Multiple Choice Questions (MCQ) with Answers on Shear Force and Bending Moment diagram
1-A beam is a structural member which is subjected to
- Axial tension or compression
- Transverse loads and couples
- Twisting moment
- No load, but its axis should be horizontal and x-section rectangular or circular
(Ans: b)
2-Which of the following are statically determinate beams?
- Only simply supported beams
- Cantilever, overhanging and simply supported
- Fixed beams
- Continuous beams
(Ans: b)
3-A cantilever is a beam whose
- Both ends are supported either on rollers or hinges
- One end is fixed and other end is free
- Both ends are fixed
- Whose both or one of the end has overhang
(Ans: b)
4-In a cantilever carrying a uniformly varying load starting from zero at the free end, the shear force diagram is
- A horizontal line parallel to x-axis
- A line inclined to x-axis
- Follows a parabolic law
- Follows a cubic law
(Ans: c)
5-In a cantilever carrying a uniformly varying load starting from zero at the free end, the Bending moment diagram is
- A horizontal line parallel to x-axis
- A line inclined to x-axis
- Follows a parabolic law
- Follows a cubic law
(Ans: d)
6-In a simply supported beam, bending moment at the end
- Is always zero if it does not carry couple at the end
- Is zero, if the beam has uniformly distributed load only
- Is zero if the beam has concentrated loads only
- May or may not be zero
(Ans: a)
7-For any part of the beam, between two concentrated load Shear force diagram is a
- Horizontal straight line
- Vertical straight line
- Line inclined to x-axis
- Parabola
(Ans: a)
8-For any part of a beam between two concentrated load, Bending moment diagram is a
- Horizontal straight line
- Vertical straight line
- Line inclined to x-axis
- Parabola
(Ans: c)
9-For any part of a beam subjected to uniformly distributed load, Shear force diagram is
- Horizontal straight line
- Vertical straight line
- Line inclined to x-axis
- Parabola
(Ans: c)
10-For any part of a beam subjected to uniformly distributed load, bending moment diagram is
- Horizontal straight line
- Vertical straight line
- Line inclined to x-axis
- Parabola
(Ans: d)
11-A sudden jump anywhere on the Bending moment diagram of a beam is caused by
- Couple acting at that point
- Couple acting at some other point
- Concentrated load at the point
- Uniformly distributed load or Uniformly varying load on the beam
(Ans: a)
12-In a simple supported beam having length = l and subjected to a concentrated load (W) at mid-point.
- Maximum Bending moment = Wl/4 at the mid-point
- Maximum Bending moment = Wl/4 at the end
- Maximum Bending moment = Wl/8 at the mid-point
- Maximum Bending moment = Wl/8 at the end
(Ans: a)
13-In a simply supported beam subjected to uniformly distributed load (w) over the entire length (l), total load=W, maximum Bending moment is
- Wl/8 or wl2/8 at the mid-point
- Wl/8 or wl2/8 at the end
- Wl/4 or wl2/4
- Wl/2
(Ans: a)
14-In a cantilever subjected to a concentrated load (W) at the free end and having length =l, Maximum bending moment is
- Wl at the free end
- Wl at the fixed end
- Wl/2 at the fixed end
- Wl at the free end
(Ans: b)
15-An axle is subjected to loads as shown
Maximum bending moment is
(Ans: c)
16-At a point in a simply supported or overhanging beam where Shear force changes sign and = 0, Bending moment is
- Maximum
- Zero
- Either increasing or decreasing
- Infinity
(Ans: a)
17-In a cantilever subjected to a combination of concentrated load, uniformly distributed load and uniformly varying load, Maximum bending moment is
- Where shear force=0
- At the free end
- At the fixed end
- At the mid-point
(Ans: c)
18-Point of contra-flexure is a
- Point where Shear force is maximum
- Point where Bending moment is maximum
- Point where Bending moment is zero
- Point where Bending moment=0 but also changes sign from positive to negative
(Ans: d)
19-Point of contra-flexure is also called
- Point of maximum Shear force
- Point of maximum Bending moment
- Point of inflexion
- Fixed end
(Ans: c)
20-The slope of shear force line at any section of the beam is also called
- Bending moment at that section
- Rate of loading at that section
- Maximum Shear force
- Maximum bending moment
(Ans: b)
I was to prepare the Shear force diagram and bending moment diagram for simply supported beam with UDL acting throughout the beam and two Point Loads anywhere on the beam. I was able to determine the Shear Force Diagram, but currently I'm struggling with the Bending moment diagram. I'm not too familiar with the Matlab I have to say, so if someone could assist me I will be grateful. In this file user inputs the values for: Span UDl Point Load 1 Point Load 1 distance from LHS Point Load 2 Point Load 2 distance from LHS I have it worked out on paper, but I cannot do it in Matlab. Span=input('Please enter the length of the beam in meteres: '); UDL=input('Please specify the magniture of UDL in kN/m: '); P1=input('Please enter a magnitude of the Point Load 1 in kN: '); PL1=input('Please enter the distance at which Point load 1 is acting from left hand side in meteres: '); P2=input('Please enter a magnitude of the Point Load 2 in kN: '); PL2=input('Please enter the distance at which Point load 2 is acting from left hand side in meteres: '); R1 = (P1*(Span-PL1)/Span)+(P2*(Span-PL2)/Span)+(UDL*Span/2) R2 = ((P1*PL1)+(P2*PL2)+(UDL*Span*(Span/2)))/10 else SF(i)=R1-(UDL*D*(i))-P1-(UDL*(PL2-PL1))-(UDL*(Span-PL2))+R2; BMO(i)=R1*x(i)-UDL*((x(i)-PL1)^2)/2; ; else BM(i)=(R1*D*(i))-(UDL*D*(i))-P1-(UDL*(PL2-PL1))-(UDL*(Span-PL2))+R2; title('Span of the Beam') plot(x,SF,'m',x,SFO,'b','linewidth',1.5) title('Shear Force Diagram') ylabel('Shear Force (kN)') plot(x,BM,'m',x,BMO,'b','linewidth',1.5) title('Bending Moment Diagram') ylabel('Bending Moment (kN-m)')
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