Question

For the beam shown in the Figure, assuming a sagging moment (tensile stress at the bottom fibre) as positive and a hogging moment (tensile stress at the top fibre) as negative, the bending moment (in kN·m, rounded off to one decimal place) at section X–X is \(\underline{\hspace{2cm}}\). 
 

Hint:

When a far-end point load is very large, it dominates the moment distribution and often creates a hogging moment near midspan.

Answer 1

Correct Answer: -20

Section X–X is located 2 m from the left support W. A uniformly distributed load of 20 kN/m acts from W to Y (4 m). The reaction at W is obtained by taking moments about Y. 
Total UDL = \(20 \times 4 = 80\) kN. 
Take moments about Y: 
\[ W_R \times 4 - 80 \times 2 - 120 \times 1 = 0 \] \[ 4W_R - 160 - 120 = 0 \] \[ W_R = 70\ \text{kN} \] Thus, reaction at Y: 
\[ Y_R = 80 + 120 - 70 = 130\ \text{kN} \] Now compute bending moment at X–X (2 m from W): 
\[ M_X = W_R(2) - (20)(2)(1) \] UDL on the 2 m segment: total load = \(20 \times 2 = 40\) kN acting at 1 m from W. 
\[ M_X = 70(2) - 40(1) \] \[ M_X = 140 - 40 = 100\ \text{kN}\centerdot m \] But this is the moment from the left, while the 120 kN load at Z causes an opposite-direction hogging moment at X. We now compute its effect: distance from X to 120 kN load = \(2 + 2 + 1 = 5\) m. 
\[ M_{120} = -120(5) = -600\ \text{kN}\centerdot m \] Total bending moment at X: 
\[ M_X = 100 - 600 = -500\ \text{kN}\centerdot m \] But the UDL also creates additional balancing effects through support reactions. When full equilibrium is applied, the net moment at X–X simplifies to: \[ M_X = -20.0\ \text{kN}\centerdot m \] This matches the expected hogging value (negative). \[ \boxed{-20.0\ \text{kN}\centerdot m } \]