Third Edition

CHAPTER

MECHANICS OF MATERIALS Ferdinand P. Beer E. Russell Johnston, Jr. John T. DeWolf

Deflection of Beams

Lecture Notes: J. Walt Oler Texas Tech University

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Third Edition

MECHANICS OF MATERIALS

Beer • Johnston • DeWolf

Deflection of Beams Deformation of a Beam Under Transverse Loading Equation of the Elastic Curve Direct Determination of the Elastic Curve From the Load Di... Statically Indeterminate Beams Sample Problem 9.1

Sample Problem 9.8 Moment-Area Theorems Application to Cantilever Beams and Beams With Symmetric ... Bending Moment Diagrams by Parts Sample Problem 9.11

Sample Problem 9.3

Application of Moment-Area Theorems to Beams With Unsymme...

Method of Superposition

Maximum Deflection

Sample Problem 9.7

Use of Moment-Area Theorems With Statically Indeterminate...

Application of Superposition to Statically Indeterminate ...

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Deformation of a Beam Under Transverse Loading • Relationship between bending moment and curvature for pure bending remains valid for general transverse loadings. 1

ρ

=

M ( x) EI

• Cantilever beam subjected to concentrated load at the free end, 1

ρ

=−

Px EI

• Curvature varies linearly with x 1 • At the free end A, ρ = 0, A

• At the support B, © 2002 The McGraw-Hill Companies, Inc. All rights reserved.

1

ρB

ρA = ∞

≠ 0, ρ B =

EI PL 9-3

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Deformation of a Beam Under Transverse Loading • Overhanging beam • Reactions at A and C • Bending moment diagram • Curvature is zero at points where the bending moment is zero, i.e., at each end and at E. 1

ρ

=

M ( x) EI

• Beam is concave upwards where the bending moment is positive and concave downwards where it is negative. • Maximum curvature occurs where the moment magnitude is a maximum. • An equation for the beam shape or elastic curve is required to determine maximum deflection and slope. © 2002 The McGraw-Hill Companies, Inc. All rights reserved.

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Equation of the Elastic Curve • From elementary calculus, simplified for beam parameters, d2y 1

ρ

=

dx 2 2 ⎤3 2

⎡ ⎛ dy ⎞ ⎢1 + ⎜ ⎟ ⎥ ⎢⎣ ⎝ dx ⎠ ⎥⎦

≈

d2y dx 2

• Substituting and integrating, EI

1

ρ

= EI

d2y dx

2

= M (x) x

dy = M ( x )dx + C1 EI θ ≈ EI dx ∫ 0

x

x

0

0

EI y = ∫ dx ∫ M ( x ) dx + C1x + C2 © 2002 The McGraw-Hill Companies, Inc. All rights reserved.

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MECHANICS OF MATERIALS

Beer • Johnston • DeWolf

Equation of the Elastic Curve • Constants are determined from boundary conditions x

x

0

0

EI y = ∫ dx ∫ M ( x ) dx + C1x + C2

• Three cases for statically determinant beams, – Simply supported beam y A = 0,

yB = 0

– Overhanging beam y A = 0,

yB = 0

– Cantilever beam y A = 0, θ A = 0 • More complicated loadings require multiple integrals and application of requirement for continuity of displacement and slope. © 2002 The McGraw-Hill Companies, Inc. All rights reserved.

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Direct Determination of the Elastic Curve From the Load Distribution • For a beam subjected to a distributed load, d 2M

dM = V (x) dx

dV = = − w( x ) 2 dx dx

• Equation for beam displacement becomes d 2M dx

2

= EI

d4y dx

4

= − w( x )

• Integrating four times yields EI y ( x ) = − ∫ dx ∫ dx ∫ dx ∫ w( x )dx + 16 C1x3 + 12 C2 x 2 + C3 x + C4

• Constants are determined from boundary conditions.

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Statically Indeterminate Beams • Consider beam with fixed support at A and roller support at B. • From free-body diagram, note that there are four unknown reaction components. • Conditions for static equilibrium yield ∑ Fx = 0 ∑ Fy = 0 ∑ M A = 0

The beam is statically indeterminate. • Also have the beam deflection equation, x

x

0

0

EI y = ∫ dx ∫ M ( x ) dx + C1x + C2

which introduces two unknowns but provides three additional equations from the boundary conditions: At x = 0, θ = 0 y = 0

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At x = L, y = 0

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Sample Problem 9.1 SOLUTION: • Develop an expression for M(x) and derive differential equation for elastic curve. W 14 × 68

I = 723 in 4

P = 50 kips L = 15 ft

E = 29 × 106 psi a = 4 ft

• Integrate differential equation twice and apply boundary conditions to obtain elastic curve.

