INSTITUTE OF AERONAUTICAL ENGINEERING (AUTONOMOUS) Dundigal – 500 043, Hyderabad
Regulation: R16 (AUTONOMOUS) Course code: ACE004
STRENGTH OF MATERIALS – II B.TECH IV SEMESTER
Prepared By
SURAJ BARAIK Assistant Professor
1 Unit I Deflection in Beams
2 Deflection in Beams
• Topics Covered Review of shear force and bending moment diagram
Bending stresses in beams
Shear stresses in beams
Deflection in beams
3 Beam Deflection
Recall: THE ENGINEERING BEAM THEORY M E y I R Moment-Curvature Equation
v (Deflection) y NA x A B
A’ B’
If deformation is small (i.e. slope is “flat”):
4 1 d R dx R y and (slope is “flat”) x B’ y 1 d2y A’ R dx 2
Alternatively: from Newton’s Curvature Equation
y d2y dy2 R if 1 1 dx 2 dx 3 y f (x) R 2 dy 2 2 1 1 d y dx x R dx 2
5 From the Engineering Beam Theory: M E 1 M d2y I R R EI dx2 d2y EI M dx 2 Flexural Bending Stiffness Moment Curvature
6 Relationship
A C B Deflection = y dy Slope = dx d2y A C B Bending moment = EI 2 y dx d3y Shearing force = EI dx3 d4y Rate of loading = EI dx4
7 Methods to find slope and deflection
Double integration method
Moment area method
Macaulay’s method
8 Double integration method d2y 1 M Curvature Since, dx 2 EI dy 1 M dx C Slope dx EI 1 1 y M dx dx C1 dx C 2 Deflection EI
Where C1 and C2 are found using the boundary conditions. Curvature Slope Deflection
y R dy dx
9 Double integration method
Simple supported W Slope Deflection L/2 L/2 dy A C B Slope = Deflection = y dx c 3 yc WL WL2 48EI A B 16EI L
Uniform distributed load Slope Deflection x w/Unit length A C dy B Slope = Deflection = y dx c 3 yc 5 WL WL2 384 EI A B 24EI L
10 Macaulay’s method
The procedure of finding slope and deflection for simply supported beam with an eccentric load is very laborious.
Macaulay’s method helps to simplify the calculations to find the deflection of beams subjected to point loads.
11 Moment-Area Theorems
• Consider a beam subjected to arbitrary loading, d d2y M dx dx2 EI D xD M d dx EI C xC xD M dx D C EI xC dx CD Rd dx R d
• First Moment-Area Theorem: area under BM diagram between C and D.
9 - 11
12 M o me n t - Area Theorems
• Tangents to the elastic curve at P and P’ intercept a segment of length dt on the vertical through C. M dt xd x dx EI x x D M 1 D 1 t x dx xMdx Ax C D EI EI EI xC xC
A= total area of BM diagram between C & D x = Distance of CG of BM diagram from C
• Second Moment-Area Theorem: The tangential deviation of C with respect to D is equal to the first moment with respect to a vertical axis through C of the area under the BM diagram between C and D. 9 - 12
13 Moment Area Method
14 An Exercise- Moment of Inertia – Comparison
1 Load
Maximum distance of 4 inch to the centroid I2 2 x 8 beam Load 2
I1
Maximum distance of 1 inch to 2 x 8 beam the centroid
I2 > I1 , orientation 2 deflectsless
15 UNIT II
DEFLECTION BY ENERGY METHODS
16 Elastic Deflection Castigliano’sMethod
Complementary Energy U’
Incremental: Deflection: Castiglino’s Theorem: ∆ U Q When a body is elastically deflected by any combination of loads, the deflection at any point and in any direction is equal to the partial derivative of strain energy (computed with all loads acting) with respect to a load located at that point and acting in that direction
17 Elastic Deflection Castigliano’s M e t h o d Energy and Deflection Equations
Example: Axial Tension Stored Elastic Energy:
Case 1 from Table 5.1:
gives: For varying E
and A:
18 Elastic Deflection Castigliano’s Method
Energy and Deflection Equations
(1) Obtain expression for all components of energy Table 5.3
(2) Take partial derivative to obtain deflection
Castiglino’s Theorem: ∆ U Q
19 Elastic Deflection: Castigliano’s Method
first compute Energy, then Partial Derivative to get deflection
Here 2 types of loading: Bending and Shear magnitude @ x: Table 5.3
1. Energy: here it has two components:
(23=8)*3*4 = 96 2. Partial Derivatives for deflection:
20 Elastic Deflection: Castigliano’s Method
Table 5.3 TWO METHODS
Differentiate after Integral Differentiate under Integral
21 Elastic Deflection: Castigliano’s Method
m
m
Transverse shear contributes only <5% to deflection
22 Elastic Deflection: Castigliano’s Method
•90° bend cantilever beam •shear neglected
Use of “Dummy Load” Q=0 •Shear neglected => only 4 energy components:
1) BENDING portion a_b: Mab=Py (Tension and Compression mostly 2) BENDING portion b_c: Mbc=Qx +Ph negligible if torsion and bending 3) TENSION portion a_b: Q are present) 4) COMPRESSION portion b_c: P
:
23 Elastic Deflection: Castigliano’s Method
2 •Eccentrically Load Column 500kg x 9.8m/s •No Buckling =4900 N
Redundant Support Guy wire
•Now Deflection known (=0) •Find necessary Tension Force F •Hence partial derivative of total elastic energy with respect to F must be zero •Omit zero derivatives - all energy terms above a - compression term below a Nm •Only bending term is left: M= (4900(Nm N)(1.2m))2 -FyNm = 5880 - Fy
(Nm)2 m Nm3 m3
Nm3 m3
finite value 24 Energy Method External Work When a force F undergoes a displacement dx in the same direction as the force, the work done is Ue P/2 dUe F dx Ifthe total displacement is x the work become x Ue F dx 0
25 The work of a moment is defined by the product of the magnitude of the moment M and the angle d then if the total angle of rotation is the work become:
dUe M d
e
U M/2 Ue M d 0
26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 Maxwell–Betti Reciprocal theorem Consider a simply supported beam of span L as shown. Let this beam be loaded by two systems of forces P1and P2 separately as shown in the figure. Let u21 be the deflection below the load point P2 when only load P1is acting.
Similarly let u12 be the deflection below load P1 , when only load P2 is acting on the beam.
