Topic: Engineering Mechanics
A simple structure ABCDE is supported on a hinge at A and on rollers at B while it carries a horizontal force of 1000 kN at E as shown in the figure. Determine the force in member AC using the method of joints.
- A. 1500 kN
- B. 1400 kN
- C. 1314 kN
- D. 1414 kN
Answer: D. 1414 kN
Explanation: First, determine the reactions at the supports. Taking moments about hinge A: $V_{B} \times 1 = 1000 \times 1.87$, which yields $V_{B} = 1870\text{ kN}$. Resolving vertical forces gives $V_{A} = -1870\text{ kN}$. For the roller support at B, the horizontal reaction is zero ($H_{B}=0$), so the force in member AB is $F_{AB} = 0$. Considering joint A and resolving forces vertically: $F_{AC}\sin45^{\circ} + F_{AD} = 1870$. Resolving forces horizontally at joint A: $F_{AB} + F_{AC}\cos45^{\circ} = 1000$. Since $F_{AB} = 0$, we have $F_{AC}\cos45^{\circ} = 1000$. Solving for $F_{AC}$ provides $F_{AC} = \frac{1000}{\cos45^{\circ}} = 1414\text{ kN}$.