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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Textbook Question
Chapter 2, Problem 2.3.6P

Repeat Problem 2.3-4, but now include the weight of the bar. Sec Table 1.1 in Appendix I for the weight density of steel.

Expert Solution & Answer
Check Mark
To determine

The displacements at point B , D and E .

Answer to Problem 2.3.6P

The displacements at point B 4.22×104mm_ .

The displacements at point C 6.46×104mm_ .

The displacements at point D 7.84×104mm_ .

Explanation of Solution

Given information:

The length from point A to B is 20in , the length from point B to C is 20in , the length from point C to D is 40in , modulus of elasticity is 200GPa , area of bar from point A to B is 6000mm2 , area of bar from point B to C is 5000mm2 , area of bar from point C to D is 4000mm2 .

Write the expression for the elongation at the point B.

  δB=[ P B L 1 A 1 E+ P C L 1 A 1 E+ P E L 1 A 1 E+ γ A 1 L 1 2 2 A 1 E+ ( γ A 2 L 2 ) L 1 A 1 E+ ( γ A 2 L 3 ) L 1 A 1 E+ ( γ A 3 L 4 ) L 1 A 1 E]=L1A1E(PB+PC+PE)+γL1A1E( A 1 L 1 2+A2L2+A2L3+A3L4)  .........(I)

Here, the elongation at the point B is δB , the load acting at B is PB . The load acting at C is PC , the load acting at E is PE and the modulus of elasticity is E , the length from the point A to point B is L1 , the length from point B to C is L2 , the length from the point C to D is L3 , the length from the point D to E is L4 , the cross-section area of bar from A to B is A1 , the cross-section of bar from point B to C is A2, and the cross-section area of bar from point D to E is A3 .

Write the expression for the elongation at the point C .

  δC=δB+[ P C L 2 A 2 E+ P E L 2 A 2 E+ γ A 2 L 3 2 2 A 2 E+ ( γ A 2 L 3 ) L 1 A 2 E+ ( γ A 3 L 4 ) L 1 A 2 E]=δB+L2A2E(PC+PE)+γL2A2E( A 2 L 2 2+A2L3+A3L4)  .........(II)

Write the expression for the elongation at the point D .

  δD=δC+[ P E L 3 A 2 E+ γ A 2 L 3 2 2 A 2 E+ ( γ A 3 L 4 ) L 3 A 2 E]=δC+PEL3A2E+γL3A2E( A 2 L 3 2+A3L4)   .........(III)

Write the expression for the elongation at the point E .

  δE=δD+[PEL4A3E+(γ A 3 L 4)L4A3E]   .........(IV)

Calculation:

Refer to appendix Ι table 1.1 “Weight and mass densities” to obtain weight density of steel as 77kN/m3 .

  γ=77kN/m3=77kN/m3×( 1m 1000mm)3=7.7×105N/mm3

Substitute 500mm for L1 , 250mm for L2 , 250mm for L3 , 500mm for L4 , 50N for PB , 100N for PC and 250N for PC , 350N for PE , 200GPa for E, 6000mm2 for A1 , 5000mm2 for A2 , 4000mm2 for A3 and 7.7×105N/mm3 for γ in Equation (I).

  δB=[{ 500mm ( 6000 mm 2 )( 200GPa ) ×( 50N+250N+350N )}+{ ( 7.7× 10 5 N/mm 3 )×500mm ( 6000 mm 2 )( 200GPa ) ×( ( 6000 mm 2 )×500mm 2 +( 5000 mm 2 ×250N ) +( 5000 mm 2 ×250N ) +( 4000 mm 2 ×500N ) )}]

  =[{ 500mm ( 6000 mm 2 )( 200GPa )( 1000N/ mm 2 1GPa ) ×( 50N+250N+350N ) }+{ ( 7.7× 10 5 N/ mm 3 )×500mm ( 6000 mm 2 )( 200GPa )( 1000N/ mm 2 1GPa ) ×( ( 6000 mm 2 )×500mm 2 +( 5000 mm 2 ×250mm ) +( 5000 mm 2 ×250mm ) +( 4000 mm 2 ×500mm ) ) }]=[275000Nmm12× 10 8N+0.0385N/ mm 2 12× 10 8N×( 1500000mm+1250000mm +1250000mm+2000000mm )]=[2.291×104mm+N/ mm2×(60× 10 5mm)]=4.22×104mm

Substitute 250mm for L2 , 250mm for L3 , 500mm for L4 , 50N for PB , 100lb for PC and 250N for PC , 350N for PE , 200GPa for E, 5000mm2 for A2 , 4000mm2 for A3 and 7.7×105N/mm3 for γ in Equation (II).

   δ C =4.22× 10 4 mm+[ 250mm ( 5000 mm 2 )( 200GPa ) ( 250N+350N )+ { ( 7.7× 10 5 N/ mm 3 )×250mm ( 5000 mm 2 )( 200GPa ) ×( ( 5000 mm 2 )×250mm 2 +( 5000 mm 2 ×250N ) +( 4000 mm 2 ×500N ) ) } ]

   =4.22× 10 4 mm+[ { 250mm ( 5000 mm 2 )( 200GPa )( 1000N/ mm 2 1GPa ) ×( 250N+350N ) }+ { ( 7.7× 10 5 N/mm 3 )×250mm ( 5000 mm 2 )( 200GPa )( 1000 N/ mm 2 2 1GPa ) ×( ( 5000 mm 2 )×250mm 2 +( 5000 mm 2 ×250N ) +( 4000 mm 2 ×500N ) ) } ]

   =4.22× 10 4 mm+[ 150000Nmm 10× 10 8 N +{ 0.01925N/ mm 2 10× 10 8 N ×( 625000 mm 3 + 1250000 mm 3 + 2000000 mm 3 ) } ]

   =4.22× 10 4 mm+[ 150000Nmm 10× 10 8 N +{ 0.01925N/ mm 3 10× 10 8 N ×3875000 mm 3 } ]

  =4.22×104mm+150000Nmm10× 108N+74593.75Nmm10× 108N=6.46×104mm

Substitute 250mm for L3 , 500mm for L4 , 350N for PE , 200GPa for E, 5000mm2 for A2 , 4000mm2 for A3 and 7.7×105N/mm2 for γ in Equation (II).

  δD=6.46×104mm+[{ 250mm ( 5000 mm 2 )( 200GPa ) ×( 350N )}+{ ( 7.7× 10 5 N/mm 3 )×250mm ( 5000 mm 2 )( 200GPa ) ×( ( 5000 mm 2 )×250mm 2 +( 4000 mm 2 ×500N ) )}]

  =6.46×104mm+[{ 250mm ( 5000 mm 2 )( 200GPa )( 1000 N/mm 2 1GPa ) ×( 350N ) }+{ ( 7.7× 10 5 N/mm 3 )×250mm ( 5000 mm 2 )( 200GPa )( 1000 N/mm 2 1GPa ) ×( ( 5000 mm 2 )×250mm 2 +( 4000 mm 2 ×500mm ) ) }]=6.46×104mm+[87500Nmm10× 10 8N+{ 0.01925 N/mm 2 10× 10 8 N ×( 625000 mm 3 +2000000 mm 3 )}]=6.46×104mm+[87500Nmm10× 10 8N+50531.25Nmm10× 10 8N]

  =7.84×104mm

Conclusion:

The displacements at point B is 4.22×104mm_ .

The displacements at point C is 6.46×104mm_ .

The displacements at point D is 7.84×104mm_ .

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Chapter 2 Solutions

Mechanics of Materials (MindTap Course List)

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