43 44 The reciprocal theorem states that the work done by forces acting through displacement of the second system is the same as the work done by the second system of forces acting through the displacements of the first system. Hence, according to reciprocal theorem,
P1 u12 P2 u21
Now, u12 and u21 can be calculated using Castiglinao’s first theorem. Substituting the values of u12 and u21 in equation we get, 5P L3 5PL3 P 2 P 1 48EI 48EI
45 Hence it is proved. This is also valid even when the first system of forces is
P1, P2 ,...., Pn and the second system of forces is given by Q1, Q2 ,...., Qn . Let u1 , u2 ,...., un be the displacements caused by the forces P1, P2 ,...., Pn only and
1, 2 ,...., n be the displacements due to system of forces Q1, Q2 ,...., Qn only acting on the beam as shown in Fig
46 UNIT III
STRESSES IN CYLINDERS AND SPHERICAL SHELLS
47 THIN CYLINDER
48 Introduction
Cylindrical and spherical vessels are used in the engineering field to store and transport fluids. Such vessels are tanks ,boilers , compressed air receivers , pipe lines etc.
these vessels, when empty, are subjected to atmospheric pressure internally as well as externally and the resultant pressure on the walls of the shell is nil.
but whenever a vessel is subjected to an intenal pressure (due to air , water , steam etc.) its walls are subjected to tensile stresses.
49 Thin cylindrical shell.
When t/d <= d/10 to d/15, it is called thin cylindrical shell. t = thickness of the shell d =internal diameter of shell.
in thin cylindrical shells hoops stress and longitudinal stresses are constant over the Thickness and radial stresses are negligible.
When t/d > d/10 to d/15, it is called THICK cylindrical shell.
50 Stresses in thin cylindrical shells :
whenever, a thin cylindrical shell is subjected to an internal pressure (p). Its Walls are subjected to two types of tensile stresses.
(a) Hoop stress (circumferential stress) (b) Longitudinal stress.
51 Hoop stress (circumferential stress) : 흈풄 Consider a thin cylindrical shell subjected to an internal Pressure as shown in fig. 휎푐 = circumferential stress in the shell material. p =internal pressure d =internal diameter of shell t =thickness of the shell.
Total pressure, p = Pressure * Area =p.d.l
Resisting area = A = 2.t.l 휎푐 = P/A = p.d.l/2.t.l
흈풄 = p.d/2t. 52 (b) Longitudinal stress (휎푙 ):
Total pressure p = Pressure * pArea = p휋 4 푑2 Resisting area, A = 휋푑푡 휋푑 2 휎푙 = P/A = 4 휋푑 p 푡
흈풍 = pd/4t 53 Change in dimensions of a thin cylindrical shell due to internal pressure: Let, 훿푑=change in dia. Of shell 휎푙=change in length of shell
Circumferential strain, 훿푑 휎 푝푑 푝 휀 = 푐 − 1 푑 퐸 푙 푚퐸 − 2푡퐸푑 = 휎 = 4푡푚퐸 ퟏ 휺ퟏ = 풑풅/ퟐ풕푬(ퟏ − ) ퟐ풎
Longitudinal strain,
훿푙 휎 푝푑 푝 휀 = 푙 − 푐 = 2 퐸 푚퐸 4 2푑푡푚 휎 − = 푡퐸 퐸 푙 풑풅 ퟏ ퟏ 휺 = ퟐ ퟐ풕푬 ퟐ − 54 풎 Change in volume of a thin cylindrical shell due to internal pressure:
훿푣 푉 + 훿푣 = 휋 = [ 푑 + 훿푑2 ∗ (−푙 +푉훿푙)] − ∗ 휋 4 2 Volume of 4 푑 ∗ 푙 shell, 휋 = [ 휋 (푑2 + 2훿푑 ∗ 푑 + 훿푑 ) ∗ (푙 + 훿푙)]2 4 2 − 4 ∗ 푑 휋 2 ∗ 푙 휋 V = ∗ 푑 = 푑2푙 + 2훿푑 ∗ 푑 ∗ 푙 + 훿푑2 ∗ 푙 + 푑2 ∗ 훿푙 + 2훿푑 ∗ 푑 ∗ 훿푙 − 푑2 휋 4 4 ∗푙 + 훿푑2 ∗훿푙 ∗ 푙 = 푑2훿푙 + 2훿푑 ∗ 푑 휋 4 4 ∗ 푙 Final volume, 훿푣= 2 휋 휋 푑 훿푙 + 2훿푑 ∗ 푑 ∗ 푙 / ∗ 휋 2 푉 2 4 V+훿푣 = 푑 + 푙 + 훿푙 푑 2∗ 푙 4 = + 훿푑 ∗ 훿푙 4 훿푙푑 푑 =휀2 + 2휀1 훿푑 훿 (휀1= , 휀 = 푙 푑 Change in 휹풗 =V(흐ퟐ + ퟐ )2 volume, 휺ퟏ) 푙 55 Thin spherical shells Consider a thin spherical shell subjected to internal pressure p as shown in fig.
p =internal pressure d =internal diameter of shell t =thickness of the shell 휎 = stress in the shell material P = 휋 ∗ 푑2 Total force 4 ∗푝 Resisting section =휋푑푡 Stress in the shell 휎 = 푇표푡푎푙 force/resisting section = 휋 푑2푝/휋 4 푑푡 푝 푑 56 휎 = 4 Chenge in diameter and volume of a thin spherical shell due to internal pressure
Consider a thin spherical shell subjected to internal pressure.
p =internal pressure d =internal diameter of shell t =thickness of the shell 휎 = stress in the shellmaterial
We know that for thin spherical shell 휎 = 푝푑 4푡 Strain in any 휎 푝푑 direction휎 푝 휖= 푑 퐸 − 4푡퐸 4푡 − 푚퐸 ퟏ = 푚퐸 휺 = 풑풅/ퟒ풕푬(ퟏ − 풎 We know that, ) strain, 훿 ∈ 푑푑 훿푑 ퟏ = 풑풅/ퟒ풕푬(ퟏ ……( =푑 풎 − 1) ) 57 CLASSIFICATIONS OF SHELLS
58 SHELLS
• Shell is a type of building enclosures. • Shells belong to the family of arches . They can be defined as curved or angled structures capable of transmitting loads in more than two directions to supports. • A shell with one curved surface is known as a vault (single curvature ). • A shell with doubly curved surface is known as a dome (double curvature).
59 Classification of shells
• There are many different ways to classify shell structures but two ways are common:
1. The material which the shell is made of: like reinforced concrete, plywood or steel, because each one has different properties that can determine the shape of the building and therefore, these characteristics have to be considered in the design.
2. The shell thickness: shells can be thick or thin.
60 Thin Concrete Shells
The thin concrete shell structures are a lightweight construction composed of a relatively thin shell made of reinforced concrete, usually without the use of internal supports giving an open unobstructed interior. The shells are most commonly domes and flat plates, but may also take the form of ellipsoids or cylindrical sections, or some combination thereof. Most concrete shell structures are commercial and sports buildings or storage facilities.
