Chapter 9 Deflections of Beams

Dept. of Civil and Environmental Engineering, Seoul National University Junho Song 457.201 Mechanics of Materials and Lab. [email protected]

Chapter 9 Deflections of Beams

9.2 Differential Equations of the Deflection Curve

 Sign Conventions and Main Concepts

1. Deflection : Displacement in y-direction at a point (upward positive)

2. Angle of rotation𝑣𝑣 : Angle between x-axis and t______to the deflection curve

(counterclockwise positive)𝜃𝜃

3. Center of curvature : Intersection of the orthogonal lines at two points on the

curve 𝑂𝑂′

4. Radius of curvature : Distance between and the deflection curve (Recall

= ) 𝜌𝜌 𝑂𝑂′

5. Curvature𝜌𝜌𝜌𝜌𝜌𝜌 𝑑𝑑𝑑𝑑 : Reciprocal of the radius of curvature, i.e. = 1/ = / (∪

positive) κ 𝜅𝜅 𝜌𝜌 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑

6. Slope of the deflection curve

= tan = arctan 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝜃𝜃 𝜃𝜃 In𝑑𝑑𝑑𝑑 a similar manner 𝑑𝑑𝑑𝑑

cos = sin = 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝜃𝜃 𝜃𝜃 7. These are𝑑𝑑𝑑𝑑 all based only𝑑𝑑𝑑𝑑 on geometric considerations, and thus valid for beams of any material. No restrictions on the magnitudes of the slopes and deflections. Dept. of Civil and Environmental Engineering, Seoul National University Junho Song 457.201 Mechanics of Materials and Lab. [email protected]

 Beams with Small Angles of Rotation

1. Very small angle of rotations and deflections  approximation greatly

simplify beam analysis 𝑑𝑑𝑑𝑑 ≈ 𝑑𝑑𝑑𝑑

2. Curvature

1 =

𝑑𝑑𝑑𝑑 𝜅𝜅 ≈ 3. Angle𝜌𝜌 of 𝑑𝑑rotation (equal to the s______; angle should be given in r______)

tan = 𝑑𝑑 𝜃𝜃 ≈ 𝜃𝜃 4. Curvature (in𝑑𝑑 𝑑𝑑terms of deflection curve):

1 = = 2 𝑑𝑑𝑑𝑑 𝑑𝑑 𝑣𝑣 𝜅𝜅 ≈ 2 Note𝜌𝜌: Exact𝑑𝑑𝑑𝑑 curvature𝑑𝑑𝑥𝑥 (See textbook)

= = [1 + ( ) ] / 𝑑𝑑𝑑𝑑 𝑣𝑣′′ 𝜅𝜅 ′ 2 3 2 5. For a𝑑𝑑 𝑑𝑑linearly elastic𝑣𝑣 beam (i.e. following Hooke’s law), = 1/ = / , we now

obtain differential equation of the deflection curve as 𝜅𝜅 𝜌𝜌 𝑀𝑀 𝐸𝐸𝐸𝐸

= = 2 𝑑𝑑 𝑣𝑣 ′′ 𝑀𝑀 2 𝑣𝑣  Nonprismatic𝑑𝑑𝑥𝑥 Beams𝐸𝐸𝐸𝐸

1. From above, but with flexural rigidity varying over , i.e. ,

𝑥𝑥 𝐸𝐸𝐸𝐸𝑥𝑥 = 2 𝑑𝑑 𝑣𝑣 𝐸𝐸𝐸𝐸𝑥𝑥 2 𝑀𝑀 2. From𝑑𝑑𝑥𝑥 / = and / =

𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑉𝑉 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 −𝑞𝑞 = = 2 𝑑𝑑 𝑑𝑑 𝑣𝑣 𝑑𝑑𝑑𝑑 �𝐸𝐸𝐸𝐸𝑥𝑥 2� 𝑉𝑉 𝑑𝑑𝑑𝑑 𝑑𝑑𝑥𝑥 𝑑𝑑𝑑𝑑 = = = 2 2 2 𝑑𝑑 𝑑𝑑 𝑣𝑣 𝑑𝑑 𝑀𝑀 𝑑𝑑𝑑𝑑 2 �𝐸𝐸𝐸𝐸𝑥𝑥 2� 2 −𝑞𝑞  Prismatic𝑑𝑑𝑥𝑥 Beams𝑑𝑑𝑥𝑥 𝑑𝑑𝑥𝑥 𝑑𝑑𝑑𝑑

