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 11, Problem 11.3.27P

A space truss is restrained at joints O, A,B, and C as shown in the figure. Load F is applied at joint A and load 2F acts downward at joint C. Each member is a slender, circular pipe (E = 10,600 ksi) with an outside diameter of 3.5 in. and wall thickness of 0.25 in. Length variable L = 11 ft. Determine the critical value of load variable P (kips) at which member OB fails by Euler buckling.

  Chapter 11, Problem 11.3.27P, A space truss is restrained at joints O, A,B, and C as shown in the figure. Load F is applied at

Expert Solution & Answer
Check Mark
To determine

The critical value of load P at which member OB fails by Euler buckling.

Answer to Problem 11.3.27P

The critical value of load P at which member OB fails by Euler buckling is 109415821.2kips_ .

Explanation of Solution

Given information:

Young’s modulus of each member is 10600ksi , outside diameter of each member is 3.5in , wall thickness of each member is 0.25in and length variable is 11ft .

Write the expression for force equilibrium at joint O in x -direction.

  Ox=FOB

Here, force generated in member OB is FOB and reaction force in x -direction at point O is Ox .

Write the expression for force equilibrium at joint O in y -direction.

  Oy=FOC

Here, force generated in member OC is FOC and reaction force in y -direction at point O is Oy .

Write the expression for force equilibrium at joint O in z -direction.

  OZ=0

Here, the reaction force in z direction at point O is Oz .

Write the expression for moment equilibrium equation at point B along y -axis.

  Oz×L+(Ax×0.8L)+PL=0   ......(I)

Here, the length variable of the truss is L , reaction force in x -direction at point A is Ax and load variable of the truss is P .

Substitute 0 for Oz in equation (I).

  0×L+(Ax×0.8L)+PL=0(Ax×0.8L)+PL=0Ax×0.8L=PLAx=PL0.8L

  Ax=1.25P

Write the expression for moment equilibrium equation at point C along z -axis.

  By×L=0By=0

Here, reaction force in y -direction at point B is By .

Write the expression for moment equilibrium equation at point O along y -axis.

  (Bz×L)+(0.8L×Ax)=0   ......(II)

Here, reaction force in z -direction at point B is Bz .

Substitute 1.25P for Ax in equation (II).

  (Bz×L)+(0.8L×( 1.25P))=0Bz×L=0.8L×1.25PBz=0.8×1.25PBz=P

The figure below shows the free body diagram of joint A .

  Mechanics of Materials (MindTap Course List), Chapter 11, Problem 11.3.27P

  Figure-(1)

Write the expression for force equilibrium at joint A in x -direction.

  Ax+FABcos(θABO)=0   ......(III)

Here, the force generated in member AB is FAB and the angle between member AB and member OB is θABO .

Write the expression for the angle between member AB and member OB .

  θABO=cos1(L L 2 + ( 0.8L ) 2 )

Substitute cos1(L L 2 + ( 0.8L ) 2 ) for θABO and 1.25P for Ax in equation (III).

  1.25P+FABcos( cos 1( L L 2 + ( 0.8L ) 2 ))=01.25P+FAB×L L 2 + ( 0.8L ) 2 =01.25P+FAB×L L 2 +0.64 L 2 =01.25P+FAB×L1.28L=0

  FAB=1.25P×1.28LLFAB=1.6P

Write the expression for force in member OB .

  FOB=FABcosθABO   ......(IV)

Substitute 1.6P for FAB and cos1(L L 2 + ( 0.8L ) 2 ) for θABO in equation (IV).

  FOB=1.6P×cos( cos 1( L L 2 + ( 0.8L ) 2 ))=1.6P×L L 2 + ( 0.8L ) 2 =1.6P×L1.28L=1.25P

Write the expression for moment of inertia for circular pipe.

  I=π64(d4di4)   ......(V)

Here, outer diameter of circular pipe is d and inner diameter of circular pipe is di .

Write the expression for inner diameter of circular pipe.

  di=d2t   ......(VI)

Here, the wall thickness of the circular pipe is t .

Write the expression for critical load in member OB .

  Pcr=π2EIL2   ......(VII)

Here, the young’s modulus of members is E .

Substitute FOB for Pcr in equation (VII).

  FOB=π2EIL2   ......(VIII)

Substitute 1.25P for FOB in equation (VIII).

  1.25P=π2EIL2P=π2EI1.25×L2   ......(IX)

Calculation:

Substitute 0.25in for t and 3.5in for d .

  di=3.5in2×0.25indi=3in

Substitute 3.5in for d and 3in for di in equation (V).

  I=π64( ( 3.5in )4 ( 3in )4)=π64( ( 150.0625in )4 ( 81in )4)=π64×(464047556.5)=22778881.16in4

Substitute 10600ksi for E , 22778881.16in4 for I and 11ft for L in equation (IX).

  P=π2×10600ksi×22778881.16 in41.25× ( 11ft )2=π2×10600ksi( 10 3 lb/ in 2 1ksi )×22778881.16 in41.25× ( 11ft×( 12in 1ft ) )2=π2×10600× 103lb/ in 2×22778881.16 in41.25× 1322 in2=2.383076585× 10 15lb in221780 in2

  =1.094158×1011lb=1.094158×1011lb×( 10 3 kips 1lb)=109415821.2kips

Conclusion:

The critical value of load P at which member OB fails by Euler buckling is 109415821.2kips_ .

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Chapter 11 Solutions

Mechanics of Materials (MindTap Course List)

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