There are two important factors in the development of the thin concrete shell structures: • The first factor is the shape which was was developed along the history of these constructions. Some shapes were resistant and can be erected easily. However, the designer’s incessant desire for more ambitious structures did not stop and new shapes were designed. • The second factor to be considered in the thin concrete shell structures is the thickness, which is usually less than 10 centimeters. For example, the thickness of the Hayden planetarium was 7.6 centimeters.
61 Types of Thin Concrete Shells
1. Barrels shells The cylindrical thin shells, also called barrels, should not be confused with the vaults even with the huge similarity in the shape of both structures, because each of these structures has a different structural behavior as well as different requirements in the minimum thickness and the shape.
62 • On one hand, the structural behavior of the vault is based on connected parallel arches, which transmit the same effort to the supports . Therefore, the materials used in these structures have to be able to resists compressions (e.g. stone) and the thickness is usually higher. Furthermore, the shape of the vaults must be as similar as possible to the arch in order to achieve the optimum structural behavior.
• On the other hand, the structural behavior of the barrels shell is that it carries load longitudinally as a beam and transversally as an arch. and therefore, the materials have to resist both compression and tension stresses. This factor takes advantage of the bars of the reinforced concrete, because these elements can be placed where tension forces are needed and therefore, the span to thickness Ratios can be increased. Furthermore, the shape has fewer requirements than the vaults and therefore, new curves like the ellipse or the parabola can be used improving the aesthetic quality of the structure.
63 64 Types of Thin Concrete Shells
2. Folded plate A thin-walled building structure of the shell type.
Advantages of Folded Plate Roofs over Shell Roofs are: (a) Movable form work can be employed. (b) Form work required is relatively simpler. (c) Design involves simpler calculations.
Disadvantages of Folded Plate Roofs over Shell Roofs are: (a) Folded plate consumes more material than shells. (b) Form work may be removed after 7 days whereas in case of shells it can be little earlier.
65 Folded plate types
66 Folded Plates system
67 68 69 Folded-Plate Hut in Osaka
70 Folded Plates Library
71 72 73 74 Types of Thin Concrete Shells
3. Hyperbolic Paraboloid (Hypar)
A Hypar is a surface curved in two directions that can be designed as a shell or warped lattice.
A hypar is triangular, rectangular or rhomboidal in plan, with corners raised to the elevation desired for use and/or appearance. The edges of Hypars are typically restrained by stiff hollow beams that collect & transfer roof loads to the
foundations. Rhomboi d
75 76 77 4. Various Double Curvature
Types of shells
78 79 Types of Thin Concrete Shells
5. Dome A rounded roof, with a circular base, shaped like an arch in all directions.. First used in much of the Middle East and North Africa whence it spread to other parts of the Islamic world, because of its distinctive form the dome has, like the minaret, become a symbol of Islamic architecture.
Dome has double curvature and the resulting structure is much stiffer and stronger than a single curved surface, such as a barrel shell.
80 81 82 83 84 Types of Thin Concrete Shells
6. Translation Shells A translation shell is a dome set on four arches. The shape is different from a spherical dome and is generated by a vertical circle moving on another circle. All vertical slices have the same radius. It is easier to form than a spherical dome.
85 • Advantages of Concrete Shells: The curved shapes often used for concrete shells are naturally strong structures. Shell allowing wide areas to be spanned without the use of internal supports, giving an open, unobstructed interior. The use of concrete as a building material reduces both materials cost and the construction cost. As concrete is relatively inexpensive and easily cast into compound curves.
• Disadvantages of Concrete Shells Since concrete is porous material, concrete domes often have issues with sealing. If not treated, rainwater can seep through the roof and leak into the interior of the building. On the other hand, the seamless construction of concrete domes prevents air from escaping, and can lead to buildup of condensation on the inside of the shell. Shingling or sealants are common solutions to the problem of exterior moisture, and ventilation can address condensation.
86 UNIT IV
INDETERMINATE BEAMS: PROPPED CANTILEVER AND FIXED BEAMS
87 Contents: • Concept of Analysis -Propped cantilever and fixed beams-fixed end moments and reactions
• Theorem of three moments – analysis of continuous beams – shear force and bending moment diagrams.
88 Indeterminate beams • St a t i c a lly determinatebe a m s: – Cantilever beams – Simple supported beams – Overhanging beams
• Statically indeterminate beams: – Propped cantilever beams – Fixed beams – Continuous beams
89 Indeterminate beams
• Propped cantilever Beams:
Degree of static indeterminacy= N0. of unknown reactions – static equations=3-2 = 1
90 Indeterminate beams • Fixed beam: A fixed beam is a beam whose end supports are such that the end slopes remain zero (or unaltered) and is also called a built-in or encaster beam.
Degree of static indeterminacy= N0. of unknown reactions – static equations=4-2 =2
91 Continuous beam: Continuous beams are very common in the structural design. For the analysis, theorem of three moments is useful. A beam with more than 2 supports provided is known as continuous beam.
Degree of static indeterminacy= N0. of unknown reactions – static equations=3-2 =1
Degree of static indeterminacy= N0. of unknown reactions – static equations=5-2 =3
92 •B.M. diagram for a fixed 푊 푊 beam : Figure shows a fixed 1 2 beam AB carrying an external load system. Let V and V be A B 푊 푊 the vertical reactions at the 1 2 supports A and B. 푀 푀 Let MA and MB be the fixed end 퐴 퐵 Moments. 푉퐵 푉퐴
93 Fixed Beams The beam may be analyzed in the following stages. (i) Let us first consider the beam as Simply supported.
Let va and vb be the vertical reactions at the supports A and B. Figure (ib) shows the bending moment diagram for this condition. At any section the bending moment Mx is a sagging moment.
푊1 푊2
v a vb (ia) Freely supportedcondition
푀 푥
(ib) FreeB.M.D.
94 • (ii) Now let us consider the effect of end couples MA and MB alone.
Let v be the reaction at each 푀 end due to this condition. 푀 v 퐴 퐵 v Suppose 푀퐵 > 푀퐴. (iia) Effect of endcouples 푀 퐵 −푀 퐴 푀 Then 푉 = . ′ 퐿 푥 If 푀퐵 > 푀퐴 the reaction V is upwards at B and downwards at A. (iib) FixedB.M.D. Fig Shows the bending moment diagram for this condition.
At any section the bending moment Mx’ is hogging moment.
