If a beam is loaded with two point loads on the span, the bending moment diagram of the beam is

Multiple Choice Questions (MCQ) with Answers on Shear Force and Bending Moment diagram

1-A beam is a structural member which is subjected to

1. Axial tension or compression
2. Transverse loads and couples
3. Twisting moment
4. 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?

1. Only simply supported beams
2. Cantilever, overhanging and simply supported
3. Fixed beams
4. Continuous beams

(Ans: b)

3-A cantilever is a beam whose

1. Both ends are supported either on rollers or hinges
2. One end is fixed and other end is free
3. Both ends are fixed
4. 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

1. A horizontal line parallel to x-axis
2. A line inclined to x-axis
3. Follows a parabolic law
4. 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

1. A horizontal line parallel to x-axis
2. A line inclined to x-axis
3. Follows a parabolic law
4. Follows a cubic law

(Ans: d)

6-In a simply supported beam, bending moment at the end

1. Is always zero if it does not carry couple at the end
2. Is zero, if the beam has uniformly distributed load only
3. Is zero if the beam has concentrated loads only
4. May or may not be zero

(Ans: a)

7-For any part of the beam, between two concentrated load Shear force diagram is a

1. Horizontal straight line
2. Vertical straight line
3. Line inclined to x-axis
4. Parabola

(Ans: a)

8-For any part of a beam between two concentrated load, Bending moment diagram is a

1. Horizontal straight line
2. Vertical straight line
3. Line inclined to x-axis
4. Parabola

(Ans: c)

9-For any part of a beam subjected to uniformly distributed load, Shear force diagram is

1. Horizontal straight line
2. Vertical straight line
3. Line inclined to x-axis
4. Parabola

(Ans: c)

10-For any part of a beam subjected to uniformly distributed load, bending moment diagram is

1. Horizontal straight line
2. Vertical straight line
3. Line inclined to x-axis
4. Parabola

(Ans: d)

11-A sudden jump anywhere on the Bending moment diagram of a beam is caused by

1. Couple acting at that point
2. Couple acting at some other point
3. Concentrated load at the point
4. 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.

1. Maximum Bending moment = Wl/4 at the mid-point
2. Maximum Bending moment = Wl/4 at the end
3. Maximum Bending moment = Wl/8 at the mid-point
4. 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

1. Wl/8 or wl2/8 at the mid-point
2. Wl/8 or wl2/8 at the end
3. Wl/4 or wl2/4
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

1. Wl at the free end
2. Wl at the fixed end
3. Wl/2 at the fixed end
4. 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

1. Maximum
2. Zero
3. Either increasing or decreasing
4. 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

1. Where shear force=0
2. At the free end
3. At the fixed end
4. At the mid-point

(Ans: c)

18-Point of contra-flexure is a

1. Point where Shear force is maximum
2. Point where Bending moment is maximum
3. Point where Bending moment is zero
4. Point where Bending moment=0 but also changes sign from positive to negative

(Ans: d)

19-Point of contra-flexure is also called

1. Point of maximum Shear force
2. Point of maximum Bending moment
3. Point of inflexion
4. Fixed end

(Ans: c)

20-The slope of shear force line at any section of the beam is also called

1. Bending moment at that section
2. Rate of loading at that section
3. Maximum Shear force
4. 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)')

Translated by