Item 4 Determine the displacement of A The A709 steel rod is
Solution
Axial load in part AB=P=8 kN=8000 N
Axial load in part BC = 8+2*6*sin45=16485.28 N
Axial load in part CD=16.48+2*6*sin45=24970.56 N
Modulus of elasticity of A709 steel = 200 GPa=2*105 N/mm2
area of cross section=80 mm2
Stress in part AB=8000/80=100 N/mm2
Stress in part BC=16485.28/80=206.066 N/mm2
Stress in part CD=24970.56/80=312.132 N/mm2
length of CD=0.75 m=750 mm
length of BC=1.5 m = 1500 mm
length of AB=1 m = 1000 mm
Strain in AB=100/2*105 = 0.0005
Strain in BC=206.066/2*105=0.00103
Strain in CD=312.132/2*105 = 0.00156
displacement of C with respect to D=0.00156*750=1.17 mm
displacement of B with respect to C = 0.00103*1500=1.545 mm
displacement of A with respect to B=0.0005*1000=0.5 mm
total displacement of A=1.17+1.545+0.5=3.215 mm
total displacement of B=1.17+1.545=2.715 mm