95 • Now the final bending moment 푊1 푊2 diagram can be drawn by combining the above two B.M. 푀 푀 diagrams as shown in Fig. (iiib) 퐵 퐴 푉 푉퐴 + 퐵 Now the final reaction VA =va- v - - and VB = vb + v The actual bending moment at any (iiib) ResultantB.M.D. section X, distance 푥 from the end A is given by, 푑2푦 퐸퐼 = 푀 − 푀 ′ 푑푥2 푥 푥
96 Fixed Beams 푀 푊 푊 푥 1 2
푉 푉 푎 푏 (ib) FreeB.M.D. (ia) Freely supportedcondition 푀 푥′ 푀 푀 퐴 퐵 (iib) FixedB.M.D. 푉 (iia) Effect of end couples 푉 푊1 푊2 + - - 푀 푀 퐴 (iiia) Fixedbeam 퐵 푉 (iiib) ResultantB.M.D. 푉퐴 퐵 97 푑2푦 퐸퐼 = 푀 − 푀 ′ 푑푥2 푥 푥 • Integrating, we get, 푑푦 푙 푙 푙 • 퐸퐼 = 푀 푑푥− 푀′ 푑 푑푥 0 0 푥 0 푥 푥 • But at x=0, 푑푦 =0 푑푥 푑푦 and at 푥 = 푙, 푑푥 = 0 푙 Further0 푀푥푑푥 = area of the Free BMD=푎 푙 ′ ′ 푀푥 푑푥 = area of the fixed B. M. D = 푎 0 Substituting in the above equation, we get, 0 = 푎 − 푎′ ∴ 푎 = 푎′
98 Fixed Beams
푎 = 푎′ ∴Area of the free B.M.D. =Area of the fixedB.M.D. Again consider the relation, 푑2푦 퐸퐼 = 푀 −푀 ′ 푑푥2 푥 푥 푀푢푙푡푦푖푛푔푏푦 푥 푤푒 푔푒푡, 푑2푦 퐸퐼푥 = 푀 푥 − 푀 ′ 푥 푑푥2 푥 푥 • Integrating we get, 푙 푑2푦 푙 푙 • 퐸퐼푥 = 푀 푥푑푥 − 푀′ 푥 푑 0 푑푥2 0 푥 0 푥 푥 • ∴ 퐸퐼 푥 푑푦 −푦 푙=a푥-a’푥′ 푑푥 0 • Where 푥= distance of the centroid of the free B.M.D. from A. and 푥′= distance of the centroid of the fixed B.M.D. fromA.
99 Fixed Beams • Further at x=0, y=0 and 푑푦 = 0 푑푥 • and at x=l, y=0 and 푑푦 = 0. 푑푥 • Substituting in the above relation, we have 0 = 푎푥-푎′푥′ 푎푥=푎′푥′ or 푥 = 푥 ∴ The distance of the centroid of the free B.M.D . From A= The distance of the centroid of the fixed B.M.D. from A.
∴ 푎 = 푎′ 푥 = 푥
100 Fixed Beams
푊 푊 푀푥 1 2
푉 푉 푎 푏 (ib) Free B.M.D. (ia) Freely supported condition
푀푥′
푀퐴 푀퐵 (iib) Fixed B.M.D. 푉 (iia) Effect of end couples 푉 푊1 푊2 + - - 푀퐴 푀퐵 (iiib) Resultant B.M.D. 푉퐴 (iiia) Fixed beam 푉 퐵 101 Fixed beam problems
• Find the fixed end moments of a fixed beam subjected to a point load at the center.
W
A B l/2 l/2
102 Fixed beam problems • 퐴′ =퐴 W 1 푊푙 A B 푀 ×푙= ×푙× 푙/2 푙/2 2 4 푊 푙 4 푊 푙 푀 = = 푀 = 푀 + 8 퐴 퐵
FreeBMD
- M M FixedBMD 푊 푙 4 + - - ResultantBMD
103 Fixed beam problems
• Find the fixed end moments of a fixed beam subjected to a eccentric pointloadW.
W
A B a b 푙
104 Fixed beam problems
W • 퐴′ =퐴 푎 푏 A B 푀 +푀 1 푊푎푏 푙 퐴 푊푎푏 퐵 ×푙= ×푙× 2 2 푙 푙 푊푎푏 푀 +푀 = −−−−(1) + 퐴 푙 • 퐵푥′ = 푥 푀 Free-BMD 퐴 푀 - 퐵 푀 +2푀 푙 푙+푎 퐴 FixedBMD 퐵 × = 푀퐴 +푀 3 3 푎 퐵 푀 = 푀 × 퐵 퐴 푙−푎 푎 푀 퐵 = −− −(2) 푏 10519 Fixed beam problems
W 푊푎푏 푎 푏 푀퐴 +푀 = −−−−(1) A 푙 B 푙 퐵 푎 푀 퐵 = 푀퐴 × −− −(2) 푊 푎푏 푊 푎 푏 2 2 푙2 푏푙 푊 푏 푎 푙2 By substituting (2) in (1), - + -
푊 푎 푏 ResultantBMD 푀퐴 = 2 푙2 From(2), 푊 푏푎 푀 퐵 = 2 푙2
106 Clapeyron’s theorem of three moments A B푀 C 푀 퐵 푀 푙 푙 퐴 1 2 퐶
• As shown in above Figure, AB and BC are any two successive spans of a continuous beam subjected to an external loading.
• If the extreme ends A and C fixed supports, the support moments 푀퐴, 푀퐵and 푀 퐶 at the supports A, B and C aregiven by therelation,
6푎1푥1 6푎2푥2 푀 퐴 1푙 +2푀 퐵 1푙 + 2푙 +푀 퐶 2푙 = + 1푙 2푙
107 Clapeyron’s theorem of three moments (contd…)
6푎1푥1 6푎2푥2 푀 퐴 1푙 +2푀 퐵 1푙 +2푙 +푀 퐶 2푙 = + 1푙 2푙 • Where,
• 푎1 =area of the free B.M. diagram for the spanAB. • 푎2 =area of the free B.M. diagram for the spanBC.
• 푥1= Centroidal distance of free B.M.D on AB fromA.
• 푥2= Centroidal distance of free B.M.D on BC fromC.
108 Clapeyron’s theorem of three moments (contd…)
A B C (a) 1푙 2푙
푑푥 푀 푥 + + (b) 푥 푥2 푥1 FreeB.M.D
푑푥 푀 푀 퐵 푥 ’ 푀 푥 푀 + (c) 퐴 + 퐶 ′ ′ 푥1 푥2 FixedB.M.D
푀 퐵 푀 +ve -ve +ve 푀퐶 퐴 (d) 109 Clapeyron’s theorem of three moments (contd…)
A B C (a)The givenbeam
푑푥 푀 푥 + + (b) FreeB.M.D. 푥 푥2 푥1 FreeB.M.D
푑푥 푀 푀 퐵 푥 ’ 푀 푥 푀 - (c) Fixed B.M.D. 퐴 - 퐶 ′ ′ 푥1 푥2 FixedB.M.D
110 Clapeyron’s theorem of three moments (contd…) • Consider the span AB: • Let at any section in AB distant 푥 from A the free and fixed ′ bending moments be 푀푥 and 푀푥 respectively.
