Question

A three-member isosceles right-angle triangular truss is hinged at A and roller support at B such that the vertical member AB is perpendicular to the horizontal member AC. If a vertical force of 100 kN is acting at C, the force in the member AC is equal to

• A. zero
• B. 100 kN
• C. $100\sqrt{2}$ kN
• D. $100/\sqrt{2}$ kN

Hint:

For simple trusses, the Method of Joints is often the fastest approach.
At joint C, the external vertical load must be balanced by the vertical component of the diagonal member.
The horizontal force in the horizontal member must then balance the horizontal component of the diagonal member.
Since the angles are $45^\circ$, the horizontal and vertical components of the diagonal force are equal.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question asks for the internal axial force in the horizontal member AC of a specific triangular truss under a vertical load at joint C.

Step 2: Key Formula or Approach:
The most direct way to solve for the forces in members connected to a loaded joint is the Method of Joints. This method involves applying the equations of static equilibrium ($\Sigma F_x = 0$ and $\Sigma F_y = 0$) to the joint.

Step 3: Detailed Explanation:
Let's analyze the forces acting on joint C.

• There is an external vertical downward load of 100 kN.

• There is an internal force from the horizontal member AC, which we'll call $F_{AC}$.

• There is an internal force from the diagonal member BC, which we'll call $F_{BC}$.


Geometry: The truss is an isosceles right-angle triangle with the right angle at A. This means the angles at B and C are both $45^\circ$. Therefore, the member BC is inclined at $45^\circ$ to the horizontal.

Equilibrium at Joint C:
We resolve the forces into horizontal (x) and vertical (y) components. Let's assume tensile forces (pulling away from the joint) are positive.
1.

Sum of vertical forces ($\Sigma F_y = 0$):
The downward external load of 100 kN must be balanced by the upward vertical component of the force in member BC. This implies that member BC must be in tension (pulling joint C upwards).
\[ F_{BC} \sin(45^\circ) - 100 \text{ kN} = 0 \] \[ F_{BC} = \frac{100}{\sin(45^\circ)} = \frac{100}{1/\sqrt{2}} = 100\sqrt{2} \text{ kN (Tension)} \] 2.

Sum of horizontal forces ($\Sigma F_x = 0$):
The horizontal component of the force in member BC pulls joint C to the left. To balance this, the force in member AC, $F_{AC}$, must pull to the right. This implies that member AC is also in tension.
\[ F_{AC} - F_{BC} \cos(45^\circ) = 0 \] \[ F_{AC} = F_{BC} \cos(45^\circ) \] Substitute the value of $F_{BC}$ we just found:
\[ F_{AC} = (100\sqrt{2} \text{ kN}) \times \cos(45^\circ) \] Since $\cos(45^\circ) = 1/\sqrt{2}$:
\[ F_{AC} = (100\sqrt{2}) \times \frac{1}{\sqrt{2}} = 100 \text{ kN} \] Since the result is positive, our assumption that AC is in tension is correct. The magnitude of the force is 100 kN.

Step 4: Final Answer:
The force in the member AC is equal to 100 kN.