Atruss is composed of members AB, BC, CD, AD and BD, as shown in the figure. A vertical load of 10 kN is applied at point D. The magnitude of force (in kN) in the member BC is _______.

Correct Answer: The magnitude of force in member BC is 5 kN.

To find the force in the member BC, we can analyze the support reactions and then use the method of joints at joint C.

Step 1: Calculate the support reactions
The truss is symmetrical with respect to the vertical line passing through member BD. A vertical downward load of 10 kN is applied at node D.
Let $R_A$ be the vertical reaction at support A and $R_C$ be the vertical reaction at support C.
By symmetry, the vertical reactions at the two supports are equal:
$R_A=R_C=\frac{10\text{kN}}{2}=5\text{kN}$
Thus, the upward vertical reaction at support C is $R_C=5\text{kN}$.

Step 2: Analyze Joint C
Let us consider the equilibrium of joint C. At joint C, the forces acting are:
1. The vertical support reaction $R_C=5\text{kN}$ acting upwards.
2. The force in member CD, denoted as $F_{\mathrm{CD}}$, acting at an angle of 45° to the horizontal.
3. The force in member BC, denoted as $F_{\mathrm{BC}}$, acting horizontally along the member BC.
Using the equations of equilibrium for joint C:

Sum of vertical forces:
$\sum F_y=0$
$R_C+F_{\mathrm{CD}}sin(45°)=0$
$5+F_{\mathrm{CD}}\left(\frac{1}{\sqrt{2}}\right)=0$
$F_{\mathrm{CD}}=-5\sqrt{2}\text{kN}$ (Compression)

Sum of horizontal forces:
$\sum F_x=0$
Assuming tensile forces are positive (acting away from the joint):
$-F_{\mathrm{BC}}-F_{\mathrm{CD}}cos(45°)=0$
$F_{\mathrm{BC}}=-F_{\mathrm{CD}}cos(45°)$
Substitute the value of $F_{\mathrm{CD}}$ into the equation:
$F_{\mathrm{BC}}=-(-5\sqrt{2})\times \frac{1}{\sqrt{2}}$
$F_{\mathrm{BC}}=5\text{kN}$ (Tension)

Therefore, the magnitude of the force in member BC is 5 kN.