• Hence the net bending moment at the section is given by 푑2푦 퐸퐼 = 푀 − 푀 ′ 푑푥2 푥 푥
• Multiplying by 푥, we get 푑2푦 퐸퐼푥 = 푀 푥 − 푀 ′ 푥 푑푥 2 푥 푥
111 Clapeyron’s theorem of three moments (contd…)
푑2푦 • 퐸퐼푥 = 푀 푥−푀 ′ 푥 푑푥2 푥 푥
• Integrating from 푥 = 0 푡표푥 = 푙1, we get, 1푙 1푙 1푙 푑2푦 퐸퐼 푥 = 푀 푥 푑푥 − 푀 ′푥 푑푥 푑푥2 푥 푥 0 0 0
푙1 푙1 푑푦 1푙 퐸퐼 푥. −푦 = 푀 푥 푑푥 − 푀 ′푥 푑푥 −− −(1) 푑푥 푥 푥 0 0 0
112 Clapeyron’s theorem of three moments (contd…) A B C
푥 1푙 2푙
• But it may be suchthat At 푥 = 0, deflection 푦 =0 • At 푥 = 푙, 푦 = 0; 푎푛푑 푠푙표푝푒푎푡퐵 푓표푟퐴퐵, 푑푦 =휃 1 퐵푑퐴푥
푙1 • 0 푀푥 푥 푑푥 =푎 1푥 1 =Momentof thefreeB.M. D.onAB aboutA.
푙1 ′ ′ • 0 푀푥 푥 푑푥 =푎 1 푥1 = Momentof thefixedB.M. D.onAB aboutA. ′
113 Clapeyron’s theorem of three moments (contd…)
푙1 푙1 푑푦 1푙 퐸퐼 푥. −푦 = 푀 푥 푑푥 − 푀 ′푥 푑푥 −−(1) 푑푥 푥 푥 0 0 0 • Therefore the equation (1) is simplifiedas, ′ ′ 퐸퐼 푙1휃퐵퐴 −0 =푎1푥1 −푎1푥1 .
푀 + But 푎′ =area of the fixed B.M.D. on AB = 퐴 푙 1 푀 퐵2 1
′ 푀 퐴 +2푀 퐵 1푙 푥1 =Centroidof thefixedB.M. D.fromA= 푀 퐴 +푀 퐵 3
114 Clapeyron’s theorem of three moments (contd…) • Therefore, (푴 +푴 ) 푴 + ퟐ 푴 풍 풍ퟐ ′ ′ 푨 푩 푨 ퟏ 풂ퟏ풙ퟏ = 푩 풍ퟏ× = (푴 푨 + ퟐ푴 푩 ) ퟐ 푴 푨 +푴 푩 ퟏ ퟔ ퟑ ퟐ ퟏ풍 푬푰풍ퟏ휽푩푨 = 풂ퟏ풙ퟏ − (푴 푨 + ퟐ푴 푩 ) ퟔ
ퟔ풂ퟏ풙ퟏ ퟔ푬푰 휽푩푨 = −(푴 푨 +ퟐ푴푩)풍ퟏ − − −−(ퟐ) ퟏ풍 Similarly by considering the span BC and taking C as origin it can be shownthat,
ퟔ풂ퟐ 풙ퟐ ퟔ푬푰휽푩푪 = − (푴 푪 + ퟐ푴푩)풍ퟐ − − −−(ퟑ) 풍ퟐ
휃퐵퐶 = slope for span CB atB
115 Clapeyron’s theorem of three moments (contd…)
• But휃퐵퐴 = −휃퐵퐶 as the direction of 푥 from A for the span AB, and from C for the span CB are in oppositedirection.
• And hence,휃퐵퐴+ 휃퐵퐶 =0
ퟔ풂ퟏ풙ퟏ ퟔ푬푰 휽푩푨 = −(푴 푨 +ퟐ푴푩)풍ퟏ − − −−(ퟐ) 풍ퟏ
ퟔ풂ퟐ풙 ퟐ ퟔ푬푰휽푩푪 = − (푴 푪 + ퟐ푴푩)풍ퟐ − − −−(ퟑ) ퟐ풍
• Adding equations (2) and (3), we get
ퟔ풂ퟏ풙ퟏ ퟔ풂ퟐ풙ퟐ 푬푰휽푩푨 +ퟔ푬푰 휽푩푪 = + −( 푴 푨 +ퟐ푴푩)풍ퟏ −(푴 푪 + ퟐ푴푩)풍ퟐ 풍ퟏ 풍ퟐ
ퟔ풂ퟏ풙ퟏ ퟔ풂ퟐ풙ퟐ ퟔ푬푰(휽푩푨 + 휽푩푪) = + − 푴 푨 풍ퟏ +ퟐ 푴푩 (풍 ퟏ + 풍) ퟐ+ 푴 풍 푪 ퟏ풍 ퟐ풍 ퟐ
116 Clapeyron’s theorem of three moments (contd…)
ퟔ풂ퟏ풙ퟏ ퟔ풂ퟐ풙ퟐ ퟎ = + −푴푨풍ퟏ +ퟐ푴푩(풍ퟏ + 풍ퟐ) +푴푪풍ퟐ 풍ퟏ 풍ퟐ
ퟔ풂ퟏ풙ퟏ ퟔ풂ퟐ풙ퟐ 푴푨풍ퟏ +ퟐ푴푩(풍ퟏ + 풍ퟐ) + 푴푪풍ퟐ = + 풍ퟏ 풍ퟐ
117 P r o b l e ms
• Acontinuous beam of three equal span is simply supported over two supports. It is loaded with a uniformly distributed load of w/unit length, over the two adjacent spans only. Using the theorem of three moments, find the support moments and sketch the bendingmoment diagram.Assume EI constant.
