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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.
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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.
- v=−2Px729EI(19L2−27x2) When 0≤x≤L3
- v=−P1458(−13L3+175PL2x−243Lx2+81x2) When L3≤x≤L
- θA=38PL2729EI
- θC=34PL2729EI
- 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:
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:
Now, applying differential equation,
E(IAB)d2vdx2=ME(1.5I)d2vdx2=2Px3 0≤x≤L3EId2vdx2=4Px9 0≤x≤L3.....(1)EId2vdx2=PL3−Px3 L3≤x≤L.....(2)
Now doing integration of equation (1),
EId2vdx2=4Px9 0≤x≤L3EIdvdx=4Px218+C1......(3)
Performing integration for equation (2),
EId2vdx2=PL3−Px3 EIdvdx=PL3x−Px26+C2........(4)
We are applying boundary conditions at x=L3
(dvdxB)left=(dvdxB)right4Px218+C1=PL3x−Px22+C24P18(L3)2+C1=PL3(L3)−P(L3)22+C2C2=C1+4PL2162−PL29+PL218C2=C1−11PL2162......(5)
For equation (3), taking integration,
EIv=4Px354+C1x+C3 0≤x≤L3
For equation (4), taking integration,
EIdvdx=PL3x−Px26+C2EIv=PL6x4−Px318+C2x+C4 L3≤x≤L
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=−PL39−C2L........(6)
For boundary conditions at x=L3 ,
(vB)left=(vB)right4P54+C1(L3)=PL6(L3)2−P(L3)318+C2(L3)+C44Px354+C1(L3)=PL354−PL3486+C2(L3)+C4C1L=10PL3243+C2L+3C4.........(7)
With the equation (6) and (7), we will get
C1L=10PL3243+C2L+3C4C1L=10PL3243+C2L−PL33−3C2LC1L=−71PL3243-2C2LC1=−71PL3243-2C2........(8)
From equation (5) and (8), we will get
C1=−71PL3243-2C2+11PL2813C1=−71PL2+33PL22433C1=−38PL2243C1=−38PL2729
So,
C1=−71PL2243-2C2−38PL2729=−71PL2243-2C22C2=−71PL2243+38PL27292C2=−213PL2+PL2729C2=−175PL21458C3=0C4=13PL31458
So,
dvdx=1EI[4Px218+C1]dvdx=1EI[4Px218−38PL2729]dvdx=1EI[324Px2−76PL21458]dvdx=−4P1458EI[−81Px2+19L2]dvdx=−2P729EI[19L2−81Px2] 0≤x≤L3
When,
0≤x≤L3dvdx=1EI[PLx3−Px26+C2]dvdx=1EI[PLx3−Px26−175PL21458]dvdx=1EI[486PLx−243Px2−175PL21458]dvdx=−P1458EI[175L2−486Lx+243x2]
Angle of rotation at point A,
θA=−(dvdx)x=0θA=2P729EI(19L2)θA=38PL2729EI
Angle of rotation at point C,
θC=+(dvdx)x=LθC=P1458EI(175L2−486L2+243L2)θC=68PL21458EIθC=34PL2729EI
Deflection of the beam is as follows:
v=−2Px729EI(19L2−27x2) When 0≤x≤L3When L3≤x≤Lv=1EI(PL6x2−Px318+C2x+C4) v=1EI(PL6x2−Px318+−175PL21458x+13PL31458) v=−P1458(−13L3+175PL2x−243Lx2+81x2) When L3≤x≤L
Deflection at point B,
δB=(v)x=L3δB=2P(L3)729EI[19L2−27L29]δB=2PL3(729EI)[19L2−3L2]δB=2PL3(729EI)[16L2]δB=32PL32187EI
Conclusion:
The answers are calculated by integration and boundary conditions.
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