Question

For the frame as shown in the figure below, the final end moment \(M_{BC}\) has been calculated as \(-40\) kN-m. What is the end moment \(M_{CD}\)?

• A. \(+40\) kN-m
• B. \(-40\) kN-m
• C. \(+30\) kN-m
• D. \(-30\) kN-m

Hint:

In the Moment Distribution Method, the moment distributed to connected members is proportional to their stiffness. For a member with the far end fixed, \[ k=\frac{4EI}{L}. \] Hence, shorter members receive a larger share of the joint moment.

Answer 1

The Correct Option is D

Concept: According to the Moment Distribution Method, the balancing moment at a joint is distributed among the connected members in proportion to their stiffness. The stiffness of a member fixed at the far end is \[ k=\frac{4EI}{L}. \] Hence, the distribution factor (DF) is \[ \text{DF}=\frac{k}{\sum k}. \]

Step 1: Determine the stiffness of the members meeting at joint \(C\). Member \(BC\): \[ L_{BC}=4\,\text{m} \] \[ k_{BC}=\frac{4EI}{4}=EI. \] Member \(CD\): \[ L_{CD}=3\,\text{m} \] \[ k_{CD}=\frac{4EI}{3}. \] Therefore, \[ k_{BC}:k_{CD} = 1:\frac43 = 3:4. \]

Step 2: Use the ratio of end moments. The end moments at joint \(C\) are proportional to the stiffnesses of the connected members. Thus, \[ \frac{M_{BC}}{M_{CD}} = \frac{3}{4}. \] Given, \[ M_{BC}=-40\ \text{kN-m}. \] Hence, \[ M_{CD} = -\frac{3}{4}\times40 = -30\ \text{kN-m}. \] Therefore, \[ \boxed{M_{CD}=-30\ \text{kN-m}.} \] Thus, the correct option is \[ \boxed{(D)\;-30\ \text{kN-m}.} \]