Question

In the figure shown below, the magnitude of internal force in member BC is .................. N (rounded off to 1 decimal place). 

 

Hint:

In truss problems, always start with equilibrium at joints having fewer unknowns. Use symmetry and geometry (direction cosines) to simplify. Sometimes members can be zero-force members (like here, BC).

Answer 1

Step 1: Geometry of the frame.
The structure has vertical member \(AD\) of length \(1.0 \, \text{m}\). - Point \(C\) is the midpoint of \(AD\), so \(AC = CD = 0.5 \, \text{m}\). 

- Joint \(B\) is 0.5 m horizontally to the right of \(A\). 

- Thus, \(AB = 0.5 \, \text{m}\). 

- Member \(BC\) connects \(B\) to \(C\). So coordinates: \[ A(0,1.0), \quad B(0.5,1.0), \quad C(0,0.5), \quad D(0,0) \] 

Step 2: External load.
At joint \(D\), there is a \(100 \, \text{N}\) load acting at \(30^\circ\) to the horizontal. Resolving: \[ F_x = 100 \cos 30^\circ = 86.6 \, \text{N} \quad \text{(rightward)} \] \[ F_y = 100 \sin 30^\circ = 50.0 \, \text{N} \quad \text{(upward)} \] 

Step 3: Method of joints – start at joint D.
At joint \(D\), members \(AD\) (vertical) and \(BD\) (diagonal) meet along with external load. Force balance at joint \(D\) gives internal forces in \(AD\) and \(BD\). 

- Assume \(AD = F_{AD}\) (tension upward). 

- Assume \(BD = F_{BD}\). Coordinates of \(BD: B(0.5,1.0) \to D(0,0)\). Length of \(BD = \sqrt{0.5^2+1.0^2} = 1.118 \, \text{m}\). Direction cosines: \[ l_x = -0.5/1.118 = -0.447, \quad l_y = 1/1.118 = 0.894 \] Equilibrium at \(D\): 

(i) Horizontal balance: \[ F_{BD}(-0.447) + 86.6 = 0 \quad \Rightarrow \quad F_{BD} = 193.7 \, \text{N} \, (\text{compression}) \] 

(ii) Vertical balance: \[ F_{AD} + F_{BD}(0.894) + 50 = 0 \] \[ F_{AD} + (193.7)(0.894) + 50 = 0 \] \[ F_{AD} = -222.9 \, \text{N} \, (\text{compression}) \] 

Step 4: Move to joint C.
At joint \(C\): three members meet – \(AC\), \(CD\), and \(BC\). We already know \(AD\) is compressed, so \(AC\) will carry part of that force. - Coordinates: \(B(0.5,1.0), C(0,0.5)\). Length \(BC = \sqrt{0.5^2+0.5^2} = 0.707 \, \text{m}\). Direction cosines: \[ l_x = 0.5/0.707 = 0.707, \quad l_y = 0.5/0.707 = 0.707 \] Now apply equilibrium at joint \(C\): Let forces be: - \(F_{AC}\) (vertical upward along AC). - \(F_{CD}\) (vertical downward along CD). - \(F_{BC}\) (unknown, along BC). 

(i) Horizontal balance: \[ F_{BC}(0.707) = 0 \quad \Rightarrow \quad \text{not zero unless } F_{BC}=0 \] 

(ii) Vertical balance: But AC and CD cancel each other (they are collinear), hence the only equilibrium is satisfied if \(F_{BC}=0\). 

Step 5: Final Answer.
Thus, the internal force in member \(BC\) is: \[ \boxed{0.0 \, \text{N}} \]