Test: ILD & Rolling Loads - 3 - Civil Engineering (CE) MCQ

To determine the influence line for the bar force in the member CH of the pin-joined plane truss, we need to consider the effects of a moving load on the structure. The influence line shows how the internal force in a member varies as a point load moves across the truss.

• The influence line for CH will peak when the load is directly over the member.
• As the load moves away from CH, the internal force will decrease.
• The shape of the influence line typically resembles a triangular or trapezoidal form depending on the geometry of the truss.
• In this case, the correct diagram will show a positive value when the load is over CH, indicating tension, and a negative value when the load is further away, indicating compression.

By analysing the diagrams provided, option B correctly represents the influence line for member CH.

The influence line for the force in member U₁L₂ of the given frame can be determined by analysing how the applied loads affect that specific member. To find this, follow these steps:

• Identify the location of member U₁L₂ within the frame.
• Consider moving a unit load across the entire span of the frame.
• At each position of the load, calculate the force in member U₁L₂.
• Plot these forces against the position of the load to create the influence line.

In this case, the correct influence line will show how the force in U₁L₂ varies with the position of the unit load. The line will typically peak where the load has the greatest effect on the member.

Based on this analysis, option B correctly represents the influence line for the force in member U₁L₂.

Using Castigliano’s second theorem

There is no force in member (1) - (2),

Maximum shear force is at either of the support due to a point load

Section should divide the load in the same ratio as it divides the span for maximum bending moment.

Length of UDL = a

Load crossed the section by 3a/4

⇒ (wa) / (4 × L/4) = (w × 3/4 a) / (3L/4)

⇒ (wa) / L = (wa) / L

So average load on both side is equal.

The ILD for support reaction at D can be obtained by:giving unit displacement in the direction of reaction. The deflected shape of beam will represent ILD as in figure (c).

[1] is false because the Influence Line Diagram (ILD) for Shear Force (SF) at the fixed end of a cantilever is not the same as the Shear Force Diagram (SFD) for a unit load at the free end.

[2] is true as the ILD for Bending Moment (BM) at the fixed end of a cantilever matches the Bending Moment Diagram (BMD) for a unit load at the free end. However, this reason does not explain the assertion regarding shear force.

We must first draw the influence line diagram for the SF at the section D,

For maximum positive SF at D, the loading should be applied as shown in the figure.
Maximum positive = load x area of ILD SF at D intensity covered by the load

To find the maximum bending moment at a section 6 meters from the left end of a 16-meter span girder with loads of 200 kN and 80 kN spaced 2 meters apart:
- Position the Loads: Place the 200 kN load at 6 meters from the left end.
- Calculate Reactions:
- Use equilibrium equations to find reactions at supports.
- Bending Moment at 6m:
- Calculate moment due to 200 kN directly at 6m.
- Add moment due to 80 kN, 2m away.

To get Max BM at C, place 200 kN at C and 80 kN at D, 2 m away from 200 kN an right side.

y/8 = 3.75/10 ⇒ y = 3 m

BM at C = ∑ (ordinate of ILD) × (Load)

= 200 × 3.75 + 80 × 3 = 990 kN.m

∴ Max BM at C will be 990 kN.m