1. Differential equations for prismatic beams, i.e. =

𝐸𝐸𝐸𝐸𝑥𝑥 𝐸𝐸𝐸𝐸 Dept. of Civil and Environmental Engineering, Seoul National University Junho Song 457.201 Mechanics of Materials and Lab. [email protected]

1) Bending-moment equation = ′′ 2) Shear-force equation 𝐸𝐸𝐸𝐸=𝑣𝑣 𝑀𝑀 ′′′ 3) Load equation =𝐸𝐸𝐸𝐸𝑣𝑣 𝑉𝑉 ′′′′ 𝐸𝐸𝐸𝐸𝑣𝑣 −𝑞𝑞 9.3 Deflections by Integration of the Bending-Moment Equation

 Conditions needed for Solving Bending-Moment Equations by Method of Successive Integrations (i.e. integrate the differential equation and find the undetermined coefficients by given conditions)

1. Boundary conditions: Deflection and slope at boundaries

2. Continuity conditions: At a given point, the deflections (or slopes) obtained for the left- and right-hand parts should be equal

3. Symmetry conditions: For example, the slope of the deflection curve at the midpoint is zero (for a symmetric beam under symmetric loads)

Dept. of Civil and Environmental Engineering, Seoul National University Junho Song 457.201 Mechanics of Materials and Lab. [email protected]

 Example 9-1: Determine the equation of the deflection curve for a simple beam supporting a uniform load of intensity . Also determine the maximum deflection

at the𝑞𝑞 midpoint of the beam, and the angles of 𝛿𝛿rotationmax at the supports, i.e. and . The beam has length and constant flexural𝜃𝜃𝐴𝐴 rigidity𝜃𝜃 𝐵𝐵 .

𝐿𝐿 𝐸𝐸𝐸𝐸

Dept. of Civil and Environmental Engineering, Seoul National University Junho Song 457.201 Mechanics of Materials and Lab. [email protected]

 Example 9-2: Determine the equation of the deflection curve for a cantilever beam subjected to a uniform load of intensity . Also determine the angle

of rotation and the deflection𝑞𝑞 at the free end. The beam ha𝜃𝜃𝐵𝐵s length and constant𝛿𝛿𝐵𝐵 flexural rigidity . 𝐿𝐿

𝐸𝐸𝐸𝐸

Dept. of Civil and Environmental Engineering, Seoul National University Junho Song 457.201 Mechanics of Materials and Lab. [email protected]

 Example 9-3: Determine the equation of the deflection curve for a simple beam subjected to a concentrated load acting at distances and

from the left- and right𝑃𝑃 -had supports, respectively.𝑎𝑎 𝑏𝑏 Also determine the angles of rotation and , the

maximum deflection , and the deflection𝜃𝜃𝐴𝐴 𝜃𝜃𝐵𝐵 at the midpoint of the beam𝛿𝛿max. The beam has length𝛿𝛿𝐶𝐶 and constant flexural rigidity . 𝐿𝐿

𝐸𝐸𝐸𝐸 Dept. of Civil and Environmental Engineering, Seoul National University Junho Song 457.201 Mechanics of Materials and Lab. [email protected]

9.4 Deflections by Integration of the Shear-Force and Load Equations

 Solving Differential Equations for Deflections

1. Among three differential equations for deflections, i.e.

1) Bending-moment equation = (______unknowns) ′′ 2) Shear-force equation 𝐸𝐸𝐸𝐸=𝑣𝑣 (______𝑀𝑀 unknowns) ′′′ 3) Load equation =𝐸𝐸𝐸𝐸𝑣𝑣 (______𝑉𝑉 unknowns) ′′′′ One can solve a diffe𝐸𝐸𝐸𝐸rential𝑣𝑣 equation−𝑞𝑞 based on available conditions

