Answer in Mechanical Engineering for Evans #213049

The beam shown in Figure P2 is a simply supported beam with a rectangular

cross-section.

A. a. Draw the free body diagram for the beam.

B. b. Calculate all the reaction forces at the supports.

C. c. Draw the shearing force diagram for the beam on a graph sheet.

D. d. Draw the bending moment diagram for the beam on a graph sheet.

E. e. What is the maximum normal stress in the beam?

Part a

Part b

“u03a3F_x = 0: H_A = 0\nu03a3F_y = 0: – (U_1^{left} *7)0.5- P_1 + R_A = 0\nu03a3M_A = 0: (U_1^{left} *7*0.5) * (9 – 8 + (2/3)*7) + 5*P_1 + M_A = 0;”

“H_A = 0 (kN)\nRA = (U_1 ^{left} *7)/2 + P_1 = (30*7)*0.5 + 40 = 145.00 (kN)\nMA = – (U_1^{left} *7/2) * (9 – 8 + (2/3)*7) – 5*P_1 = – (U_1^{left} *7/2) * (9 – 8 + (2/3)*7) – 5*40 = -795.00 (kN*m)”

Part c

Part d

Part e

“Q(x_2) = – ([(U_1^{left} – U_1^{left} *(8 – x)/7)*(x – 1)]/2 + U_1^{left} *(8 – x)/7*(x – 1))\nQ2(1) = – ([(30 – 30*(8 – 1)/7)*(1 – 1)]/2 + 30*(8 – 1)/7*(1 – 1)) = 0 (kN)\nQ2(4) = – ([(30 – 30*(8 – 4)/7)*(4 – 1)]/2 + 30*(8 – 4)/7*(4 – 1)) = -70.71 (kN)”

“Q_3(4) = – ([(30 – 30*(8 – 4)/7)*(4 – 1)]/2 + 30*(8 – 4)/7*(4 – 1)) – 40 = -110.71 (kN)\nQ_3(8) = – ([(30 – 30*(8 – 8)/7)*(8 – 1)]/2 + 30*(8 – 8)/7*(8 – 1)) – 40 = -145 (kN)”

“Q_4(8) = – ([(30 – 30*(8 – 8)/7)*(8 – 1)]/2 + 30*(8 – 8)/7*(8 – 1)) – 40 = -145 (kN)\nQ_4(9) = – (-30*7)/2 – 40 = -145 (kN)”