Question

1. The circular shaft is subjected axial forces as shown below. Calculate the average normal stress at points A, B, and C. 1 in. 0.5 in. 0.5 in. Bo Bo skip 3 kip z kip 8 kip | C A 19 kip 2 kip

Answer 1

The normal stresswat the respected points are given by where is the internal pressuke. At the paint A, the internal pressure is 3x1000 lb PAPE 25)2 = fanya cuma - PA = 15278.87 lb/in2 (A radius of the shaft at A). 1 kip - 1000 (6) Then normal stress at A is - 7639. 4 lbsin? The internal pressure at B is P. Fr . 19x1000 lb II (ro)2 - (0.52 in2 PB = 11459.15 Absint.

Therefore, the internal stress at Bis. OBí - PB - - 5729.5 lbsint. This is due to the force 9 kip. at the point of the internal pressure at 2x1000 lb Tre) TX (0.25)2 in2 => Pc = 10185.9 lbsint. Then the normal stress at ces Oo - - Pel = - 5092.9 lb/in2.

The internal pressure due to the force 8 kip at Bis PB2 - 2 2 = 8x1000 lb 132 T(B)2 - ITX (0.52 in2 PB2 = 10185.9 ebline. Then the normal stress is Pod 10185-9/2 Abhina >> On 5-5092.9 26/cm2 Therefore, resultant normal stress es, Da = 00:- o - -699.6 lb/in2. These are the required normal stresses.