2. Boundary, continuity and symmetry conditions are available in terms of

1) Deflection ( ) and slope ( ) for bending-moment equation (Section 9.3) ′ 2) In addition, 𝑣𝑣moment ( ) conditions𝑣𝑣 can be used for the shear-force equation = because 𝑀𝑀 ′′ 3) In addition,𝐸𝐸𝐸𝐸 shear𝑣𝑣 𝑀𝑀force ( ) conditions can be used for solving the load equation = because 𝑉𝑉 ′′′ 𝐸𝐸𝐸𝐸𝑣𝑣 𝑉𝑉

Load Equation = ′′′′ 𝐸𝐸𝐸𝐸𝐸𝐸 −𝑞𝑞

= [ ( )] + 1 Conditions on ′′′ 𝐸𝐸𝐸𝐸𝐸𝐸 ∫ −𝑞𝑞 𝑥𝑥 𝑑𝑑𝑑𝑑 𝐶𝐶 𝑉𝑉 Shear-Force Equation = ′′′ 𝐸𝐸𝐸𝐸𝐸𝐸 𝑉𝑉

= ( ) + 2 Conditions on ′′ 𝐸𝐸𝐸𝐸𝐸𝐸 ∫𝑉𝑉 𝑥𝑥 𝑑𝑑𝑑𝑑 𝐶𝐶 𝑀𝑀 Bending-Moment Equation = ′′ 𝐸𝐸𝐸𝐸𝐸𝐸 𝑀𝑀

= ( ) + 3 = ( ) Conditions on ′ ′ 𝐸𝐸𝐸𝐸𝐸𝐸 ∫𝑀𝑀 𝑥𝑥 𝑑𝑑𝑑𝑑 𝐶𝐶 𝑓𝑓 𝑥𝑥 𝑣𝑣 = ( ) + Conditions on 4 𝐸𝐸𝐸𝐸𝐸𝐸 ∫ 𝑓𝑓 𝑥𝑥 𝑑𝑑𝑑𝑑 𝐶𝐶 𝑣𝑣 Deflection Curve ( )

𝑣𝑣 𝑥𝑥

Dept. of Civil and Environmental Engineering, Seoul National University Junho Song 457.201 Mechanics of Materials and Lab. [email protected]

 Example 9-4: Determine the equation of the deflection curve for a cantilever beam AB supporting a triangularly distributed load of maximum intensity . Also determine the

deflection and angle𝑞𝑞0 of rotation at the free end. Use the𝛿𝛿𝐵𝐵 load equation. 𝜃𝜃𝐵𝐵

Dept. of Civil and Environmental Engineering, Seoul National University Junho Song 457.201 Mechanics of Materials and Lab. [email protected]

 Example 9-5: Determine the equation of the deflection curve for a simple beam with an overhang under concentrated load at the end. Also determine the deflection . Use

the third-order differential equation, 𝑃𝑃i.e. shear-force equation. The beam has constant𝛿𝛿𝐶𝐶 flexural rigidity .

𝐸𝐸𝐸𝐸

Dept. of Civil and Environmental Engineering, Seoul National University Junho Song 457.201 Mechanics of Materials and Lab. [email protected]

9.5 Method of Superposition

 Method of Superposition

1. Under suitable conditions, the deflection of a beam produced by several different loads acting simultaneously can be found by s______the deflections produced by the same loads acting separately.

