Question

The simple truss structure is shown below. The forces acting in the members AB and BC respectively are

• A. \( 1.732\text{ W} \text{ compression \& } 2\text{ W} \text{ tensile} \)
• B. \( 1.414\text{ W} \text{ tensile and } 1.732\text{ W} \text{ compression} \)
• C. \( 3.464\text{ W} \text{ compression \& } 2\text{ W} \text{ tensile} \)
• D. \( 1.732\text{ W} \text{ tensile and } 2\text{ W} \text{ compression} \)

Hint:

To quickly solve simple right-triangle trusses, use the load to solve the vertical component first. Since the downward load \( W \) must be balanced by the vertical component of the angled strut, \( F_{BC}\sin(30^\circ) = W \implies F_{BC} = 2W \) instantly!

Answer 1

The Correct Option is D

Concept: Truss analysis can be efficiently conducted at an individual junction point using the Method of Joints. Since the entire truss structure is in static equilibrium, each individual joint must also be in a state of perfect equilibrium. This means the sum of all horizontal and vertical forces acting on a joint must equal zero: \[ \sum F_x = 0 \quad \text{and} \quad \sum F_y = 0 \] We assume unknown member forces are pulling away from the joint (tensile force). If our final calculated value is negative, it indicates that the member is actually under compression.

Step 1: Setting up the force vectors at joint B.
Let us analyze Joint B. The forces acting directly on this joint are:

• An external load \( W \) acting vertically downward.

• An internal horizontal member force \( F_{AB} \) acting along member AB towards the left wall.

• An internal inclined member force \( F_{BC} \) acting along member BC down and to the left towards joint C.
The angle between member BC and the horizontal member AB is given as \( 30^\circ \).

Step 2: Applying the vertical equilibrium condition \( \sum F_y = 0 \).
Let us resolve the forces acting at Joint B along the vertical \( y \)-axis (treating upwards as positive): \[ \sum F_y = 0 \quad \Rightarrow \quad -W + F_{BC}\sin(30^\circ) = 0 \] Isolating the vertical component of the member force: \[ F_{BC}\sin(30^\circ) = W \] Since \( \sin(30^\circ) = \frac{1}{2} \), we substitute this value into the equation: \[ F_{BC} \cdot \left(\frac{1}{2}\right) = W \quad \Rightarrow \quad F_{BC} = 2W \] Since our calculated value is positive, member BC is pushing toward joint B, confirming it is under compression. Thus: \[ F_{BC} = 2W \text{ (Compression)} \]

Step 3: Applying the horizontal equilibrium condition \( \sum F_x = 0 \).
Now, let us resolve the forces acting at Joint B along the horizontal \( x \)-axis (treating rightwards as positive). Both member forces pull to the left: \[ \sum F_x = 0 \quad \Rightarrow \quad -F_{AB} - F_{BC}\cos(30^\circ) = 0 \] Isolating the force term \( F_{AB} \): \[ F_{AB} = -F_{BC}\cos(30^\circ) \] Substitute our value \( F_{BC} = 2W \) and the geometric value \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \): \[ F_{AB} = -(2W) \cdot \left(\frac{\sqrt{3}}{2}\right) = -\sqrt{3}W \] The negative sign indicates that our initial direction assumption was reversed, meaning the force is actually pulling away from the joint. This confirms it is a tensile force. Using the numerical value \( \sqrt{3} \approx 1.732 \): \[ F_{AB} = 1.732W \text{ (Tensile)} \] Our results are \( F_{AB} = 1.732W \text{ (Tensile)} \) and \( F_{BC} = 2W \text{ (Compression)} \), which matches Option (D).