118 P r o b l e ms w/ unitlength • Solution: A B C D 푙 푙 푙
푤 푙2 푤 푙2 8 8
• The theorem of three FreeB.M.D. moments equation for two spansis, 6푎1푥1 6푎2푥2 푀 퐴 (1푙) +2푀 퐵 1푙 +2푙 +푀 퐶 푙2 = + 푙1 푙2 Apply the theorem of three moment equation for spans AB and BC is,
6푎1푥1 6푎2푥2 푀 퐴 푙 +2푀 퐵 푙 = + 1푙 2푙 +푙 +푀 퐶 푙
119 P r o b l e ms w/ unitlength • Solution: A B C D
2 • 푎 = 2 ×푙× 푤푙 푙 푙 푙 1 3 8 푤 푙2 푤 푙2 1 3 = 푤 푙 8 12 8 • 푥 = 푙 1 2 FreeB.M.D. 2 푤 푙2 • 푎 = ×푙× 2 3 8 • 푥 = 푙 2 2 0 6×1 푤푙3×푙 6×1 푤푙3×푙 12 2 12 • 푀 퐴 푙 +2푀 퐵 푙 = + 2 푙 푙 + + 푙 푀 퐶 푙2 • 4푀 +푀 = 푤푙 −−−− −(1) 퐵 퐶 2
120 w/ unitlength • Solution: A B C D 푙 푙 푙
푤 푙2 푤 푙2 8 8
• The theorem of three FreeB.M.D. moments equation for two spansis, 6푎1푥1 6푎2푥2 푀 퐴 (1푙) +2푀 퐵 1푙 +2푙 +푀 퐶 푙2 = + 푙1 푙2 Apply the theorem of three moment equation for spans BC and CD is,
6푎1푥1 푀 퐵 푙 +2푀 퐶 푙 = +0 1푙 +푙 +푀 퐷 푙
121 P r o b l e ms w/ unitlength • Solution: A B C D
2 • 푎 = 2 ×푙× 푤푙 푙 푙 푙 1 3 8 푤 푙2 푤 푙2 1 3 = 푤 푙 8 12 8 • 푥 = 푙 1 2 FreeB.M.D.
0 6×1 푤푙3×푙 12 2 • 푀 퐵 푙 +2푀 퐶 푙 = +0 푙 +푙 +푀 퐷 푙2 • 푀 +4푀 = 푤푙 −−−−−(2) 퐵 퐶 4
122 P r o b l e ms 2 • 4푀퐵 + 16푀퐶 =w푙 −−−− −(2) ×4 푤 푙2 4푀 + 푀 = −−−− −(1) 퐵 퐶 2 푤 푙2 15푀 = −−−− − 2 ×4 −(1) 퐶 2 푤 푙2 푀 = 퐶 30
푤 푙2 Substitute푀 퐶 = 30 in equation(2), 푤 푙2 푤 푙2 푀 +4× = 퐵 30 4 7푤 푙2 푀 = 퐵 60
123 Fixed beam Problems
• A fixed beam AB of span 6 m carries uniformly varying load of intensity zero at A and 20 kN/m at B. Find the fixed end moments and draw the B.M. and S.F. diagrams for the beam.
124 P r o b l e ms 20×푥 20 kN/m 푑푥 6 X A B 푥 X 6m Consider any section XX distant 푥 from the end A, the intensityof 푤 푥 20푥 loading at XX = = 퐿 6 20푥 Hence the load acting for an elemental distance 푑푥 = 푑푥 6 Due to this elemental load the fixed moments are as follows: Wa푏2 푑푀 = (Formula is derived from firstprinciples) 푎 퐿2 20푥 푑푥 ×푥 × 6 − 푥 2 20푥2 6 − 푥 2푑푥 = 6 = 62 63
125 P r o b l e ms 20×푥 20 kN/m 푑푥 6 X A B 푥 X 6m and W푏푎2 푑푀 = Formula is derived from basicprinciples 푏 퐿2 20푥 2 푑푥× 6−푥 ×푥 20푥3 6−푥 푑푥 = 6 = 62 63 Taking fixing moment at A, 푙 6 20 푀 = 푑푀 = 푥2 6 − 푥 2푑푥 퐴 푎 216 0 0
126 Problems 6 20 푀 = 푥2 36 + 푥2 − 12푥 푑푥 퐴 216 0 6 20 36푥3 푥5 12푥4 = + − 216 3 5 4 0 20 36 ×63 65 12 ×64 = + − 216 3 5 4
∴ 푀퐴 = 24 kNm
127 P r o b l e ms
푙 6 20 푀 = 푑푀 = 푥3 6 − 푥 푑푥 퐵 퐵 63 0 0 6 20 푥4 푥5 = ×6 − 216 4 5 0 20 64 ×6 65 = − 216 4 5
∴ 푀 퐵 = 36 kNm
128 P r o b l e ms 20×푥 20 kN/m 푑푥 6 FreeBMD: X 푤 푙2 20 ×62 A 푀 푚 푎 푥 = = 푥 9 3 9 3 X B = 46.18 kNm (Cubic paraboliccurve) 6m 46.18kNm Will occurat 6 3 m from left endA. +
6 3 FreeBMD
FixedBMD
129 P r o b l e ms
ResultantBMD: 20×푥 20 kN/m 푑푥 6 X A 푥 X B 6m
46.18kNm 36 kNm + 24 kNm - -
6 3 Resultant BMD
130 P r o b l e ms Calculation of supportreactions: 20×푥 20 kN/m 푑푥 6 X A 푥 X B 푀 퐴 =0 24 6m 36 푅 푅 1 퐴 2 퐵 푅 ×6 + 24 = 36 + ×6 ×20 × ×6 퐵 2 3 252 푅 = = 42 kN 퐵 6
1 푅 + 42 = ×6 × 20 퐴 2 푅퐴 = 60 − 42 = 18kN
131 P r o b l e ms SFD: 20×푥 20 kN/m 푑푥 6 S.F. @ A =+18kN X A 푥 X B S.F. @ B =-42 kN 24 6m 36 SFD between A and B is 18 kN 42kN aparabola. 18 kN Parabola S.F. @ XX =0 1 푥 18 − ×푥 ×20 × = 0 3.29m 2 6 10푥2 SF 18 = 6 D 42 kN 푥 = 3.29 푚
132 P r o b l e ms ResultantBMD 20×푥 20 kN/m 푑푥 6 & SFD: X A 푥 X 6m B
46.18kNm 36 kNm + 24 kNm - - Resultant BMD
6 3 =3.46m 18 kN
3.29m SF D 42 kN 133 UNIT V
INDETERMINATE BEAMS: CONTINUOUS BEAMS
134 Continuous beam with supports at different levels A B푀 C 푀 퐵 푀 푙 훿1 푙 퐴 1 2 퐶 • Consider the continuous beam shown in above Figure. Let the support B be 훿1 below A and below C. • Consider the spanAB: • Let at any section in AB distant 푥 from A the free and fixed bending ′ moments be 푀푥 and 푀 푥 respectively. • Hence the net bending moment at the section is given by 푑2푦 퐸퐼 = 푀 −푀 ′ 푑푥2 푥 푥 • Multiplying by 푥, we get 푑2푦 퐸퐼푥 = 푀 푥−푀 ′ 푥 푑푥2 푥 푥
135 Continuous beam with supports at different levels
푑2푦 • 퐸퐼푥 = 푀 푥−푀 ′ 푥 푑푥2 푥 푥
• Integrating from 푥 = 0 푡표푥 = 푙1, we get, 1푙 1푙 1푙 푑2푦 퐸퐼 푥 = 푀 푥 푑푥 − 푀 ′푥 푑푥 푑푥2 푥 푥 0 0 0
푙1 푙1 푑푦 1푙 퐸퐼 푥. −푦 = 푀 푥 푑푥 − 푀 ′푥 푑푥 −− −(1) 푑푥 푥 푥 0 0 0
136 Continuous beam with supports at different levels
A B푀 C 푀 퐵 푀 푥 푙 훿1 푙 퐴 1 2 퐶
• But it may be suchthat At 푥 = 0, deflection 푦 =0 • At 푥 = 푙, 푦 = −훿 ; 푎푛푑 푠푙표푝푒푎푡퐵 푓표푟퐴퐵, 푑푦 =휃 1 1 퐵푑퐴푥
푙1 • 0 푀푥 푥 푑푥 =푎 1푥 1 =Momentof thefreeB.M. D.onAB aboutA.
