Engineering Questions with Answers - Multiple Choice Questions
MCQs on Shear Force and Bending Moment
MCQs on _______ is a horizontal structural member subjected to transverse loads perpendicular to its axis. a) Strut b) Column c) Beam d) Truss
Example for cantilever beam is ______ a) Portico slabs b) Roof slab c) Bridges d) Railway sleepers
The diagram depicts _______ kind of beam.
c) Over hanging
d) Propped cantilever
Fixed beam is also known as __________ a) Encastered beam b) Built on beam c) Rigid beam d) Tye beam
U.D.L stands for? a) Uniformly diluted length b) Uniformly developed loads c) Uniaxial distributed load d) Uniformly distributed loads
Given below diagram is ______ load.
a) Uniformly distributed load
b) Uniformly varying load
c) Uniformly decess load
d) Point load
Moving train is an example of ____ load. a) Point load b) Cantered load c) Rolling load d) Uniformly varying load
Continuous beams are _________ a) Statically determinate beams b) Statically indeterminate beams c) Statically gravity beams d) Framed beams
A beam which extends beyond it supports can be termed as __________ a) Over hang beam b) Over span beam c) Isolated beams d) Tee beams
Explanation: A Beam extended beyond its support. And the position of extension is called as over hung portion.
Units of U.D.L? a) KN/m b) KN-m c) KN-m×m d) KN
MCQs on Introduction to Shear Force and Bending Moment
Shear force is unbalanced _____ to the left or right of the section.
a) Horizontal force
b) Vertical force
c) Inclined force
d) Conditional force
Explanation: The shear force at the cross section of a beam may also be defined as the unbalanced vertical force to the left or right of the section. It is also the algebraic sum of all the forces I get to the left to the right of the section.
SI units of shear force is _______________
Explanation: As shear force at any section is equal to the algebraic sum of the forces, the units of the shear force are also in kilo newtons and it is denoted by kN.
Determine the moment at fixed end.
a) 40 kNm
b) 50 kNm
c) 60 kNm
d) 80 kNm
Explanation: Let the fixed end be “A” Reaction at A = 10×4 = 40 kN Moment at A = (10×4)×4/2 = 80 kNm. advertisement
Shear force is diagram is _______ representation of shear force plotted as ordinate.
Explanation: Shear Force diagram is a graphical representation of the shear force plotted as ordinate on baseline representing the axis of the Beam.
a) Negative bending moment
b) Positive shear force
c) Positive bending moment
d) Negative shear force
Explanation: The bending moment at a section is considered to be negative when it causes convexity upwards or concavity at bottom, such bending moment is called hogging bending moment or negative bending moment.
At the point of contraflexure, the value of bending moment is ____________
c) Can’t be determined
Explanation: A point at which bending moment changes its sign from positive to negative and vice versa. Such point is termed as point of contraflexure. At this point, the value of bending moment is zero (0).
_________ positive/negative bending moments occur where shear force changes its sign.
d) Remains same
Explanation: If shear force and bending moment values obtained are thus plotted as a diagram, the SF & BM relationship always behaves vice versa.
Shear force of following diagram
Explanation: SF @ AB is 10 kN
FA = 10 kN
FB = 10 kN.
SI units of Bending moment is ___________
Explanation: Moment is a product of force and perpendicular distance and the bending moment is the algebraic sum of moments taken away from the left or the right of the section hence the SI units of bending moment is same as the moment i.e kNm.
What is the other name for a positive bending moment?
Explanation: The bending moment at a section is considered to be positive when it causes convexity downwards such bending moment is called sagging bending moment positive bending moment.
MCQs on Types of Supports
A simple support offers only _______ reaction normal to the axis of the beam.
Explanation: In a simple support there will not be any resistance to horizontal loads, moment or rotation. In fact, it only offers a vertical reaction normal to the axis of the beam.
To avoid _____ stresses in beams, one end of the beam is placed on the rollers.
Explanation: Roller support reaction is normal to the axis of the beam. In case the beam subjected to trust or to avoid temperature stresses in the beam, one end of the beam is placed on roller because it facilitate free horizontal movement of end. It is similar to simple support.
________ support develops support moment.
Explanation: A fixed support offers resistance against horizontal and vertical movement and against the rotation of the member and that in turn developers support moment. advertisement
Hinge support is called as __________
a) Socket joint
b) Swivel joint
c) Ball joint
d) Pin joint
Explanation: Hinge support is one, in which the position is fixed but not the direction. In their words hinged support offers resistance against vertical and horizontal moments.it is fixed in such a way that it resembles like a pin joint.
Name the support from following figure.
a) Hinge support
b) Fixed support
c) Free support
d) Roller support
Explanation: In the above figure we can observe that the beam is supported at both the ends so the beam is fixed at both ends. Hence the support is a fixed support.
For a simply supported beam, the moment at the support is always __________
d) Cannot be determined
Explanation: As the moment is a product of force and perpendicular distance, the flexural moment at the support is zero because there is no distance at the support.