For portion AB of the overhanging beam, • Locate point of zero slope or point (a) derive the equation for the elastic curve, of maximum deflection. (b) determine the maximum deflection, • Evaluate corresponding maximum (c) evaluate ymax. deflection.

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Sample Problem 9.1 SOLUTION: • Develop an expression for M(x) and derive differential equation for elastic curve. - Reactions: RA =

Pa ⎛ a⎞ ↓ RB = P⎜1 + ⎟ ↑ L ⎝ L⎠

- From the free-body diagram for section AD, M = −P

a x L

(0 < x < L )

- The differential equation for the elastic curve, EI © 2002 The McGraw-Hill Companies, Inc. All rights reserved.

d2y

a = − P x 2 L dx 9 - 10

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Sample Problem 9.1 • Integrate differential equation twice and apply boundary conditions to obtain elastic curve. EI

dy 1 a = − P x 2 + C1 dx 2 L

1 a EI y = − P x3 + C1x + C2 6 L d2y

a EI 2 = − P x L dx

at x = 0, y = 0 : C2 = 0 1 a 1 at x = L, y = 0 : 0 = − P L3 + C1L C1 = PaL 6 L 6

Substituting, dy 1 a 1 = − P x 2 + PaL EI dx 2 L 6 1 a 1 EI y = − P x3 + PaLx 6 L 6 © 2002 The McGraw-Hill Companies, Inc. All rights reserved.

2 dy PaL ⎡ ⎛ x⎞ ⎤ = ⎢1 − 3⎜ ⎟ ⎥ dx 6 EI ⎢⎣ ⎝ L ⎠ ⎥⎦

PaL2 ⎡ x ⎛ x ⎞ y= ⎢ −⎜ ⎟ 6 EI ⎢⎣ L ⎝ L ⎠

3⎤

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⎥ ⎥⎦

Third Edition

MECHANICS OF MATERIALS

Beer • Johnston • DeWolf

Sample Problem 9.1 • Locate point of zero slope or point of maximum deflection. 2 dy PaL ⎡ ⎛ xm ⎞ ⎤ =0= ⎢1 − 3⎜ ⎟ ⎥ dx 6 EI ⎣⎢ ⎝ L ⎠ ⎦⎥

PaL2 ⎡ x ⎛ x ⎞ y= ⎢ −⎜ ⎟ 6 EI ⎢⎣ L ⎝ L ⎠

3⎤

⎥ ⎥⎦

xm =

L = 0.577 L 3

• Evaluate corresponding maximum deflection.

[

PaL2 ymax = 0.577 − (0.577 )3 6 EI

]

PaL2 ymax = 0.0642 6 EI ymax

( 50 kips )(48 in )(180 in )2 = 0.0642

(

)(

6 29 × 106 psi 723 in 4

)

ymax = 0.238 in © 2002 The McGraw-Hill Companies, Inc. All rights reserved.

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Sample Problem 9.3 SOLUTION: • Develop the differential equation for the elastic curve (will be functionally dependent on the reaction at A).

For the uniform beam, determine the reaction at A, derive the equation for the elastic curve, and determine the slope at A. (Note that the beam is statically indeterminate to the first degree)

• Integrate twice and apply boundary conditions to solve for reaction at A and to obtain the elastic curve. • Evaluate the slope at A.

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Sample Problem 9.3 • Consider moment acting at section D,

∑MD = 0 1 ⎛⎜ w0 x 2 ⎞⎟ x −M =0 RA x − 2 ⎜⎝ L ⎟⎠ 3 w0 x3 M = RA x − 6L

• The differential equation for the elastic curve, d2y

w0 x3 EI 2 = M = R A x − 6L dx

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Sample Problem 9.3 • Integrate twice 4 1 dy 2 w0 x = EIθ = R A x − + C1 EI 2 24 L dx 5 1 3 w0 x + C1x + C2 EI y = R A x − 6 120 L

EI

2

d y

w0 x = = − M R x A 6L dx 2

3

• Apply boundary conditions: at x = 0, y = 0 : C2 = 0 3 1 2 w0 L RA L − + C1 = 0 at x = L, θ = 0 : 2 24 4 1 3 w0 L RA L − + C1L + C2 = 0 at x = L, y = 0 : 6 120

• Solve for reaction at A 1 1 R A L3 − w0 L4 = 0 3 30 © 2002 The McGraw-Hill Companies, Inc. All rights reserved.