2. At a given location, the deflection by a load  (alone) and by a load (alone)𝑣𝑣1 the 𝑞𝑞1 deflection under𝑣𝑣2 loads 𝑞𝑞and2 is ______𝑞𝑞1 𝑞𝑞2

3. Justification: the differential equation for deflection is l______~ “Principle of Superposition” (mathematics)

4. “Principle of Superposition” is valid under the following conditions:

(1) H______’s law for the material

(2) Deflections and rotations are s______

(3) Presence of deflection does not alter the actions of the applied loads

 Three sources of n______in general structural engineering problems: nonlinear material property (constitutive law, large deformation, load-displacement interactions)

 Tables of Beam Deflections

1. Superpose the deflection equations or deflections and slopes at specific locations determined for each type of the loads

2. If a given pattern of distributed loads is not available in the table, one can use the results about concentrated loads as shown in the following example Dept. of Civil and Environmental Engineering, Seoul National University Junho Song 457.201 Mechanics of Materials and Lab. [email protected]

3. From Appendix G, the midpoint deflection caused by a concentrated load at = ( ) is

𝑃𝑃 𝑥𝑥 𝑎𝑎 𝑎𝑎 ≤ 𝑏𝑏 (3 4 ) 48 𝑃𝑃𝑃𝑃 2 2 𝐿𝐿 − 𝑎𝑎 4. The𝐸𝐸𝐸𝐸 midpoint deflection caused by at location

is thus 𝑞𝑞𝑞𝑞𝑞𝑞 𝑥𝑥 ( ) (3 4 ) 48 𝑞𝑞𝑞𝑞𝑞𝑞 𝑥𝑥 2 2 𝐿𝐿 − 𝑥𝑥 5. In the𝐸𝐸𝐸𝐸 example, the distributed load is given as

2 ( ) = 𝑞𝑞0𝑥𝑥 𝑞𝑞 𝑥𝑥 6. Substituting𝐿𝐿 this into the equation in “4”,

(3 4 ) 24 2 𝑞𝑞0𝑥𝑥 2 2 𝐿𝐿 − 𝑥𝑥 𝑑𝑑𝑑𝑑 7. Finally,𝐸𝐸𝐸𝐸 the deflection at is obtained by the integral

𝐶𝐶

= 𝐿𝐿 (3 4 ) = 2 4 2 24 240 0 2 2 0 𝐶𝐶 𝑞𝑞 𝑥𝑥 𝑞𝑞 𝐿𝐿 𝛿𝛿 �0 𝐿𝐿 − 𝑥𝑥 𝑑𝑑𝑑𝑑  Example 9-6: A𝐸𝐸𝐸𝐸 cantilever beam AB supports𝐸𝐸𝐸𝐸 a uniform load of intensity acting over part of the span and a

concentrated load𝑞𝑞 acting at the free end. Determine the deflection and𝑃𝑃 the angle of rotation at end B. 𝐵𝐵 𝐵𝐵 𝛿𝛿 𝜃𝜃 Dept. of Civil and Environmental Engineering, Seoul National University Junho Song 457.201 Mechanics of Materials and Lab. [email protected]

 Example 9-7: A cantilever beam AB supports a uniform load of intensity acting on the right-hand

half of the beam. Determine𝑞𝑞 the deflection and the angle of rotation at the free end. 𝛿𝛿𝐵𝐵 𝐵𝐵 𝜃𝜃 Dept. of Civil and Environmental Engineering, Seoul National University Junho Song 457.201 Mechanics of Materials and Lab. [email protected]

9.6 Moment-Area Method

 Moment-Area Method

1. Refers to two theorems related to the area of the b______-m______diagram ( )  Find the angle and deflection of a beam

2. Assumptions:𝑀𝑀 𝑥𝑥 L______e______materials and s_____ deformation

 First Moment-Area Theorem

1. The angle between the two tangents at points and

𝐴𝐴 𝐵𝐵 / =

𝐵𝐵 𝐴𝐴 𝐵𝐵 𝐴𝐴 2. The𝜃𝜃 angle𝜃𝜃 −between𝜃𝜃 two orthogonal lines at and

𝑚𝑚1 𝑚𝑚2 = 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 ≅ 3. For a beam𝜌𝜌 𝜌𝜌consisting of a linear elastic material 1/ = /

