Question

The diameter of each segment is indicated in the figure below. Take P₁ = 1 kip , P₂ = 6 kip , P₃ = 9 kip , and P₄ = 4 kip . (Figure 1)

Part A Determine the average normal stress developed at point A. 

Part B Determine the average normal stress developed at point B. 

Part C Determine the average normal stress developed at point C. 

Answer 1

In this question, the unit of forces is converted into Newton and area are converted into mm square. The correct axial loading diagram is key point in solution of this question. Please go through the detailed solution provided in the image for clear understanding. Please don’t forget to like the answer if you find it useful. Have a good day....

1 in 0.5 inch 0.5in k P3 P4 1 1 T чRip 4 kip + Rip + 3 4 2 1 -skip Axial loading cliagram P =lRip P2 = 6 kip, P3=qRip, Py = 4 kip. we know- I kip= 4448.22 16 Newton linch = 25.4 mm

Stress in Section (1-2) 1-2 = TRip Hoea of (8058 section Asea) = A 2 - #x (65x2504) *= 126:67 mm2 35:11 N/mm2. ( Tenzile) 5.2 = 1x 4448.2216 126.67 Stress in Section (2-3) 5 kip Area of cross section Az3 = *(25.4x1) = 506.70 mm? -5x4448.2216 -43.89 N/mm2 (compressive) 623 = 506.70 Stres in Section (3-4) ► 33-4 4Rip Asea oj (8088 Section = 126.67 mm2 A345 7x10.5*25.4) ? = 126-67 mm2 63-4 4 x 4448.2216 = 140.46 N/mm2 126.67