behaves as free are so small that the end. approximate curvature can be For a cantilever beam subjected to used in place of true curvature. concentrated load at free end, the 3EI stiffness is l3 Stiffness of given system 3EI 3EI 6EI =+= hhh333 So displacement of point Ph3 B = 6EI

Q.5 (a)

Before starting the chapter of matrix method of analysis, we need to acquaint 13− 2 our self with the basics of determinant and  35 7 matrix. 43 1 8.1 Definition and Notations Is a square matrix of dimension (or order) 3. Here, the elements 1, 5 and 1 are called • Matrix is defined as an array of the diagonal elements and the diagonal is quantities, usually called elements, named as principal diagonal. grouped together for some specific • If any matrix (square type) has the purpose arranged systematically in diagonal elements non-zero and the rows and columns. non-diagonal elements are all zero, it is • For demonstration purpose, these called a diagonal matrix. elements are enclosed by either square • If any square matrix has the elements brackets, { }, or parentheses ( ). symmetrical about the principal • In general, the entire matrices are diagonal, it is called a symmetric i.e = denoted by boldface letters A, B,C, etc. matrix, aaij ji The matrix A of order m x n consisting 143  of m rows and n columns of elements aij 427 situated at the intersection of the i-th  row and j-th column may be 3− 61 represented as follows : Is a 3 X 3 symmetric matrix. • If any diagonal matrix has the entire a11 a 12 - - - - a 1n diagonal element same, it is called a a a - - - - a A= [A] = [a ] = 21 22 2n scalar matrix. mn×× ijmn ------x00 am1 a m2 - - - - a mn  0x0 00x 8.2 Types of Matrix 33× • If any scalar matrix has all the diagonal 8.2.1 Square Matrix elements one (unit), it is called an A matrix which has equal number of rows identity or unit matrix and represented and columns is called a square matrix i.e by [I], Thus, m = n Identity matrix of 2 dimensions 10 =I2 =  and For example : 01 Identity matrix of 3 dimensions

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 100 • The value of a determinant changes sign  =I = 010 due to interchange of two rows or two 3  columns. 001  • The value of a determinant is zero if all 8.2.2 Row and Column Matrix the elements in any rows or column are zero. • A matrix which has only one row, i.e. m • The value of a determinant is zero if = 1, is called row matrix of order 1 x n. two rows or two columns are • A matrix which has only one column, i.e proportional or same. Therefore, in the .n = 1, is called column matrix of order case of linearly dependent rows or m x 1 columns, IAI =0. • The value of a determinant is multiplied 8.2.3 Null OR Zero Matrixes by a factor K if all the element of a row or column of the determinant is A matrix which has all the elements zero is multiplied by the same factor K. called null or zero matrix and is represented by [0]. 8.5 Basic Matrix Operations 000 Thus,  is null matrix of 2 X 3 000 Addition and Subtraction of Matrices order Addition and subtraction of matrices can be done only if they have the same order. 8.2.4 Singular Matrix Thus, ±= [][][]A× BC ×× A singular matrix is the one when the mn mn mn  ±=± determinant of it is zero. All rectangular Aij  b ij() ab ij ij matrices are singular. • They follows commutative law, i.e. A + B = B + A and associative law, i.e. 8.3 Definition and Properties of A + (B + C) = (A + B) + C Determinants

1. The determinant is a unique scalar Example 1 quantity associated with each square Find (i) A + B, (ii) A – B, and (iii) B – A, matrix. It can be expanded and has a When value. Generally, the determinant is 45 11− shown by a pair of vertical lines. A =  and B =  53 32 So, the determinant Solution 135 45 1− 1  A= A = 311 i) AB+= +  63×× 3 2  203 22 22 ()()41+− 51 54 12++31 22 15 32 + 15 = = = − +− + − ++  31() 11() 01() ()()63 32 9222× 23 23 31 22× 45 1− 1  =−[]3 × 7 + 1 ×−() 7 + 0 =− 28 ii) AB−= −   63 3 2 22××  22 8.4 Properties of Determinants ()()41−+ 51 36 • The value of a determinant is = =  ()()63−− 32 31 unchanged due to interchange of all the 22× 22× rows and columns. So |A | = |AT | 1− 1  45 iii) BA−=  − 3 2 22×× 63 22

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission ()()14− −− 15 −−36 C13 =×+×= 13 02 3 = =  36− +− 23 −−31 C21 = 21 ×+ 40 × = 2 ()() 22× 22× C22 = 22 ×+×= 41 8 • Here, it is to be noted that C=×+×= 2 3 4 2 14 (B – A) = - (A – B) 23

8.5.1 Scalar Multiplication of a Matrix 8.5.3 Properties of Matrix Multiplication If we consider KA = B Then,  =  • Multiplication is not commutative, i.e. Kaij  b ij AB # BA, except in identity matrix Where K is a scalar constant. • Multiplication is associative, i.e. A (BC) = (AB) C; where A, B and C are three 8.5.2 Multiplication of Matrix matrices of the order mXn, n X p, p X q respectively. The following procedures are adopted for • Multiplication is distributive, i.e. A (B + the multiplication of two matrices: C) = AB + AC, where A, B and C are [][][]Lead matrix x lag matrix= product (m x k) (k x n) (m x n) three matrices of the order mn× , np× , np× respectively. i.e., [][][]Amk×× BC kn mn ×  8.5.4 Transpose of Matrix and = Cij∑ a ik b kj k A new matrix obtained by interchanging (a) The element of a row of the lead matrix rows and columns of the original matrix is A should be multiplied by the called transposed matrix. corresponding elements of a column of Thus, if we have a matrix, Aa=  the lag matrix B. ij mn× (b) The summation of the product is put in Then, AaT =  the new product matrix C at the ji mn× corresponding row of A and column of B. Example 3 Find the transpose of the matrix Example 2 123  10  123  []A= 213 Find  ×   24  012  678 Solution Solution Here, we know Transpose of the matrix A 10  123  ×= 126   []C × 24  012  23 T  22×× 23 A=  217 Thus we get, 338 12 3 = []C ×  23 2()() 224×+ 2342 ×+× 8.5.5 Properties of Transposed Matrices 12 3 ∴ = []C 23×  • Transpose of a product of two matrices 2 8 14 is the product of transposes of Here, individual matrices in the reverse C11 =×+ 11 0 × 0 = 1 order.