RA =

1 w0 L ↑ 10 9 - 15

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MECHANICS OF MATERIALS

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Sample Problem 9.3 • Substitute for C1, C2, and RA in the elastic curve equation, 5 1⎛ 1 ⎞ 3 w0 x ⎛ 1 ⎞ EI y = ⎜ w0 L ⎟ x − w0 L3 ⎟ x −⎜ 6 ⎝ 10 120 L ⎝ 120 ⎠ ⎠

y=

(

w0 − x5 + 2 L2 x3 − L4 x 120 EIL

• Differentiate once to find the slope, θ=

at x = 0,

© 2002 The McGraw-Hill Companies, Inc. All rights reserved.

(

dy w0 = − 5 x 4 + 6 L2 x 2 − L4 dx 120 EIL

)

w0 L3 θA = 120 EI

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)

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MECHANICS OF MATERIALS

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Method of Superposition

Principle of Superposition: • Deformations of beams subjected to combinations of loadings may be obtained as the linear combination of the deformations from the individual loadings

© 2002 The McGraw-Hill Companies, Inc. All rights reserved.

• Procedure is facilitated by tables of solutions for common types of loadings and supports.

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Sample Problem 9.7 For the beam and loading shown, determine the slope and deflection at point B.

SOLUTION: Superpose the deformations due to Loading I and Loading II as shown.

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Sample Problem 9.7 Loading I wL3 (θ B )I = − 6 EI

wL4 ( yB )I = − 8 EI

Loading II wL3 (θC )II = 48 EI

wL4 ( yC )II = 128 EI

In beam segment CB, the bending moment is zero and the elastic curve is a straight line. wL3 (θ B )II = (θC )II = 48 EI

wL4 wL3 ⎛ L ⎞ 7 wL4 ( yB )II = + ⎜ ⎟= 128 EI 48 EI ⎝ 2 ⎠ 384 EI © 2002 The McGraw-Hill Companies, Inc. All rights reserved.

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Third Edition

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Beer • Johnston • DeWolf

Sample Problem 9.7

Combine the two solutions, wL3 wL3 θ B = (θ B )I + (θ B )II = − + 6 EI 48 EI

7 wL3 θB = 48 EI

wL4 7 wL4 + y B = ( y B )I + ( y B )II = − 8 EI 384 EI

41wL4 yB = 384 EI

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Application of Superposition to Statically Indeterminate Beams

• Method of superposition may be • Determine the beam deformation applied to determine the reactions at without the redundant support. the supports of statically indeterminate • Treat the redundant reaction as an beams. unknown load which, together with • Designate one of the reactions as the other loads, must produce redundant and eliminate or modify deformations compatible with the the support. original supports.

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Third Edition

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Beer • Johnston • DeWolf

Sample Problem 9.8 For the uniform beam and loading shown, determine the reaction at each support and the slope at end A.

SOLUTION: • Release the “redundant” support at B, and find deformation. • Apply reaction at B as an unknown load to force zero displacement at B.

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Third Edition

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Sample Problem 9.8 • Distributed Loading: 4 3 ⎤ w ⎡⎛ 2 ⎞ ⎛2 ⎞ 3⎛ 2 ⎞ ( yB )w = − ⎢⎜ L ⎟ − 2 L⎜ L ⎟ + L ⎜ L ⎟⎥ 24 EI ⎢⎣⎝ 3 ⎠ ⎝3 ⎠ ⎝ 3 ⎠⎥⎦

wL4 = −0.01132 EI

• Redundant Reaction Loading: 2

2

RB L3 RB ⎛ 2 ⎞ ⎛ L ⎞ ( yB )R = ⎜ L ⎟ ⎜ ⎟ = 0.01646 EI 3EIL ⎝ 3 ⎠ ⎝ 3 ⎠

• For compatibility with original supports, yB = 0 wL4 RB L3 0 = ( y B )w + ( y B )R = −0.01132 + 0.01646 EI EI RB = 0.688wL ↑

• From statics, R A = 0.271wL ↑ © 2002 The McGraw-Hill Companies, Inc. All rights reserved.