푙1 ′ ′ • 0 푀푥 푥 푑푥 =푎 1 푥1 = Momentof thefixedB.M. D.onAB aboutA. ′
137 Continuous beam with supports at different levels
푙1 푙1 푑푦 1푙 퐸퐼 푥. −푦 = 푀 푥 푑푥 − 푀 ′푥 푑푥 −−(1) 푑푥 푥 푥 0 0 0 • Therefore the equation (1) is simplifiedas, ′ ′ 퐸퐼 푙1휃퐵퐴 − −훿1 =푎1푥1 −푎 푥1 . 1
푀 + But 푎′ =area of the fixed B.M.D. on AB = 퐴 푙 1 푀 퐵2 1
′ 푀 퐴 +2푀 퐵 1푙 푥1 =Centroidof thefixedB.M. D.fromA= 푀 퐴 +푀 퐵 3
138 Continuous beam with supports at different levels • Therefore, (푴 +푴 ) 푴 + ퟐ 푴 풍 풍ퟐ ′ ′ 푨 푩 푨 ퟏ 풂ퟏ풙ퟏ = 푩 풍ퟏ× = (푴 푨 + ퟐ푴 푩 ) ퟐ 푴 푨 +푴 푩 ퟏ ퟔ ퟑ 푙 2 ∴ 퐸퐼(푙휃 +훿 ) =푎 푥 − 푀 +2푀 1 1 퐵퐴 1 1 1 퐴 6 퐵 6푎1푥 1 6퐸퐼훿 1 6퐸퐼휃 퐵 = − − 푀퐴 +2푀퐵 푙1 −− −(2) 퐴 1푙 1푙 Similarly by considering the span BC and taking C as origin it can be shown that,
6푎2푥 2 6퐸퐼훿 2 6퐸퐼휃퐵퐶 = − − 푀퐶 +2푀퐵 푙2 −− −(3) 2푙 2푙
휃퐵퐶 = slope for span CB at B
139 Continuous beam with supports at different levels
• But휃퐵퐴 = −휃퐵퐶 as the direction of 푥 from A for the span AB, and from C for the span CB are in oppositedirection.
• And hence,휃퐵퐴+ 휃퐵퐶 =0
6푎1푥 1 6퐸퐼훿 1 6퐸퐼휃퐵퐴 = − − 푀퐴 +2푀퐵 푙1 −− −(2) 1푙 1푙
6푎2푥 2 6퐸퐼훿 2 6퐸퐼휃퐵퐶 = − − 푀 퐶 +2푀 푙2 −− −(3) 2푙 퐵 2푙 Adding equations (2) and (3), we get
6퐸퐼(휃퐵퐴+휃퐵퐶) 6푎 푥 6푎 푥 6퐸퐼훿 6퐸퐼훿 = 1 1 + 2 2 − 1 − 2 푙1 푙2 푙1 푙2 − 푀퐴푙1 +2푀퐵 푙1 +푙2 + 푀푐푙2
140 Continuous beam with supports at different levels 6퐸퐼(휃퐵퐴+휃퐵퐶) 6푎 푥 6푎 푥 6퐸퐼훿 6퐸퐼훿 = 1 1 + 2 2 − 1 − 2 푙1 푙2 푙1 푙2 − 푀퐴푙1 +2푀퐵 푙1 +푙2 + 푀푐푙2
6푎1푥 1 6푎2푥 2 6퐸퐼훿 1 6퐸퐼훿 2 0= + − − − 푀퐴푙1 +2푀퐵 푙1 +푙2 + 푀 푐푙2 1푙 2푙 1푙 2푙
6푎1푥1 6푎2푥2 훿1 훿2 푀퐴푙1 +2푀퐵 푙1 +푙2 + 푀 푐푙2 = + − 6퐸퐼 + 푙1 푙2 푙1 푙2
141 P r o b l e m s
• The following Figure shows a continuous beam carrying an external loading. If the support B sinks by 0.25 cm below the level of the other supports find support moments. Take I for section= 15000 cm4 and E=2x103 t/cm2.
A 4 t/m B C 2t/m D
4m 4m 4m
142 P r o b l e ms A 4 t/m B C 2t/m D 0.25cm 4m 4m 4m • The theorem of three moments for two spans AB and BC is as follows,
6푎1푥1 6푎2푥2 훿1 훿2 푀퐴푙1 +2푀퐵 푙1 +푙2 + 푀푐푙2 = + −6퐸퐼 + 1푙 2푙 1푙 2푙 • Consider the spans AB and BC,
• 푀퐴 = 0 • 훿1 = +0.25cm • 훿2 = +0.25cm
143 P r o b l e ms A 4 t/m B C 2t/m D 4m 0.25cm 4m 4m
8tm 8tm 4tm
FreeBMD 2 • 푎1 = 3 ×4 ×8 2 6 ×2 ×103 ×15000 • 6푎1푥1 6× ×4×8×2 = 3 =64 6퐸퐼 = 2 푙 4 100 1 = 18000 t푚2 2 6푎 푥 6× ×4×8×2 • 2 2 = 3 =64 2푙 4
144 P r o b l e ms A 4 t/m B C 2t/m D 4m 0.25cm 4m 4m
8 tm 8 tm 4tm
6푎1푥1 6푎2푥2 훿1 훿2 푀 푙 +2푀 푙 +푙 + 푀 푙 = + − 6퐸퐼 + 퐴 1 퐵 1 2 푐2 푙 푙 1 2 0.25 푙10.25푙2 • ∴ 0+2푀 4+4 + 4푀 = 64 + 64 − 18000 + 퐵 퐶 400 400 • ∴ 16푀퐵+4푀퐶 = 128 −22.5 • 16푀퐵+4푀퐶 =105.5 • 4푀퐵+푀퐶 = 26.375 −−−−(1)
145 P r o b l e m s A 4 t/m B C 2t/m D 0.25cm 4m 4 m 4m 8tm 8tm 4tm
• Now consider the spans BC andCD,
• 푀 푑 =0, • 훿1 = −0.25 cm • 훿2 =0
146 P r o b l e m s A 4 t/m B C 2t/m D 4m 0.25cm 4m 4m
8 tm 8 tm 4tm
6푎1푥1 6푎2푥2 훿1 훿2 푀 푙 +2푀 푙+푙 + 푀 푙 = + − 6퐸퐼 + 퐴 1 퐵 1 2 푐2 푙1 푙2 1푙 2푙 −0. 25 0 푀 ×4+2푀 4+4 + 0 = 64 + 32 − 18000 + 퐵 퐶 400 400
• ∴ 4푀퐵+16푀퐶 = 96 +11.25 • 4푀퐵+16푀퐶 = 107.25 −−− −(2)
147 P r o b l e m s • 4푀퐵+푀퐶 = 26.375 −−−−(1) • 4푀퐵+16푀퐶 = 107.25 −−− −(2) • Solving (1) and (2), we get,
• 푀 퐵 =5.24tm hogging . • 푀 퐶 = 5.39tm hogging .