C12 =×+×= 12 01 2 For example,

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission T TT If we define a quantity , where = ()AB= B . A ; fij fij T displacement in the direction T due to unit ()ABC= CTT B A T and so on. load in the direction ‘j’. Then • Transpose of algebraically sum of two LL33 matrices is the algebraically sum of the ff11= 12 = transpose of the individual matrices. 3EI 2EI 2 T TT LL For example, (A ±=+ B) A B ff= = 212EI 22 EI • Transpose of a transpose matrix is the = + original matrix. δ1 fP 11 fM 12 T T δ= fP + fM For example, ()A= A 2 21 22   δ1 ff 11 12 P =  8.6 Development of Flexibility Matrix δ2 ff 21 22 M

ff11 12 Development of Flexibility matrix can be  is called flexibility matrix illustrated with the following example. Let ff21 22 us take a cantilever beam as shown below [][][]δ= fR th In the flexibility matrix fij = element of i row and jth column.

• If unit load is acting in the direction of 1 Important Point It will produce deflection along 1L× 3 • Order of matrix is the no. of co-ordinate 1 = chosen for solution of problem. 3EI • Elements of flexibility matrix are 1L× 2 It will be proud rotation along 2 = displacements. 2EI • Flexibility Matrix will always be a Similarly, it will produce deflection square matrix. 1L× 2 • Elements along the diagonal will always along 1 = 2EI be positive. Other elements can be zero 1L× or negative. It will produce rotation along 2 = EI • Matrix will always be a symmetric Hence δ and δ produced along 1 and matrix about its main diagonal. 1 2 [This follows from Maxwell reciprocal 2 due to system of force P and M along 1 theorem, ff= ] and 2 respectively are ij ji Flexibility matrix can be calculated only for stable structure in which there are no rigid body displacements.

8.7 Development of Stiffness Matrix

LL33   δ1 =  PM +  3EI  2EI  LL2  δ2 = PM +  Let 1 and 2 is two Co-ordinate 2EI EI directions in which force are P and M.

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission • When unit displacement is given in the {[k] is stiffness matrix in which direction of ‘1’ without any element} Kij = element of ith row and displacement in the direction ‘2’. jth column. 12EI 6EI KK= = − 11ll32 12 6EI 4EI KK=−= 21ll2 22 Hence • Force developed in the direction PK=11 ∆+ K 12θ 12EI 1 = MK= ∆+ Kθ l2 21 22 ∆ • Force developed in the direction P KK11 12   =   6EI M KK θ = −  21 22  2 2 l [][][]PK= ∆ • When unit displacement is given in the direction of ‘2’ without any displacement in the direction 1 Important Point Hence forces developed along 1 and 2 due to displacements A and 0 along 1 • All the properties discussed for and 2 respectively are flexibility matrix holds for Stiffness matrix also except that elements of Stiffness matrix are forces. • Thus, to find out j th column of the Stiffness matrix, unit displacement is given in the direction of 7 without any displacement at other co-ordinates and forces developed at all the co-ordinates are determined. These forces constitute • Force developed in the direction the elements of jth column. 6EI 1 = − l2 • Force developed in the direction 4EI 2 = − l 4×× E 2I Stiffness of BA = = 2EI 4 12EI  6EI  P =  ∆+ − θ ll32   6EI  4EI  M = −  ∆+ θ ll2   If we define a quantity ‘Kij’, where Kij = Force developed in the direction of i due to unit displacement in the direction of j without any displacement in other direction.

Q.1 The stiffness K. of a beam deflecting in a symmetric mode, as shown in the figure, is d) force at j due to a unit a) EI /L b) 2El/ L deformation at i c) 4El/ L d) 6El/ L [GATE-2007]

Q.3 For the beam shown below, the stiffness coefficient K22 can be written as

[GATE – 2003]

Q.2 The stiffness coefficient kij indicates a) force at i due to a unit deformation at j b) deformation at j due to a unit 6EI 12E1 a) b) force at i 2 3 L L c) deformation at i due to a unit 3E1 EI force at j c) d) L 6L2 [GATE-2015]

ANSWER KEY:

1 2 3 (b) (a) (b)

If unit rotation at both ends, as shown 4EI 2EI Q.1 (b) M = − AB LL 2EI = (clockwise) L 4EI 2EI M = − BA LL 2EI = (anticlockwise) We know that moment required to L 2EI produce a unit rotation is called Hence, KM= = stiffness. L slope = 1 at both ends Q.2 (a) θ Q.3 (b)

Initially for θ=1 (clockwise) At A, Keeping ‘B’ fixed. 4EI M= (clockwise) AB L 2EI M= (clockwise) BA L The allow θ=1 (anticlockwise) At B, keeping ‘A’ as fixed. Now, 4EI M= (anticlockwise) BA L 2EI M= (anticlockwise) AB L