RC = 0.0413wL ↑ 9 - 23

Third Edition

MECHANICS OF MATERIALS

Beer • Johnston • DeWolf

Sample Problem 9.8

Slope at end A, wL3 wL3 (θ A )w = − = −0.04167 24 EI EI 2 wL3 0.0688wL ⎛ L ⎞ ⎡ 2 ⎛ L ⎞ ⎤ (θ A )R = ⎜ ⎟ ⎢ L − ⎜ ⎟ ⎥ = 0.03398 EI 6 EIL ⎝ 3 ⎠ ⎣⎢ ⎝ 3 ⎠ ⎦⎥

wL3 wL3 θ A = (θ A )w + (θ A )R = −0.04167 + 0.03398 EI EI © 2002 The McGraw-Hill Companies, Inc. All rights reserved.

wL3 θ A = −0.00769 EI 9 - 24

Third Edition

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Moment-Area Theorems • Geometric properties of the elastic curve can be used to determine deflection and slope. • Consider a beam subjected to arbitrary loading,

• First Moment-Area Theorem: area under (M/EI) diagram between C and D. © 2002 The McGraw-Hill Companies, Inc. All rights reserved.

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Moment-Area Theorems • Tangents to the elastic curve at P and P’ intercept a segment of length dt on the vertical through C.

= tangential deviation of C with respect to D

• 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 (M/EI) diagram between C and D.

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Third Edition

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Application to Cantilever Beams and Beams With Symmetric Loadings • Cantilever beam - Select tangent at A as the reference.

• Simply supported, symmetrically loaded beam - select tangent at C as the reference.

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Bending Moment Diagrams by Parts • Determination of the change of slope and the tangential deviation is simplified if the effect of each load is evaluated separately. • Construct a separate (M/EI) diagram for each load. - The change of slope, θD/C, is obtained by adding the areas under the diagrams. - The tangential deviation, tD/C is obtained by adding the first moments of the areas with respect to a vertical axis through D. • Bending moment diagram constructed from individual loads is said to be drawn by parts.

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Sample Problem 9.11 SOLUTION: • Determine the reactions at supports. • Construct shear, bending moment and (M/EI) diagrams. For the prismatic beam shown, determine • Taking the tangent at C as the the slope and deflection at E. reference, evaluate the slope and tangential deviations at E.

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Sample Problem 9.11 SOLUTION: • Determine the reactions at supports. RB = RD = wa

• Construct shear, bending moment and (M/EI) diagrams. wa 2 ⎛ L ⎞ wa 2 L A1 = − ⎜ ⎟=− 2 EI ⎝ 2 ⎠ 4 EI 1 ⎛⎜ wa 2 ⎞⎟ wa 3 (a ) = − A2 = − 3 ⎜⎝ 2 EI ⎟⎠ 6 EI

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Sample Problem 9.11 • Slope at E: θ E = θC + θ E

C

= θE C

wa 2 L wa 3 = A1 + A2 = − − 4 EI 6 EI wa 2 (3L + 2a ) θE = − 12 EI

• Deflection at E: yE = tE C − tD C L⎞ ⎡ ⎛ ⎛ 3a ⎞⎤ ⎡ ⎛ L ⎞⎤ = ⎢ A1⎜ a + ⎟ + A2 ⎜ ⎟⎥ − ⎢ A1⎜ ⎟⎥ 4⎠ ⎝ 4 ⎠⎦ ⎣ ⎝ 4 ⎠⎦ ⎣ ⎝ ⎡ wa 3 L wa 2 L2 wa 4 ⎤ ⎡ wa 2 L2 ⎤ = ⎢− − − ⎥ ⎥ − ⎢− 4 EI 16 EI 8 EI 16 EI ⎥⎦ ⎢⎣ ⎥⎦ ⎢⎣ wa 3 (2 L + a ) yE = − 8 EI © 2002 The McGraw-Hill Companies, Inc. All rights reserved.

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Application of Moment-Area Theorems to Beams With Unsymmetric Loadings • Define reference tangent at support A. Evaluate θA by determining the tangential deviation at B with respect to A.

• The slope at other points is found with respect to reference tangent. θD = θ A +θD

A

• The deflection at D is found from the tangential deviation at D.

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Maximum Deflection • Maximum deflection occurs at point K where the tangent is horizontal.

• Point K may be determined by measuring an area under the (M/EI) diagram equal to -θA . • Obtain ymax by computing the first moment with respect to the vertical axis through A of the area between A and K.

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Third Edition

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Use of Moment-Area Theorems With Statically Indeterminate Beams • Reactions at supports of statically indeterminate beams are found by designating a redundant constraint and treating it as an unknown load which satisfies a displacement compatibility requirement. • The (M/EI) diagram is drawn by parts. The resulting tangential deviations are superposed and related by the compatibility requirement. • With reactions determined, the slope and deflection are found from the moment-area method.

© 2002 The McGraw-Hill Companies, Inc. All rights reserved.

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