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 11.3.23P

The truss ABC shown in the figure supports a vertical load W at joint B. Each member is a slender circular steel pipe (E = 30,000 ksi) with an outside diameter of 4 in. and wall thickness 0.25 in. The distance between supports is 23 ft. Joint B is restrained against displacement perpendicular to the plane of the truss.

Determine the critical value Wcr of the load.

  Chapter 11, Problem 11.3.23P, The truss ABC shown in the figure supports a vertical load W at joint B. Each member is a slender

Expert Solution & Answer
Check Mark
To determine

The critical of the load.

Answer to Problem 11.3.23P

The critical load is 14.41k .

Explanation of Solution

Given information:

The modulus of elasticity of the steel pipe is 30000ksi , the outside diameter of the steel pipe is 4in , the thickness of the pipe is 0.25in, and the distance between the two supports is 23ft .

Write the expression for the inside diameter of steel pipe.

  di=do2t   .........(I)

Here, the outside diameter is do and the thickness is t .

Write the expression for the moment of inertia of hollow of steel pipe.

  I=π64(( d o )4( d i )4)   .........(II)

Here, the outside diameter is do and the inside diameter is di .

The following figure shows the free body diagram.

  Mechanics of Materials (MindTap Course List), Chapter 11, Problem 11.3.23P , additional homework tip  1

  Figure-(1)

Write the expression for the length of the steel pipe AB .

  LAB=L(sinθ2sin( 180° θ 1 θ 2 ))   .........(III)

Here, the distance between two supports is L , the angle at support C is θ2, and the angle at support A is θ1 .

Write the expression for the length of the steel pipe BC .

  LBC=L(sinθ1sin( 180° θ 1 θ 2 ))   .........(IV)

Here, the distance between two supports is L , the angle at support C is θ2, and the angle at support A is θ1 .

Write the expression for critical load of pipe of both end pinned.

  PcrAB=π2EILAB2   .........(V)

Here, modulus of elasticity is E and moment of inertia is I .

Write the expression for critical load of pipe of both end pinned joined.

  PcrBC=π2EILBC2   .........(VI)

Here, modulus of elasticity is E and moment of inertia is I .

The following figure shows the free body diagram taking joint at B .

  Mechanics of Materials (MindTap Course List), Chapter 11, Problem 11.3.23P , additional homework tip  2

  Figure-(2)

Write the expression for force acting along y-direction.

  Fy=0FABsin(θ1)FBCsin(θ2)=WFABsin(θ1)FBCsin(θ2)W=0   .........(VII)

Write the expression for force acting along x-direction.

  Fx=0FABcos(θ1)FBCcos(θ2)=FABcos(θ1)FBCcos(θ2)=0   .........(VIII)

Write the expression for critical load for member AB.

  Wcr=min(WcrAB,WcrBC)   .........(IX)

Calculation:

Substitute 4in for do and 0.25in for t in Equation (I).

  di=4in(2×0.25in)=4in0.5in=3.5in

Substitute 4in for do and 3.5in for di in Equation (II).

  I=π64( ( 4in )4 ( 3.5in )4)=0.049(256 in4150.0625 in4)=5.200in4

Substitute 40° for θ1 and 55° for θ2 in Equation (III).

  LAB=(23ft)( sin55° sin( 180°40°55° ))=(23ft)( sin55° sin85°)=(23ft)(0.822)=18.912ft

Substitute 40° for θ1 and 55° for θ2 in Equation (IV).

  LBC=(23ft)( sin40° sin( 180°40°55° ))=(23ft)( sin40° sin85°)=(23ft)(0.64)=14.841ft

Substitute 30000ksi for E , 5.200in4 for I and 18.912ft for LAB in Equation (V).

  PcrAB=π2( 30000ksi)( 5.200 in 4 ) ( 18.912ft )2=π2( 30000ksi( 1k/ in 2 1ksi ))( 5.200 in 4 ) ( 18.912ft( 12in 1ft ) )2=π2( 30000k/ in 2 )( 5.200 in 4 ) ( 226.944in )2=29.59k

Substitute 30000ksi for E , 5.200in4 for I and 14.841ft for LBC in Equation (VI).

  PcrBC=π2( 30000ksi)( 5.200 in 4 ) ( 14.841ft )2=π2( 30000ksi( 1k/ in 2 1ksi ))( 5.200 in 4 ) ( 14.841ft( 12in 1ft ) )2=π2( 30000k/ in 2 )( 5.200 in 4 ) ( 178.092in )2=48.55k

Substitute 40° for θ1 and 55° for θ2 in Equation (VII).

  FABsin(40°)FBCsin(55°)W=00.642FAB0.819FBCW=0   .........(X)

Substitute 40° for θ1 and 55° for θ2 in Equation (VIII).

  FABcos(40°)FBCcos(55°)=00.766FAB0.5735FBC=00.766FAB=0.5735FBCFBC=1.33FAB

  FAB=0.75FBC

Substitute 1.33FAB for FBC in Equation (X).

  0.642FAB0.819(1.33F AB)W=0W=0.642FAB(1.12917F AB)W=0.48717FAB

Neglecting the negative sign as it shows the compression.

  W=0.48717FAB   .........(XI)

Substitute PcrAB for FAB for member AB in Equation (XI).

  WcrAB=0.48717PcrAB   .........(XII)

Substitute 0.75FBC for FAB in Equation (X).

  0.642(0.75F BC)0.819FBCW=0W=0.48FBC(0.819F BC)W=0.3375FBC

Neglecting the negative sign as it shows the compression.

  W=0.3375FBC   .........(XIII)

Substitute PcrBC for FBC in for member BC in Equation (XIII).

  WcrBC=0.3375PcrBC   .........(XIV)

Substitute 29.59k for PcrAB in Equation (XII).

  WcrAB=0.48717(29.59k)=14.41k

Substitute 48.55k for PcrBC in Equation (XIV).

  WcrBC=0.3375(48.55k)=16.38k

Substitute 14.41k for WcrAB and 16.38k for WcrBC in Equation (IX).

  Wcr=min(14.41k,16.38k)=14.41k

Conclusion:

The critical load is 14.41k .

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Chapter 11 Solutions

Mechanics of Materials (MindTap Course List)

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