A steel rod of 20mm diameter passes centrally through a copper tube 40mm external diameter and 30mm internal diameter. The tube is closes at each end by rigid plates of negligible thickness. The nuts are tightened lightly home on the projected parts of the rod. If the temperature of the assembly is raised by 60°C, calculate the stresses developed in copper and steel. Take E for steel and copper as 200GPa and 100GPa and α for steel and copper as 12 x 10-6 per°C and 18 x 10-6 per°C.
Answers
Answer:
The steel bolt is under tension and the copper tubing under compression.
The nut is fastened by one quarter of a turn.
Allowed Deformation = PithchofNut4
Allowed Deformation = 24=0.5mm
The elongation for the bolt and the tube is the same.
(PIAE)bolt=(PIAE)tubing=0.5mm........(i)
Steel Bolt – Tension
d=20mm dl=0.5mm
l=600mm Es=200GPa
As=π4(20)2=314.159mm2
From (I),
Ps(600)(314.159)(200×103)=0.5
Ps=52.36kN
Stress Induced (δ)s=PsAs=52.36×103314.159=166.67N/mm2
This is the value of the tensile stress induced in the steel bolt.
Copper Tubing – Compression
d=25mm t=10mm
D=d+2t=45mm Ec=100GPa
Ac=π4[452−252]=109.56mm2
From (I),
Pc(600)(10099.56)(100×103)=0.5
Pc=91.63kN
Stress Induced δc=PcAc=91.63×1031099.56=83.33N/mm2
This is the value of the compressive stress induced in the copper tubing.
Steel rod area = Asr = 4 (0.022) = 0.0003142 m2 Steel tube area = Ast = /4 (0.042 – 0.0252) = 0.000766 m2 Load is applied evenly to steel tube and steel rod. st1.Ast = sr1.Asr = 2000.
Explanation- For example Since steel tube compression and steel rod tension 0003142 sr1 = 0.000766 st1 = 2000 sr1 = 63.6 MN/m2 (T) st1 = 26.1 MN/m2 Let sr2 and st2 be the extra stresses generated in the rod and tube, respectively, when the nut is tightened by one-quarter.
Distance travelled by the nut = 1/4(1/0.4) = 0.625mm = 0.000625m =
LTotal = LRod + Ltub = (sr2 /E). + (st2 /E) Lsr Lst
Lst = Lsr 0.000625 = L/E(sr2 + st2)
Once again, the load is equal to st2. sr2 = ast. st2 = asr sr2. Ast/Asr = st2.0.000766/0.0003142 = 2.44st2
When we plug this number into the equation we obtain 0.000625 = (0.75/200).