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 10, Problem 10.4.21P

Uniform load q = 10 lb/ft acts over part of the span of fixed-end beam AB (see figure). Upward load P = 250 lb is applied 9 ft to the right of joint A. Find the reactions at A and B.

  Chapter 10, Problem 10.4.21P, Uniform load q = 10 lb/ft acts over part of the span of fixed-end beam AB (see figure). Upward load

Expert Solution & Answer
Check Mark
To determine

The reaction at fixed joint A and Bpoint.

Answer to Problem 10.4.21P

The reaction at A is 35.68lbf .

The moment at Ais 261.6lbfft .

The reaction at B is 154.32lbf .

The moment at B is 506.4lbfft .

Explanation of Solution

Given information:

The length of the beam is 15ft, uniform load of 10lb/ft act over beam up to 6ft, and a upward load of 250lb is applied at 9ft from the fixed and A.

   Figure below shows the free body diagram of fixed beam.

  Mechanics of Materials (MindTap Course List), Chapter 10, Problem 10.4.21P

   Figure-(1)

Write the expression for the reactions forces at point A.

  RAp=(pbpL+MALMBL) …... (I)

Here, the point load is p, the length of the beam is L, the moment s acting at A and B is MAP and Mp respectively.

Write the expression for the reactions forces at point B .

  RBp=(papL+MApLMBpL)  .......(II)

Write the expression for the compatibility Equation at B .

  θA=(θA)1(θA)2(θA)3=0  .......(III)

Here, the angles at the corresponding ends are represented by the subscript for θ .

Write the expression for the compatibility Equation at B .

  θB=(θB)1(θB)2(θB)3=0  .......(IV)

Write the expression for the values of slopes of type 1 loading at point A.

  (θA)1=papbp(L+bp)6LEI

Write the expression for the value of slopes of type 1 loading at point B .

  (θB)1=papbp(L+ap)6LEI

Write the expression for the value of slopes for type 2 loading at point A .

  (θA)2=MAPL3EI

Write the expression for the values of slopes for type 2 loading at point B .

  (θB)2=MAPL6EI

Write the expression for the values of slopes for type 2 loading at point A(θA)3=MBPL6EI

Write the expression for the values of slopes for type 2 loading at point B.

  (θB)4=MBPL3EI

Substitute papbp(L+bp)6LEI for (θA)1

  MAPL3EI for

  (θA)2

  MBPL6EI for (θA)3 in Equation (III).

  papbp(L+ap)6LEI+MAPL3EI+MBPL6EI=0MAPL3EI+MBPL6EI=papbp(L+ap)6LEI….. (V)

Substitute papbp(L+ap)6LEI for (θB)1

  MbPL6EI for (θB)2

  MaPL3EI in Equation (IV).

  papbp(L+ap)6LEI+MAPL6EI+MBPL3EI=0MAPL6EI+MBPL3EI=papbp(L+ap)6LEI  .......(VI)

Substitute MAPL6EI+MBPL3EI for papbp(L+ap)6LEI in equation (V) to obtain the values of MAP .

  MAP=Papbp2L2

Substitute MAPL6EI+MBPL3EI for papbp(L+ap)6LEI in Equation (V).

  MBP=Papbp2L2

Substitute Pap2bpL2 for MA and Papbp2L2 for MB in Equation (I).

  RAP=pbpLPap2bpL2+Papbp2L2=Pbp2L3(L+2ap)

Substitute Papbp2L2 for MA and Pap2bpL2 for MB in Equation (II).

  RAP=papL+Papbp2L2Pap2bpL2=Pap2L3(L+2bp)

Write the expression for the fixed end moments due to the element of load Ap .

  dMAP=qx(Lx)2dxL2

Write the expression for the fixed end moments due to the element of load Bq .

  dMBq=qx2(Lx)dxL2

Here the uniform distributed load is q, the length of the beam is L, and distance of the section is from the left side is x .

Write the expression the moment at A.

  MAq=dMAq  .......(VII)

Substitute qx2(Lx)dxL2 for dMAq in Equation (VII).

  MAq=qx2( Lx)dxL2=qL20ax( Lx)2=qa212L2(6L28aL+3a2)

Write the expression for the moment at point B.