“Hinged support offers resistance against rotation”.
Explanation: A hinged support offers resistance against horizontal and vertical movement but not against rotation. It support offers a vertical and horizontal reaction only.
Find the reaction at simple support A?
a) 6.5 kN
b) 9 kN
c) 10 kN
d) 7.5 kN
Explanation: Total load = 10 kN Taking moment at A = 0 4 × R @ B – 10 = 0 R @ B = 2.5 kN Reaction at A = 10 – 2.5 = 7.5kN.
Roller support is same as _____
a) Hinged support
b) Fixed support
c) Simply support
d) Roller support
Explanation: The support reaction is normal to the axis of the beam. It facilitates the vertical support. It helps the beam to overcome the temperature stresses effectively. It is similar to simple support.
Hinged supports offers vertical and ________ reaction.
Explanation: A hinged support offers a vertical and horizontal reaction. The pin jointed support offers resistance against horizontal and vertical movements but not against rotation movement.
MCQs on Maximum Shear Force
Which of these is the correct way of sign convention for shear force?
a) R U P
b) L U P
c) R U N
d) L D P
Explanation: According to the theoretical approach, there are many sign conventions to follow but the standard one is “right upwards negative” the sign convention is thoroughly followed unanimously.
At hinge, the moments will be _________
Explanation: At the support of a member, there is no distance prevailing to take the upcoming load. As we know the moment is a product of force and perpendicular distance, but at hinge (end support) the distance is zero. Hence the moment developed is zero.
What is variation in SFD, if the type of loading in the simply supported beam is U.D.L is ____
Explanation: The shear force is defined as the algebraic sum of all the forces taken from any one of the section. If you figure out the SFD for a simply supported beam carrying U.D.L throughout its entire length, in the SFD we can observe that shear force is same at supports. In the centre, the shear force is zero. Hence the diagram varies linearly. advertisement
The rate of change of shear force is equal to _____
a) Direction of load
b) Change in BMD
c) Intensity of loading
d) Maximum bending
Explanation: Consider a simply supported beam subjected to udl for the entire span considered a free body diagram of small portion of elemental length dx.
Let the shear force at left of the section is = F
Let the increase in shear force in length of the dx = dF
Let the Indian city of load on this part of the beam = w
Total downward load in this elemental length = wdx
€V = 0
dF = -wdx
dF/dx = -w
This rate of change of shear force at any section is equal to the intensity of loading at that section.
The shear force in a beam subjected to pure positive bending is _____
d) Cannot determine
Explanation: In the determination of shear force and bending moment diagrams it is clear that shear force changes its sign when the bending moment in a beam is maximum and the shear force in a beam subjected to pure positive bending will be zero as the neutralizing effect comes under.
In SFD, vertical lines are for ______
a) Point loads
Explanation: Shear Force diagram started from left side of the m as per the load. For point loads draw vertical lines and under UDL draw slope lines.
A cantilever beam loaded with udl throughout, the maximum shear force occurs at____
a) Free end
b) Fixed end
c) At centre
d) At point of contraflexure
Explanation: In a case of a cantilever beam subjected to udl, at the free end there will be zero shear force because, we need to convert udl to load by multiplying with distance. Hence at the fixed end the shear force is w×l i.e (maximum).
A simply supported beam of span 1 m carries a point load “w” in centre determine the shear force in the half left of the beam.
Explanation: Let the two ends of the beam be A and B, the given load on a beam is symmetrical hence RA = RB= W/2. SFD at any section in the left of the beam is equal to the W/2. SFDat any section in the right half of the beam is equal to -W/2.
At the Point of contraflexure, what is the value of bending moment?
Explanation: Point of contraflexure in a beam is a point at which bending moment changes its sign from positive to negative and vice versa. At the point of contraflexure, the value of bending moment is zero.
When SF is zero, the bending moment is _____
c) Very difficult to say
Explanation: When is shear force changes its sign, the bending moment in a beam will be either maximum positive or maximum negative. This is because of the sign convention adopted.
MCQs on Maximum Bending Moment
A cantilever beam subjected to point load at its free end, the maximum bending moment develops at the ________ of the beam.
a) Free end
b) Fixed end
d) Point of inflection
Explanation: As the moment is the product of perpendicular distance and force. In cantilever beam, at its free end the moment will be zero as there is no distance, but at the fixed end the moment is maximum that is W×l.
Bending moment in a beam is maximum when the _________
a) Shear force is minimum
b) Shear force is maximum
c) Shear force is zero
d) Shear force is constant
Explanation: The maximum bending moment occurs in a beam, when the shear force at that section is zero or changes the sign because at point of contra flexure the bending moment is zero.
Positive bending moment is known as _______
Explanation: The positive bending moment in a section is considered because it causes convexity downwards. Such bending moment is called a sagging bending moment or positive bending moment. advertisement
A simply supported beam of span “x” meters carries a udl of “w” per unit length over the entire span, the maximum bending moment occurs at _____
a) At point of contra flexure
c) End supports
d) Anywhere on the beam
Explanation: As we know that BM occurs at center. Because at supports the moment is obviously zero.