𝜌𝜌 𝑀𝑀 𝐸𝐸𝐸𝐸 = 𝑀𝑀𝑀𝑀𝑀𝑀 𝑑𝑑𝑑𝑑 4. Integrating𝐸𝐸𝐸𝐸 the change in the angle from point to point ,

𝐴𝐴 𝐵𝐵 = 𝐵𝐵 / 𝐵𝐵 𝐴𝐴 �𝐴𝐴 𝑑𝑑𝑑𝑑 𝜃𝜃 = / 𝑩𝑩 𝑩𝑩 𝑨𝑨 𝑴𝑴𝑴𝑴𝑴𝑴 𝜽𝜽 �𝑨𝑨 = Area of the𝑬𝑬𝑬𝑬 / diagram between points and

𝑀𝑀 𝐸𝐸𝐸𝐸 𝐴𝐴 𝐵𝐵 5. First moment-area theorem: The angle / between the tangents to the

deflection curve at two points and is𝜃𝜃 equal𝐵𝐵 𝐴𝐴 to the area of the / diagram between those points 𝐴𝐴 𝐵𝐵 𝑀𝑀 𝐸𝐸𝐸𝐸

Dept. of Civil and Environmental Engineering, Seoul National University Junho Song 457.201 Mechanics of Materials and Lab. [email protected]

 Second Moment-Area Theorem

1. Tangent deviation of point with : respect to / 𝐵𝐵 𝐴𝐴 𝑡𝑡𝐵𝐵 𝐴𝐴 2. Tangent deviation of point

made by the element 𝐵𝐵: 𝑚𝑚1𝑚𝑚2 = = 𝑀𝑀𝑀𝑀𝑀𝑀 𝑑𝑑𝑑𝑑 𝑥𝑥1𝑑𝑑𝑑𝑑 𝑥𝑥1 3. This is the contribution𝐸𝐸𝐸𝐸 of the element to the total tangent

deviation.𝑚𝑚 Thus,1𝑚𝑚2 the total tangent deviation is

= = / 𝐵𝐵 𝑩𝑩 𝑴𝑴𝑴𝑴𝑴𝑴 𝒕𝒕𝑩𝑩 𝑨𝑨 � 𝑑𝑑𝑑𝑑 � 𝒙𝒙𝟏𝟏 𝐴𝐴 𝑨𝑨 𝑬𝑬𝑬𝑬 4. This is the first moment of the area of the / diagram between points and ,

evaluated with respect to point 𝑀𝑀 𝐸𝐸𝐸𝐸 𝐴𝐴 𝐵𝐵 𝐵𝐵 5. Second moment-area theorem: The tangential deviation / of point from the

tangent at point is equal to the first moment of the area of𝑡𝑡𝐵𝐵 the𝐴𝐴 / di𝐵𝐵agram between and 𝐴𝐴 , evaluated with respect to . 𝑀𝑀 𝐸𝐸𝐸𝐸

6. Note: The𝐴𝐴 first moment𝐵𝐵 can be computed alternative𝐵𝐵 by ( )×( )

Dept. of Civil and Environmental Engineering, Seoul National University Junho Song 457.201 Mechanics of Materials and Lab. [email protected]

 Example 9-10: Determine the angle of rotation and deflection at the free end

of a cantilever𝜃𝜃𝐵𝐵 beam 𝛿𝛿supporting𝐵𝐵 a 𝐵𝐵concentrated load using𝐴𝐴𝐴𝐴 the moment-area method. 𝑃𝑃

Dept. of Civil and Environmental Engineering, Seoul National University Junho Song 457.201 Mechanics of Materials and Lab. [email protected]

 Example 9-12: A simple beam supports

a concentrated load acting at𝐴𝐴𝐴𝐴 the𝐴𝐴 position shown in the figure. Determine𝑃𝑃 the angle of rotation at support and the deflection

under𝜃𝜃 𝐴𝐴the load using𝐴𝐴 the moment area method.𝛿𝛿𝐷𝐷 𝑃𝑃

Dept. of Civil and Environmental Engineering, Seoul National University Junho Song 457.201 Mechanics of Materials and Lab. [email protected]

