Structural Analysis Multiple Choice Questions on “Qualitative Influence Lines”.
While writing influence line equations, left most point is always considered as origin and following sign convention is followed.
1. If we require to construct ILD of vertical support at a pin joint, then according to Muller-Breslau principle, by which type of support should it be replaced?
a) Roller guide
b) Pin roller
c) Fixed support
d) Hinge
Answer: a
Clarification: We need to remove the force for which we need to construct ILD and roller guide would remove the vertical reaction.
Following figure is for Q2-Q5.
A is a pin support, B is a hinge, C and D are roller type support.
AB = BC = 1m, CD = 2m
2. For ILD of shear at a point just left to C, what will be the equation for it on BC part of beam?
a) X
b) -X
c) 2X
d) 1
Answer: d
Clarification: On applying Muller-Breslau principle, we will see that the part right to the point can’t move as point C is right next to it. So, ILD will have to remain parallel to x axis.
3. If we draw ILD of shear at appoint just right to point C, then what will be its slope at BC part of beam?
a) 0.5
b) -.5
c) 1
d) -1
Answer: b
Clarification: Slope of ILD in part CD of beam will be -0.5 as initially it will be at 1 and finally at 0. Now, due to Muller-Breslau principle, both lines will be parallel.
4. If we draw ILD for shear at a point E (lying in between points C & D), then at how many points would his curve attain its peak?
a) 1
b) 2
c) 3
d) 4
Answer: b
Clarification: At E, value will be +0.5 and since slope of two lines will be parallel, value of ILD at B too will be 0.5.
5. What will be the lowest point of ILD curve for moment at a point just left to point C?
a) -1
b) -2
c) -3
d) -4
Answer: b
Clarification: For moment, we rotate it by 1 rad. So, min. point will be -2 at hinge B as after that the curve will change its slope.
Point A is fixed support, B and D are hinges and C and E are pin roller supports.
AB = BC = CD = DE=1m
6. What will be the equation for ILD of vertical reaction at point A for AB part of beam?
a) 1X
b) 2X
c) 1
d) 2
Answer: c
Clarification: Since, point A is a fixed support, ILD will be parallel to x axis.
7. What will be the maximum point of ILD of vertical reaction at point C?
a) 1
b) 2
c) 3
d) 4
Answer: b
Clarification: It will be 1 at point C and will be 2 at point D after which it will change its slope as D is a hinge.
8. What will be the area under the ILD curve if we make it for the vertical reaction at a point just left to point C?
a) 1
b) 1.5
c) 2
d) 2.5
Answer: b
Clarification: It will be a triangle with base 2m and height 1 for CE and triangle with base 1 and height 1 for BC. It will be 0 for AB as A is a fixed support.
9. What will be the area under the ILD curve if we make it for the vertical reaction at a point just right to point C?
a) 1
b) 1.5
c) 2
d) 4
Answer: b
Clarification: It will be a triangle with 1*1 dimension for CD and a triangle with base 1 and height 1 for DE. It will be 0 for rest of the beam.