At the centre, maximum bending moment is wl2/8.
The maximum BM is ______
a) 40 kNm
b) 50 kNm
c) 90 kNm
d) 75 kNm
Explanation: Above diagram depicts cantilever beam subjected to point load at the free end. The maximum bending moment at A is W × I = 30 × 3 = 90 kNm.
Bending moment can be denoted by ____
Explanation: Bending moment is the product of force and perpendicular distance. Units are kNm It is denoted by “M”. Whereas SF is denoted by “F”.
Number of points of contra flexure for a double over hanging beam.
Explanation: Point of contraflexure in a beam is a point at which bending moment changes its sign from positive to negative and vice versa. In the case of overhanging beam, there will be two points of contraflexure.
Maximum bending moment in a cantilever beam subjected to udl (w)over the entire span (l).
Explanation: In a cantilever beam the maximum bending moment occurs at the fixed end. Moment at the free end is 0 and maximum at the fixed end. Maximum shear force is w×l.
Determine the maximum bending moment for the below figure.
Explanation: First of all, let’s assume the length between end supports be ”l” the maximum bending moment in a simply supported beam with point load at its centre is wl/4. We know that in simply supported beam the maximum bending moment occurs at the centre only.
What is the variation in the BM, if the simply supported beam carries a point load at the centre.
d) Other quadrilateral
Explanation: For simply supported beam with point load at the centre, the maximum bending moment will be at the centre i.e. wl/4. The variation in bending moment is triangular.
MCQs Shear Force and Bending Moment diagram
What is the bending moment at end supports of a simply supported beam?
Explanation: At the end supports, the moment (couple) developed is zero, because there is no distance to take the perpendicular acting load. As the distance is zero, the moment is obviously zero.
What is the maximum shear force, when a cantilever beam is loaded with udl throughout?
Explanation: In cantilever beams, the maximum shear force occurs at the fixed end. In the free end, there is zero shear force. As we need to convert the udl in to load, we multiply the length of the cantilever beam with udl acting upon. For maximum shear force to obtain we ought to multiply load and distance and it surely occurs at the fixed end (w×l).
Sagging, the bending moment occurs at the _____ of the beam.
a) At supports
b) Mid span
c) Point of contraflexure
d) Point of emergence
Explanation: The positive bending moment is considered when it causes convexity downward or concavity at top. This is sagging. In simply supported beams, it occurs at mid span because the bending moment at the supports obviously will be zero hence the positive bending moment occurs in the mid span. advertisement
What will be the variation in BMD for the diagram? [Assume l = 2m].
Explanation: At support B, the BM is zero. The beam undergoes maximum BM at fixed end.
By joining the base line, free end and maximum BM point. We obtain a right angled triangle.
What is the maximum bending moment for simply supported beam carrying a point load “W” kN at its centre?
a) W kNm
b) W/m kNm
c) W×l kNm
d) W×l/4 kNm
Explanation: We know that in simply supported beams the maximum BM occurs at the central span. Moment at A = Moment at B = 0 Moment at C = W/2 × l/2 = Wl/ 4 kNm (Sagging).
How do point loads and udl be represented in SFD?
a) Simple lines and curved lines
b) Curved lines and inclined lines
c) Simple lines and inclined lines
d) Cant represent any more
Explanation: According to BIS, the standard symbols used for sketching SFD are
Point load = ———–
Udl load = \strength-materials-questions-answers-shear-force-bending-moment-diagram-q6
________ curve is formed due to bending of over hanging beams.
Explanation: The line to which the longitudinal axis of a beam bends or deflects or deviates under given load is known as elastic curve on deflection curve. Elastic curve can also be known as elastic line or elastic axis.
The relation between slope and maximum bending moment is _________
a) Directly proportion
b) Inversely proportion
c) Relative proportion
d) Mutual incidence
Explanation: The relationship between slope and maximum bending moment is inversely proportional because, For example in simply supported beams slope is maximum at supports and zero at midspan of a symmetrically loaded beam where as bending moment is zero at supports and maximum at mid span. Hence we conclude that slope and maximum bending moment are inversely proportional to each other in a case of the simply supported beam.
What is the SF at support B?
a) 5 kN
b) 3 kN
c) 2 kN
d) 0 kN
Explanation: Total load = 2×2 = 4kN Shear force at A = 4 kN ( same between A and C ) Shear force at C = 4 kN Shear force at B = 0 kN Maximum SF at A = 4 kN.
Where do the maximum BM occurs for the below diagram.
a) -54 kNm
b) -92 kNm
c) -105 kNm
d) – 65 kNm
Explanation: Moment at B = 0 Moment at C = – (10 × 3) × (3/2) = – 45 kNm Moment at A = – (10 × 3) × (1.5 + 2 ) Maximum BM at A = – 105 kNm = 105 Nm (hogging).