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 10.5.5P

Solve the preceding problem by integrating the differential equation of the deflection curve.

  Chapter 10, Problem 10.5.5P, Solve the preceding problem by integrating the differential equation of the deflection curve.

(a)

Expert Solution
Check Mark
To determine

The reactions for beam at all supports.

Answer to Problem 10.5.5P

The reaction at support A is 3α(T1T2)L2EIK2h(36EI+L3K)(upward) .

The reaction at support B is α(T1T2)L26EIKh(36EI+L3K)(downward) .

The reaction at support C is 9α(T1T2)L2EIK2h(36EI+L3K)(upward)

Explanation of Solution

Given information:

The two spans of beam are L and L3 , flexural rigidity is EI , stiffness of spring is K .

Write the Expression for force equilibrium in vertical direction.

  RA+RB+RC=0   .........(I)

Here, reaction produced at A is RA , reaction produced at B is RB and reaction produced at C is RC .

Write the expression for moment about point C .

  MC=0RA=14RB   .........(II)

Here, MC is moment about point C .

Write the expression for moment about point A .

  MA=0RC=34RB   .........(III)

Here, MA is moment about point A .

Write the expression for double order differential equation for the deflection curve when any value between 0 to L is substituted for x .

  EIV=RAx+EIα(T2T1)h   .........(IV)

Here, double order differential of deflection curve is V , depth of the beam is h , coefficient of thermal expansion is α , flexural rigidity is EI , upper surface temperature is T1 and lower surface is T2 .

Write the expression for single order derivative of deflection curve when any value between 0 to L is substituted for x .

  EIV=12RAx2+EIα(T2T1)xh+C1   .........(V)

Here, length at which deflection has to be calculate is x , single order differential of deflection curve is V and C1 is integral constant.

Write the expression for deflection curve when any value between 0 to L is substituted for x .

  EIV=16RAx3+EIα(T2T1)x22h+C1x+C2   .........(VI)

Here, deflection at a point x is V and C2 is integral constant.

Write the expression for first boundary condition.

  V(0)=0   .........(VII)

Here, deflection when 0 is substituted for x in deflection curve is V(0) .

Write the expression for second boundary condition when L is substituted for x in Equation (V).

  (V)left=(V)right   .........(VIII)

Here, single order differential of deflection curve from the loads left to spring is (V)left and single order differential of deflection curve from the loads right to spring is (V)right .

Write the expression for third boundary condition when L is substituted for x in Equation (VI).

  (V)left=(V)right   .........(IX)

Write the expression for fourth boundary condition.

  V(43L)=0   .........(X)

Here, deflection when 43L is substituted for x in deflection curve is V(43L) .

Write the expression for differential equation for the deflection curve when any value between L to 43L is substituted for x .

  EIV=RAx+EIα(T2T1)h4RA(xL)   .........(XI)

Write the expression for single order derivative of deflection curve when any value between L to 43L is substituted for x .

  EIV=32RAx2+EIα(T2T1)hx+4RALx+C3   .........(XII)

Write the expression for deflection curve when any value between L to 43L is substituted for x .

  EIV=12RAx3+EIα(T2T1)2hx2+2RALx2+C3x+C4   .........(XIII)

Write the expression for compatibility Equation for spring when L is substituted for x in deflection curve.

  (V)left=(V)right = RBK   .........(XIV)

Substituted 0 for x and 0 for V in Equation (VI).

  EI0=16RA×0+EIα( T 2 T 1 )×02h+C1×0+C2C2=0   .........(XV)

Substitute L for x in Equation (V).

  EIV=12RAL2+EIα(T2T1)Lh+C1

Substitute L for x in Equation (XII).

  EIV=32RAL2+EIα(T2T1)hL+4RAL×L+C3   .........(XIV)

Substitute 12RAL2+EIα(T2T1)Lh+C1 for EIV in Equation (XIV).

  [ 1 2 R A L 2+ EIα( T 2 T 1 )L h+ C 1]=[ 3 2 R A L 2+ EIα( T 2 T 1 ) hL+4 R AL×L+ C 3]C1=2RAL2+C3   .........(XV)

Substitute L for x and 0 for C2 in Equation (VI).

  EIV=16RAL3+EIα(T2T1)L22h+C1L   .........(XVI)

Substitute L for x in Equation (XIII).

  EIV=12RAL3+EIα(T2T1)2hL2+2RAL×L2+C3L+C4   .........(XVII)

Substitute 12RAL3+EIα(T2T1)2hL2+2RAL×L2+C3L+C4 for EIV and 2RAL2+C3 for C1 in Equation (XVI).

  12RAL3+EIα( T 2 T 1 )2hL2+2RAL×L2+C3L+C4=16RAL3+EIα( T 2 T 1 )L22h+2RAL2+C3L

  C4=(2+1632)RAL3=23RAL3   .........(XVIII)

Substitute 43L for x , 0 for EIV and 23RAL3 for C4 in Equation (XIII).

