Let's break down the problem: You are asked for **member AD**, to determine the value of \( \mu PL \) to be used in the displacement calculation. Typically, - \( \mu \): Member force (in kN, either from the real structure or the virtual unit load case) - \( P \): Member force due to the actual load - \( L \): Length of the member (in m) Since you are only asked for \( \mu PL \) for **member AD**, let's find its length first: ### Step 1: Length of Member AD Member AD is horizontal, from A to D: - \( L_{AD} = 4\, \text{m} \) ### Step 2: Virtual Unit Load (\( \mu \)) To find the vertical displacement at D, apply a **virtual unit load (1 kN)** vertically upward at D and analyze the member force in AD due to this virtual load. However, for member AD, the force due to a vertical load at D will be **zero** because the load is perpendicular to member AD (no vertical component in a horizontal member). - \( \mu_{AD} = 0 \) (Force in AD due to a unit vertical load at D) ### Step 3: Real Load (\( P \)) You would need the force in member AD due to the actual loading. But since \( \mu_{AD} = 0 \), the product will be zero regardless of the value of \( P \). ### Step 4: Calculate \( \mu PL \) for member AD \[ \mu_{AD} \cdot P_{AD} \cdot L_{AD} = 0 \cdot P_{AD} \cdot 4 = 0 \] **Value (to 3 significant figures):** \[ \boxed{0.00\ \text{kNm}} \] **Final Answer:** > For member AD, the value of \( \mu PL \) to be used is **0.00 kNm**.