This set of Structural Analysis Multiple Choice Questions & Answers (MCQs) focuses on “Qualitative Influence Lines”.
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
View Answer
Explanation: We need to remove the force for which we need to construct ILD and roller guide would remove the vertical reaction.
2. For ILD of shear at a point just left to point C on the BC part of the beam with AB = BC = 1m and CD = 2m, where A is a pin support, B is a hinge, and C and D are roller-type supports, what will be the equation for it?
a) X
b) -X
c) 2X
d) 1
View Answer
Explanation: By applying the Muller-Breslau principle, it is observed that the part to the right of point C cannot move as point C is right next to it. Therefore, the ILD will remain parallel to the x-axis.
3. If we draw the ILD of shear at a point just right to point C on the BC part of the beam with AB = BC = 1m and CD = 2m, what will be its slope?
a) 0.5
b) -.5
c) 1
d) -1
View Answer
Explanation: The slope of the ILD on the CD part of the beam will be -0.5, as initially, the value is 1 and finally, it reaches 0. Due to the Muller-Breslau principle, the lines on both sides will be parallel.
4. If we draw the ILD for shear at a point E, which lies between points C and D on the beam with AB = BC = 1m and CD = 2m, at how many points will the curve attain its peak?
a) 1
b) 2
c) 3
d) 4
View Answer
Explanation: At point E, the value will be +0.5, and since the slopes of the lines will be parallel, the value of the ILD at point B will also be 0.5.
5. What will be the lowest point of the ILD curve for the moment at a point just left to point C on the beam with AB = BC = 1m and CD = 2m?
a) -1
b) -2
c) -3
d) -4
View Answer
Explanation: For the moment, the curve is rotated by 1 radian. The lowest point will be -2 at hinge B because after this, the curve will change its slope.
6. What will be the equation for ILD of the vertical reaction at point A for the AB part of the beam with AB = BC = CD = DE = 1m, where point A is a fixed support, B and D are hinges, and C and E are pin roller supports?
a) 1X
b) 2X
c) 1
d) 2
View Answer
Explanation: Since point A is a fixed support, the ILD will remain parallel to the x-axis.
7. What will be the maximum point of the ILD of the vertical reaction at point C on the beam with AB = BC = CD = DE = 1m?
a) 1
b) 2
c) 3
d) 4
View Answer
Explanation: At point C, the ILD will be 1, and at point D, it will reach 2. After point D, the slope will change because D is a hinge.
8. What will be the area under the ILD curve for the vertical reaction at a point just left to point C on the beam with AB = BC = CD = DE = 1m?
a) 1
b) 1.5
c) 2
d) 2.5
View Answer
Explanation: The area will consist of a triangle with a base of 2m and height 1 for the segment CE, and a triangle with a base of 1m and height 1 for BC. It will be 0 for AB since A is a fixed support.
9. What will be the area under the ILD curve for the vertical reaction at a point just right to point C on the beam with AB = BC = CD = DE = 1m?
a) 1
b) 1.5
c) 2
d) 4
View Answer
Explanation: The area will be a triangle with dimensions 1m by 1m for CD, and a triangle with a base of 1m and height 1 for DE. It will be 0 for the rest of the beam.
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