Given information:
The length from point A to B is 20 in , the length from point B to C is 20 in , the length from point C to D is 40 in , modulus of elasticity is 200 GPa , area of bar from point A to B is 6000 mm2 , area of bar from point B to C is 5000 mm2 , area of bar from point C to D is 4000 mm2 .
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 77 kN/m3 .
γ=77 kN/m3=77 kN/m3×( 1 m 1000 mm)3=7.7×10−5 N/mm3
Substitute 500 mm for L1 , 250 mm for L2 , 250 mm for L3 , 500 mm for L4 , 50 N for PB , 100 N for PC and 250 N for PC , 350 N for PE , 200 GPa for E, 6000 mm2 for A1 , 5000 mm2 for A2 , 4000 mm2 for A3 and 7.7 ×10−5 N/mm3 for γ in Equation (I).
δB=[{ 500 mm ( 6000 mm 2 )( 200 GPa ) ×( −50 N+250 N+350 N )}+{ ( 7.7× 10 −5 N/mm 3 )×500 mm ( 6000 mm 2 )( 200 GPa ) ×( ( 6000 mm 2 )×500 mm 2 +( 5000 mm 2 ×250 N ) +( 5000 mm 2 ×250 N ) +( 4000 mm 2 ×500 N ) )}]
=[{ 500 mm ( 6000 mm 2 )( 200 GPa )( 1000 N/ mm 2 1 GPa ) ×( −50 N+250 N+350 N ) }+{ ( 7.7 × 10 −5 N/ mm 3 )×500 mm ( 6000 mm 2 )( 200 GPa )( 1000 N/ mm 2 1 GPa ) ×( ( 6000 mm 2 )×500 mm 2 +( 5000 mm 2 ×250 mm ) +( 5000 mm 2 ×250 mm ) +( 4000 mm 2 ×500 mm ) ) }]=[275000 N⋅mm12× 10 8 N+0.0385 N/ mm 2 12× 10 8 N×( 1500000 mm+1250000 mm +1250000 mm+2000000 mm )]=[2.291×10−4 mm+ N/ mm2×(60× 10 −5 mm)]=4.22×10−4 mm
Substitute 250 mm for L2 , 250 mm for L3 , 500 mm for L4 , 50 N for PB , 100 lb for PC and 250 N for PC , 350 N for PE , 200 GPa for E, 5000 mm2 for A2 , 4000 mm2 for A3 and 7.7 ×10−5 N/mm3 for γ in Equation (II).
δ C =4.22× 10 −4 mm+[ 250 mm ( 5000 mm 2 )( 200 GPa ) ( 250 N+350 N )+ { ( 7.7× 10 −5 N/ mm 3 )×250 mm ( 5000 mm 2 )( 200 GPa ) ×( ( 5000 mm 2 )×250 mm 2 +( 5000 mm 2 ×250 N ) +( 4000 mm 2 ×500 N ) ) } ]
=4.22× 10 −4 mm+[ { 250 mm ( 5000 mm 2 )( 200 GPa )( 1000 N/ mm 2 1 GPa ) ×( 250 N+350 N ) }+ { ( 7.7× 10 −5 N/mm 3 )×250 mm ( 5000 mm 2 )( 200 GPa )( 1000 N/ mm 2 2 1 GPa ) ×( ( 5000 mm 2 )×250 mm 2 +( 5000 mm 2 ×250 N ) +( 4000 mm 2 ×500 N ) ) } ]
=4.22× 10 −4 mm+[ 150000 N⋅mm 10× 10 8 N +{ 0.01925 N/ mm 2 10× 10 8 N ×( 625000 mm 3 + 1250000 mm 3 + 2000000 mm 3 ) } ]
=4.22× 10 −4 mm+[ 150000 N⋅mm 10× 10 8 N +{ 0.01925 N/ mm 3 10× 10 8 N ×3875000 mm 3 } ]
=4.22×10−4 mm+150000 N⋅mm10× 108 N+74593.75 N⋅mm10× 108 N=6.46×10−4 mm
Substitute 250 mm for L3 , 500 mm for L4 , 350 N for PE , 200 GPa for E, 5000 mm2 for A2 , 4000 mm2 for A3 and 7.7×10−5 N/mm2 for γ in Equation (II).
δD=6.46 ×10−4 mm+[{ 250 mm ( 5000 mm 2 )( 200 GPa ) ×( 350 N )}+{ ( 7.7 × 10 −5 N/mm 3 )×250 mm ( 5000 mm 2 )( 200 GPa ) ×( ( 5000 mm 2 )×250 mm 2 +( 4000 mm 2 ×500 N ) )}]
=6.46 ×10−4 mm+[{ 250 mm ( 5000 mm 2 )( 200 GPa )( 1000 N/mm 2 1 GPa ) ×( 350 N ) }+{ ( 7.7 × 10 −5 N/mm 3 )×250 mm ( 5000 mm 2 )( 200 GPa )( 1000 N/mm 2 1 GPa ) ×( ( 5000 mm 2 )×250 mm 2 +( 4000 mm 2 ×500 mm ) ) }]=6.46 ×10−4 mm+[87500 N⋅mm10× 10 8 N+{ 0.01925 N/mm 2 10× 10 8 N ×( 625000 mm 3 +2000000 mm 3 )}]=6.46 ×10−4 mm+[87500 N⋅mm10× 10 8 N+50531.25 N⋅mm10× 10 8 N]
=7.84×10−4 mm
Conclusion:
The displacements at point B is 4.22×10−4 mm_ .
The displacements at point C is 6.46×10−4 mm_ .
The displacements at point D is 7.84×10−4 mm_ .