Mechanics of Materials (MindTap Course List)
Mechanics of Materials (MindTap Course List)
9th Edition
ISBN: 9781337093347
Author: Barry J. Goodno, James M. Gere
Publisher: Cengage Learning
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Chapter 9, Problem 9.7.6P

A simple beam ABC has a moment of inertia 1,5 from A to B and A from B to C (see figure). A concentrated load P acts at point B.

Obtain the equations of the deflection curves for both parts of the beam. From the equations, determine the angles of rotation 0Aand Bcat the supports and the deflection 6Bat point B.

  Chapter 9, Problem 9.7.6P, A simple beam ABC has a moment of inertia 1,5 from A to B and A from B to C (see figure). A

Expert Solution & Answer
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To determine

The equations of the deflection curve for both parts of the beam, the angle of rotation θA and θC at the supports and the deflection δB at the point for given simple beam.

Answer to Problem 9.7.6P

The equations of the deflection curve for both parts of the beam, the angle of rotation θA and θC at the supports and the deflection δB at the point are as below for given simple beam.

  1.   v=2Px729EI(19L227x2) When 0xL3
  2.   v=P1458(13L3+175PL2x243Lx2+81x2)  When L3xL
  3.   θA=38PL2729EI
  4.   θC=34PL2729EI
  5.   δB=32PL32187EI

Explanation of Solution

Given Information:

We have the simple beam ABC with moment of inertia 1.5I from point A to B and I from point B to C as per shown figure with a concentrated load P acting at point B:

  Mechanics of Materials (MindTap Course List), Chapter 9, Problem 9.7.6P , additional homework tip  1

The length of beam, L

Moment of inertia between A to B, 1.5I

Moment of inertia between B to C, I

Concentrated Load P =B AB=L3and BC=2L3

For the given simple beam, we have deflective curve as shown below:

  Mechanics of Materials (MindTap Course List), Chapter 9, Problem 9.7.6P , additional homework tip  2

Now, applying differential equation,

  E(I AB)d2vdx2=ME(1.5I)d2vdx2=2Px3  0xL3EId2vdx2=4Px9  0xL3.....(1)EId2vdx2=PL3Px3  L3xL.....(2)

Now doing integration of equation (1),

  EId2vdx2=4Px9  0xL3EIdvdx=4Px218+C1......(3)

Performing integration for equation (2),

  EId2vdx2=PL3Px3  EIdvdx=PL3xPx26+C2........(4) 

We are applying boundary conditions at x=L3

  ( dv d x B )left=( dv d x B )right4Px218+C1=PL3xPx22+C24P18( L 3)2+C1=PL3(L3)P ( L 3 )22+C2C2=C1+4PL2162PL29+PL218C2=C111PL2162......(5)

For equation (3), taking integration,

  EIv=4Px354+C1x+C3 0xL3

For equation (4), taking integration,

  EIdvdx=PL3xPx26+C2EIv=PL6x4Px318+C2x+C4  L3xL

In the above equations, applying boundary conditions at

  x =0, v=0 and C3=0x =L, v=0 

So, we will get,

  0=PL36-PL318+C2L+C4 C4=PL39C2L........(6)

For boundary conditions at x=L3 ,

  ( v B)left=( v B)right4P54+C1(L3)=PL6( L 3)2P ( L 3 )318+C2(L3)+C44Px354+C1(L3)=PL354PL3486+C2(L3)+C4C1L=10PL3243+C2L+3C4.........(7)

With the equation (6) and (7), we will get

  C1L=10PL3243+C2L+3C4C1L=10PL3243+C2LPL333C2LC1L=71PL3243-2C2LC1=71PL3243-2C2........(8)

From equation (5) and (8), we will get

  C1=71PL3243-2C2+11PL2813C1=71PL2+33PL22433C1=38PL2243C1=38PL2729

So,

  C1=71PL2243-2C238PL2729=71PL2243-2C22C2=71PL2243+38PL27292C2=213PL2+PL2729C2=175PL21458C3=0C4=13PL31458

So,

  dvdx=1EI[4P x 218+C1]dvdx=1EI[4P x 21838P L 2729]dvdx=1EI[324P x 276P L 21458]dvdx=4P1458EI[81Px2+19L2]dvdx=2P729EI[19L281Px2] 0xL3

When,

   0xL3dvdx=1EI[PLx3P x 26+C2]dvdx=1EI[PLx3P x 26175P L 21458]dvdx=1EI[486PLx243P x 2175P L 21458]dvdx=P1458EI[175L2486Lx+243x2]

Angle of rotation at point A,

  θA=( dv dx)x=0θA=2P729EI(19L2)θA=38PL2729EI

Angle of rotation at point C,

  θC=+( dv dx)x=LθC=P1458EI(175L2486L2+243L2)θC=68PL21458EIθC=34PL2729EI

Deflection of the beam is as follows:

  v=2Px729EI(19L227x2) When 0xL3When L3xLv=1EI( PL6x2 P x 3 18 +C2x+C4)  v=1EI( PL6x2 P x 3 18+ 175P L 2 1458x+ 13P L 3 1458)  v=P1458(13L3+175PL2x243Lx2+81x2)  When L3xL

Deflection at point B,

  δB=(v)x=L3δB=2P( L 3 )729EI[19L227 L 29]δB=2PL3( 729EI)[19L23L2]δB=2PL3( 729EI)[16L2]δB=32PL32187EI

Conclusion:

The answers are calculated by integration and boundary conditions.

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Chapter 9 Solutions

Mechanics of Materials (MindTap Course List)

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