Question

Consider a rectangular beam subjected to bending 15 kN 10 kN/m 20 kNm IITTI 0.15 m SIA 0.4 m 2m- Im- 1m 2m- Beam's cross-section a. (10 pts) Determine internal reaction (shear and bending moment) at point located 0.6 m to the left of B. b. (30 pts) Determine normal stress at that point A. Identify whether stress is tensile or compressive c. (30 pts) Determine shear stress at that point A. Identify whether stress is tensile or compressive. d. (30 pts) Beam is being stretched with uniform distributed load of 40 kN/m- normal to the beam's cross section. Modify answers for parts (b and/or (c) to account for this additional loading.

Answer 1

Draw the free body diagram of the beam. X 0.6 m 15 kN 10 kN/m 20 kN-m 2 m T 1mT1m 2 m 5.4 m Consider Az = 0 Ax = 0 Consider moment at A

MA=0 15*2 + 20+ 10*2*5 - B,*6=0 By = 25 kN Consider Ay = 0 Ay +By - 15 - 20 = 0 Ay + 25 - 15 - 20 = 0 Ay = 10 kN a) Shear force at 5.4 meter from A is V = 10 - 15-10*1.4 =-19 KN B.M. at 5.4 meter from A is M = 10*5.4- 15*3.4 +20 - 10*1.4*0.7= 13.2 kN-m b) Normal stress at that point A

on – My...Eq. (1) Here, M = 13.2 kN-m, y = 0.35 m I = bh3 12 0.1 *0.43 == 5.33 * 10-4 m4 12 Substitute values of I, M, and y in Eq. (1) 13.2 * 0.05 ON = 533 * 10-4 ON = 1238.27 kN/m2 Hence, normal stress at that point A is 1238.27 kN/m2 The stress is in compression. (c) Shear stress at that point A. 0,= V12 ..Eq. (2) I*t

Here, V=-19 KN, I = 5.433*10-4 m4,t=width of beam = 0.1 m Q = area above point A from neutral axis * centroid of that area from the neutral axis = 0.15*0.1*.0.125 =0.001875 m3 Substitute the values of V, Q. I and t in Eq. (2) -19 * 0.001875 Os = 5.33 * 10-4* 0.1 0, = 668.38 kN/m2 The shear stress is in compression. (d) If the beam is stressed with a uniform distributed load of 40 kN/m2 normal to the beam, Then, normal stress at that point A is 1238.27-40 = 1198.27 kN/m2 Shear stress at that point A is -668.38+ 40 = -628.38 kN/m2