9.8 Strain Energy of Bending

 Strain Energy of a Beam under Bending

1. Consider a linear elastic beam under pure bending showing small deformation, i.e. the bending moment is constant and Hooke’s law works

2. The angle of the circular arc is

= = = 𝐿𝐿 𝑀𝑀𝑀𝑀 𝜃𝜃 𝜅𝜅𝜅𝜅 3. Linear𝜌𝜌 relationship𝐸𝐸𝐸𝐸 between and  Therefore, when the bending moment

gradually increases, the work𝑀𝑀 and the𝜃𝜃 stored strain energy are

= = 2 𝑀𝑀𝑀𝑀 𝑊𝑊 𝑈𝑈 4. Using the linear relationship in Item 3, the strain energy can be expressed as

= or = 2 2 2 2 𝑀𝑀 𝐿𝐿 𝐸𝐸𝐸𝐸𝜃𝜃 𝑈𝑈 𝑈𝑈 5. If the bending𝐸𝐸𝐸𝐸 moment𝐿𝐿 varies, the angle between the side faces of a small element with length is

𝑑𝑑𝑑𝑑 = = 2 𝑑𝑑 𝑣𝑣 𝑑𝑑𝑑𝑑 𝜅𝜅𝜅𝜅𝜅𝜅 2 𝑑𝑑𝑥𝑥 6. In analogy to𝑑𝑑 Item𝑥𝑥 4, the strain energy in the element is

( ) = or = = = 22 2 2 2 2 2 2 2 2 𝑀𝑀 𝑑𝑑𝑑𝑑 𝐸𝐸𝐸𝐸 𝑑𝑑𝜃𝜃 𝐸𝐸𝐸𝐸 𝑑𝑑 𝑣𝑣 𝐸𝐸𝐸𝐸 𝑑𝑑 𝑣𝑣 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 � 2 𝑑𝑑𝑑𝑑� � 2� 𝑑𝑑𝑑𝑑 7. By integrating𝐸𝐸𝐸𝐸 the strain energy𝑑𝑑𝑑𝑑 along𝑑𝑑𝑑𝑑 the𝑑𝑑 𝑥𝑥beam, the strain𝑑𝑑𝑥𝑥 energy is derived as

= or = 22 2 2 2 𝑀𝑀 𝑑𝑑𝑑𝑑 𝐸𝐸𝐸𝐸 𝑑𝑑 𝑣𝑣 𝑈𝑈 � 𝑈𝑈 � � 2� 𝑑𝑑𝑑𝑑  Deflections Caused𝐸𝐸𝐸𝐸 by a Single Load𝑑𝑑𝑥𝑥

1. Strain energy by a concentrated load and a couple is and 𝑃𝑃𝑃𝑃 𝑀𝑀0𝜃𝜃 𝑈𝑈 𝑃𝑃 𝑀𝑀0 2 2 2. Deflection and the rotation can be computed by

2 2 = and = 𝑈𝑈 𝑈𝑈 𝛿𝛿 𝜃𝜃 𝑃𝑃 𝑀𝑀0 Dept. of Civil and Environmental Engineering, Seoul National University Junho Song 457.201 Mechanics of Materials and Lab. [email protected]

 Example 9-6: Cantilever beam is subjected to

three different loading conditions:𝐴𝐴𝐵𝐵 (a) a concentrated load at its free end, (b) a couple at its free end,

and (c)𝑃𝑃 both loads acting simultaneously.𝑀𝑀0 For each loading condition, determine the strain energy of the beam. Also, determine the vertical deflection at

end of the beam due to the load acting alone𝛿𝛿𝐴𝐴 , and 𝐴𝐴determine the angle of rotation𝑃𝑃 at end due

to the moment acting alone. 𝜃𝜃𝐴𝐴 𝐴𝐴 𝑀𝑀0