A Steel wire hangs vertically under its weight. What is the greatest length it can have if the allowable tensile stress o, =200 kN/m3.(ans:-2500 m) Solution: MPa? Density of steel y=80 O; =200 MPa= 200*103 kN/m² ; L y=80 kN/m³. Wt. of wire P=(Tt/4)(D²)(L)* y c/s area of wire A=(t/4)*D² Of = PIA solving above eq. L=2500m

Determine the state of stress acting at point D. Take P₁ = 50 kN and P2 = 10 kN. OD= B ANSWER: P₁ TD= -1.5 m 1.5 m--1.5 m- 0.5 m 40 mm 100 mm 60 mm T D E P₂ 80 mm Part A Find o D. Express your answer to three significant figures and include the appropriate units. Enter negative value in the case of compression and positive value in the case of tension. ANSWER: C Q Tap image to zoom Part B Find TD- Express your answer to three significant figures and include the appropriate units.

Urgently needed. .