  0=12RA(43L)3+EIα(T2T1)2h(43L)2+2RAL(43L)2+C343L+23RAL3

  0=6454RAL3+8EIα(T2T1)L29h+329RAL3+C3(43L)+23RAL3

  C3=[8227RAL3+89EIα( T 2 T 1)L2h][34L]   .........(XIX)

Substitute [8227RAL3+89EIα( T 2 T 1)L2h][34L] for C3 in Equation (XV).

  C1=2RAL2[8227RAL3+89EIα( T 2 T 1)L2h][34L]   .........(XX)

Substitute 2RAL2[8227RAL3+89EIα( T 2 T 1)L2h][34L] for C1 and 0 for C2 and L for x in Equation (VI).

  EIV=[16RAL3+EIα( T 2 T 1 ) L 22h+[2RAL2[ 82 27 R A L 3 + 8 9 EIα( T 2 T 1 ) L 2 h ][ 3 4L]]L]   .........(XXI)

Substitute [8227RAL3+89EIα( T 2 T 1)L2h][34L] for C3 , 23RAL3 for C4 and L for x in Equation (XVII).

  EIV=12RAL3+EIα( T 2 T 1 )2hL2+2RAL×L2+8227RAL3+89EIα( T 2 T 1 )L2h[34L]L+23RAL3   .........(XXII)

Substitute [16RAL3+EIα( T 2 T 1)L22h+[2RAL2[8227RAL3+89EIα( T 2 T 1 ) L 2h][34L]]L]1EI for (V)left in Equation (XIV).

  [16RAL3+EIα( T 2 T 1 ) L 22h+[2RAL2[ 82 27 R A L 3+ 8 9 EIα( T 2 T 1 ) L 2 h][ 3 4L]]L]1EI=RBK   .........(XXIII)

Substitute [12RAL3+EIα( T 2 T 1)2hL2+2RAL×L2+[8227RAL3+89EIα( T 2 T 1 )L2h][34L]L+23RAL3]1EI for (V)right in Equation (XIV).

  12RAL3+EIα( T 2 T 1 )2hL2+2RAL×L2+[[ 82 27RAL3+89 EIα( T 2 T 1 ) L 2 h][3 4L]L]+23RAL3=RBK   .........(XXIV)

Solve Equation (XXIII) and Equation (XXIV).

  RB=α(T1T2)L26EIKh(36EI+L3K)(downward)   .........(XXV)

  RA=3α(T1T2)L2EIK2h(36EI+L3K)(upward)   .........(XXVI)

Substitute 3α(T1T2)L2EIK2h(36EI+L3K)(upward) for RA and α(T1T2)L26EIKh(36EI+L3K)(downward) for RB considering sense of direction in Equation (I).

  3α( T 1 T 2 )L2EIK2h( 36EI+ L 3 K)α( T 1 T 2 )L26EIKh( 36EI+ L 3 K)+RC=0RC=9α( T 1 T 2 )L2EIK2h( 36EI+ L 3 K)(upward)   .........(XXVII)

Conclusion:

The reaction at support A is 3α(T1T2)L2EIK2h(36EI+L3K)(upward) .

The reaction at support B is α(T1T2)L26EIKh(36EI+L3K)(downward) .

The reaction at support C is 9α(T1T2)L2EIK2h(36EI+L3K)(upward) .

(b)

Expert Solution
Check Mark
To determine

The reactions at all the supports when value of K limits to .

Answer to Problem 10.5.5P

The reaction at support A is 3EIα(T1T2)2Lh(upward) .

The reaction at support B is 6EIα(T1T2)Lh(downward) .

The reaction at support C is 9EIα(T1T2)2Lh(upward) .

Explanation of Solution

Take limit of K upto in Equation (XXVI).

  RA=limK3α( T 1 T 2 )L2EIK2h( 36EI+ L 3 K)=3EIα( T 1 T 2 )2Lh(upward) ...... (XXVIII)

Take limit of K upto in Equation (XXV).

  RB=limKα( T 1 T 2 )L26EIKh( 36EI+ L 3 K)=6EIα( T 1 T 2 )Lh(downward) ...... (XXIX)

Take limit of K upto in Equation (XXVII).

  RC=limK9α( T 1 T 2 )L2EIK2h( 36EI+ L 3 K)=9EIα( T 1 T 2 )2Lh(upward) ...... (XXX)

Conclusion:

The reaction at support A is 3EIα(T1T2)2Lh(upward) .

The reaction ay support B is 6EIα(T1T2)Lh(downward) .

The reaction at support C is 9EIα(T1T2)2Lh(upward) .

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Chapter 10 Solutions

Mechanics of Materials (MindTap Course List)

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