148 P r o b l e m s A 4 t/m B C 2t/m D 4m 0.25cm 4m 4m
8tm 8tm 5.24tm + 5.39tm 4tm + - - +
BMD
149 Fixed beam with ends at different levels (Effect of sinking of supports) V
A B 푀 훿 퐴 푀 V 퐵
푀퐴 is negative (hogging) and 푀 퐵 is positive (sagging).Numerically 푀퐴 and 푀 퐵 areequal.
Let V be the reaction at eachsupport.
150 Fixed beam with ends at different levels (Effect of s i n k i n g of supports) V
A B 푀 훿 퐴 푥 푀 V 퐵 Consider any section distance 푥 from the end A. Since the rate of loading is zero, we have, with the usualnotations 푑4푦 퐸퐼 =0 푑푥4 Integrating, we get, 푑3푦 Shear force =퐸퐼푑푥3 =퐶1 Where 퐶1is a constant At 푥 = 0, 푆. 퐹.= +V ∴ 퐶1 =V
151 Fixed beam with ends at different l e v e l s ( E f f e c t of s i n k i n g of supports) V
A B 푀 훿 퐴 푥 푀 V 퐵 푑2푦 B.M. at any section =퐸퐼푑푥2 = 푉푥 +퐶1 At 푥 = 0, 퐵. 푀.= −푀퐴 ∴ 퐶2 =−푀퐴 푑2푦 ∴ 퐸퐼 = 푉푥 −푀 푑푥2 퐴 Integratingagain, 퐸퐼푑푦 = 푉 푥2 − 푀 푥 + 퐶 (Slopeequation) 푑푥 2 퐴 3 푑푦 But at 푥 = 0, 푑푥 =0 ∴ 퐶3 =0
152 Fixed beam with ends at different levels (Effect of s i n k i n g of supports) V
A B 푀 훿 퐴 푥 푙 푀 V 퐵 Integratingagain, 푉푥3 푀 푥2 퐸퐼푦 = − 퐴 +퐶 ------(Deflectionequation) 6 2 4 But at 푥 = 0, 푦 = 0 ∴ 퐶4 =0 At 푥 = 푙, 푦 = −훿 3 푉푙 푀 퐴 푙 −퐸퐼훿 = − 2 −−−−−−− −(i) 6 2
153 Fixed beam with ends at different l e v e l s ( E f f e c t of s i n k i n g of supports) V
A B 푀 훿 퐴 푥 푙 푀 V 퐵 푑푦 But we also know that at B, 푥 = 푙푎푛푑 푑푥 =0 And substitute in slopeEq. 퐸퐼푑푦 = 푉 푥2 −푀 푥 푑푥 2 퐴 푉푙2 ∴ 0 = −푀 푙 2 퐴 2푀 ∴ 푉 = 퐴 −−−−−−− − ii 푙 3 2 Substituting in deflection Eq.(i) i.e., − = 푉푙 − 푀퐴푙 ;we have, 퐸퐼훿 6 2 3 2 2푀 푙 푀퐴푙 −퐸퐼훿 = 퐴 × − 2 154 Fixed beam with ends at different l e v e l s ( E f f e c t of s i n k i n g of supports) V
A B 푀 훿 퐴 푙 푀 V 퐵 푀 푙2 퐸퐼훿 = 퐴 6 6퐸퐼훿 ∴ = 퐴 푙2 푀 Hence the law for the bending moment at any section distant x from A is givenby, 푑2푦 푀 = 퐸퐼 = V푥 −푀 푑푥2 퐴 2푀 6퐸퐼훿 ∴ 푀 = 퐴 푥 − 푙 푙2 155 Fixed beam with ends at different levels (Effect of sinking of supports) V
A B 푀 훿 퐴 푙 푀 V 퐵 Butfor B.M. at B, put x= l, 2푀퐴 6퐸퐼훿 12퐸퐼훿 6퐸퐼훿 6퐸퐼훿 ∴ = ×푙− = − = 퐵 푙 푙2 푙2 푙2 푙2 푀 Hence when the ends of a fixed beam are at different levels, 6퐸퐼훿 The fixing moment at each end = numerically. 푙2
At the higher end this moment is a hogging moment and at the lower end this moment is a saggingmoment.
156 P r o b l e m s • A fixed beam of span 5 metres carries a concentrated load of 20 t at 3 meters from the left end. If the right end sinks by 1 cm, find the fixing moments at the supports. For the beam section take I=30,000 cm4 and E=2x103 t/cm2. Find also the reaction at the supports.
20 t 3m 2m A B
157 P r o b l e ms • A fixed beam of span 5 metres carries a concentrated load of 20 t at 3 meters from the left end.
20 t 3m 2m A B 1 cm
• The right end sinks by 1 cm, find the fixing moments at the supports.
158 P r o b l e ms 20 t
Wa푏2 6퐸퐼훿 3m 2m • 푀 = − − A B 퐴 푙2 푙2 푀 1cm 20×3×22 6×2×103퐴×30,000×1 • =− + tm 52 52×1002 • = − 9.6+0.48 tm=-10.08 tm (hogging) 푀 퐵
W푏푎2 6퐸퐼훿 • 푀 = − + 퐵 푙2 푙2 20×2×32 6×2×103×30,000×1 • = − + tm 52 52×1002 • = −14.4+0.48 tm= -13.92 tm (hogging)
159 P r o b l e m s 20 t 3m 2m 10.08 A B
• Reaction at A: 푉 푉 퐵 13.92 • 푀퐵=0, 퐴
• 푉퐴 × 5 + 13.92 − 10.08 − 20 × 2 = 0
• ∴ 푉퐴 = 7.232 t • Reaction at B:
• ∴ 푉퐵 = 20 − 7.232 = 12.768 t.
160