Bundle: Mechanics Of Materials, Loose-leaf Version, 9th + Mindtap Engineering, 1 Term (6 Months) Printed Access Card
Bundle: Mechanics Of Materials, Loose-leaf Version, 9th + Mindtap Engineering, 1 Term (6 Months) Printed Access Card
9th Edition
ISBN: 9781337594318
Author: Barry J. Goodno; James M. Gere
Publisher: Cengage Learning
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Chapter 9, Problem 9.7.5P

A beam ABC has a rigid segment from A to B and a flexible segment with moment of inertia / from B to C(see figure). A concentrated load P acts at point B.

Determine the angle of rotation SAof the rigidsegment, the deflection 8Bat point ß, and the maximum deflection 8.

  Chapter 9, Problem 9.7.5P, A beam ABC has a rigid segment from A to B and a flexible segment with moment of inertia / from B to

Expert Solution & Answer
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To determine

The angle of rotation θA of the rigid segment, the deflection δB at point B and the maximum deflection δmax should be determined for given beam.

Answer to Problem 9.7.5P

The angle of rotation θA of the rigid segment, the deflection δB at point B and the maximum deflection δmax are   θA=8PL2243EI , δB=8PL3729EI and δmax=0.01363PL3EI for given beam.

Explanation of Solution

Given Information:

We have the beam ABC with rigid segment from point A to B and flexible segment with moment of inertia I from point B to C. A concentrated load P acts at Point B as shown in below figure.

  Bundle: Mechanics Of Materials, Loose-leaf Version, 9th + Mindtap Engineering, 1 Term (6 Months) Printed Access Card, Chapter 9, Problem 9.7.5P , additional homework tip  1

We have,

Length of the beam as L

Moment of inertia as BC=I

Concentrated load at Point B = P

  AB=L3 and BC=2L3

The shear force acting over beam can be shown below diagram.

  Bundle: Mechanics Of Materials, Loose-leaf Version, 9th + Mindtap Engineering, 1 Term (6 Months) Printed Access Card, Chapter 9, Problem 9.7.5P , additional homework tip  2

The shear force working from point A to B is v=3δBxL  0xL3

We are taking derivative on both sides, we will get

  v=3δBxLdvdx=3δBL

Now for shear force working from point B to C is below.

  EId2vdx2=PL3Px3 Taking integration ,EIdvdx=PL3xPx26+C1Again taking integration on both sides,EIv=PLx26Px218+C1x+C2

According to boundary condition at x=L3 ,

  dvdx=3δBLSo,EI(3 δ BL)=PL( L 3)3PL2/96+C1(3EI δ BL)=PL29PL254+C1C1=PL254PL293(EI δ BL)C1=5PL2543(EI δ BL)So,EIdvdx=PL3xPx265PL2543EIδBL

In above equation taking integration again,

  EIdvdx=PLx26Px3185PL254x3EIδBL+C2

And at 2nd Boundary conditions, x = L and v = 0.

  0=PL36PL3185PL2543EIδB+C20=9PL33PL35PL3543EIδB+C2C2=PL254+3EIδBSo,EIdvdx=PLx26Px3185PL254x3EIδBLPL254+3EIδB

And at 3rd Boundary conditions, x=L3

  (vB)left=(vB)right3δBxL=1EI(PL x 26P x 3185P L 254x3EI δ BLP L 254+3EIδB)3δBxL=PL x 26P x 3185P L 254xP L 254EI3δBxL++3δB3δB=1EI(PL ( L 3 ) 26+P ( L 3 ) 318+5P L 254( L 3)+P L 254)3δB=1EI(P L 354+P L 3468+5P L 3162+P L 354)3δB=1EI(P L 3+15P L 3486)3δB=16PL3486EIδB=16PL31458EIδB=8PL3729EI

  EIv=P486(7L361L2x+81Lx227x3)v=P486EI(7L361L2x+81Lx227x3)dvdx=P486EI(61L2+162Lx+81x2)

For maximum deflection,

  dvdx=061L2+162Lx81x2=081x2+162LX+61L2=0x=162L± ( 162L ) 24( 81)( 61 L 2 )2(81)x=162L±( 6480 L 2 )162x=162L±365L162x=L9(925)x=0.5030962

So we will calculate maximum deflection as below,

  v=P486EI(C361L2(0.503096L)+81L( 0.503096L)227( 0.503096L)3)v=0.01363PL3EIδB=8PL3729EI

The angle of rotation,

  A1θA=δBL3        =8P L 3729EIL3        =24PL2729EI  θA=8PL2243EI

Deflection,

  EIv=PLx26Px3185PL254x3EIδBxLPL254+3EIδBEIv=PLx26Px3185PL254x3EI( 8P L 3 729EI)xLPL254+3EI(8P L 3729EI)EIv=PLx26Px3185PL254x8PL2243xPL254+PL3243

But the maximum deflection would be,

  δmax=vmaxδmax=0.01363PL3EI

Conclusion:

The angle of rotation θA of the rigid segment, the deflection δB at point B and the maximum deflection are calculated by diagram and integration equation.

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Chapter 9 Solutions

Bundle: Mechanics Of Materials, Loose-leaf Version, 9th + Mindtap Engineering, 1 Term (6 Months) Printed Access Card

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