Question 5.1: Determine the axial force, shear, and bending moment at poin...
Determine the axial force, shear, and bending moment at point B of the beam shown in Fig. 5.3(a).
Reactions Considering the equilibrium of the free body of the entire beam (Fig. 5.3(b)), we write
+\longrightarrow \sum{F_x} = 0 A_x – (\frac{4}{5})(110) = 0 A_x = 88 kN\longrightarrow
+\circlearrowleft \sum{M_c} = 0 -A_y(12) + 135(8) + (\frac{3}{5})(110)(4) = 0 A_y = 112 kN\uparrow
+\uparrow \sum{F_y} = 0 112 – 135 – (\frac{3}{5})(110) + C_y = 0 C_y = 89 kN\uparrow
Section bb A section bb is passed through point B, cutting the beam into two portions, AB and BC (see Fig. 5.3(b)). The portion AB, which is to the left of the section, is used here to compute the internal forces.
Axial Force Considering the external forces acting to the left as positive, we write
Q = -88 kNShear Considering the external forces acting upward as positive, we write
S = 112 – 135 = -23S = -23 kN
Bending Moment Considering the clockwise moments of the external forces about B as positive, we write
M = 112(6) – 135(2) = 402M = 402 kN-m
Checking Computations To check our calculations, we compute the internal forces using portion BC, which is to the right of the section under consideration.
By considering the horizontal components of the external forces acting to the right on portion BC as positive, we obtain
Q = -(\frac{4}{5})(110) = -88 kN Checks
By considering the external forces acting downward as positive, we obtain
S = -89 + (\frac{3}{5})(110) = -23 kN Checks
Finally, by considering the counterclockwise moments of the external forces about B as positive, we obtain
M = 89(6) – (\frac{3}{5})(110)(2) = 402 kN-m Checks