  MBq=dMBq  .......(VIII)

Substitute qx2(Lx)dxL2 for dMB in Equation (VIII).

  MBq=qx2( Lx)dxL2=qL20ax2(Lx)dx=qa312L2(4L3a)

Write the expression for the reaction force in term of elemental load Aq .

  dRAq=q(Lx)2(L+2x)dxL3

Write the expression for the reaction force in term of elemental load Bp .

  dRBq=qx2(3L2x)dxL3

Write the expression for the reaction at point A in term of element load.

  RAq=dRAq….. (IX)

Substitute q(Lx)2(L+2x)dxL3 for dRAq in Equation (IX)

  dRAq=q ( Lx )2( L+2x)dxL3=qL30a(Lx)2(L+2x)dx=qa2L3(2L32La2+a3)

Write the expression for the reaction at point B in term of elemental load.

  RBq=dRBq  .......(X)

Substitute qx2(3L2x)dxL3 for dRBq in Equation (IX).

  RBq=qx2( 3L2x)dxL3=qL30ax2(3L2x)dx=qa3(2La)2L3

Write the expression for the reaction at point A.

  RA=RAP+RAq  .......(XI)

Substitute Pbp2L3(L+2ap) for RAP and qa2L3(2L32La2+a3) for RAq in Equation (XI).

  RA=Pbp2L3(L+2ap)+qa2L3(2L32La2+a3)  .......(XII)

Write the expression for the reactions at point B.

  RB=Pap2L3(L+2bp)+qa32L3(2La)  .......(XIII)

Write the expression for the moment at point A.

  MA=MAp+MAq  .......(XIV)

Substitute Papbp2L2 for MAp and qa212L2(6L28aL+3a2) for MAq in equation in Equation (XIII).

  MA=Papbp2L2+qa212L2(6L28aL+3a2)….. (XV)

Write the expression for the reaction moment at point B.

  MB=MBp+MBq  .......(XVI)

Substitute Papbp2L2 for MBp and qa312L2(4L3a) for MBq in Equation (XVI)

  MB=Papbp2L2+qa312L2(4L3a)……. (XVII)

Calculation:

Substitute 250lbf for p , 6ft for bp , 9ft for ap , 15ft for L , 10lbf/ft for q and 6ft for a in Equation (XII).

  RA=250lbf×(6ft)2(15ft)3(15ft+2(9ft))+{10lbf/ft×6ft2×( 15ft)3(2( 15ft)32( 15ft)( 6ft)2+( 6ft)3)}=2.667lbf/ft×33ft+8.88×10-3(lbf/ft3)×5886(ft3)=88lbf+52.32lbf=35.68lbf

Substitute 250lbf for p, 6ft for bp , 9ft for ap , 15ft for L , 10lbf/ft for q and 6ft for a in Equation (XIII).

  RB=[250lbf× ( 9ft )2 ( 15ft )3(15ft+2( 6ft))+{10 lbf/ ft× ( 6ft ) 32× ( 15ft ) 3(2( 15ft))(6ft)}]=(6lbf/ft)×(27ft)+(0.32lbf/ft)×(24ft)=162lbf+7.68lbf=154.32lbf

Substitute 250lbffor p , 6ft for bp

  9ft for ap , 15ft for L

  10lbf/ft for q ,and 6ft for a .

  MA=250lbf/ft×9ft×(6ft)2(15ft2)+[10lbf/ft×( 6ft)212×( 15ft)2(6ft( 15ft)28( 6ft)( 15ft)+3( 6ft)2)]=360lbfft+98.4lbfft=261.6lbfft

Substitute 250lbf for p , 6ft for bp

  9ft for ap , 15ft for L

  10lbf/ft for q and 6ft for a .

  MB=250lbf×(9ft)2×6ft(15ft)2+[10lbf/ft×( 6ft)312×( 15ft)2(4(15ft)3(6ft))]=540lbfft+33.6lbfft=506.4lbfft

Conclusion:

The reaction at point Ais35.68lbf .

The reaction at point B is 154.32lbf .

The moment at point Ais261.6lbfft .

The moment at point B is 506.4lbfft .

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Chapter 10 Solutions

Mechanics of Materials (MindTap Course List)

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