2- handouts lecture-33 to 36

COURSE:COURSE: CE 201 (STATICS)CE 201 (STATICS)

LECTURE NO.:LECTURE NO.: 33 to 3633 to 36

FACULTY:FACULTY: DR. SHAMSHAD AHMADDR. SHAMSHAD AHMAD

DEPARTMENT:DEPARTMENT: CIVIL ENGINEERINGCIVIL ENGINEERING

UNIVERSITY:UNIVERSITY: KING FAHD UNIVERSITY OF PETROLEUM KING FAHD UNIVERSITY OF PETROLEUM & MINERALS, DHAHRAN, SAUDI ARABIA& MINERALS, DHAHRAN, SAUDI ARABIA

TEXT BOOK:TEXT BOOK: ENGINEERING MECHANICSENGINEERING MECHANICS--STATICS STATICS by R.C. HIBBELER, PRENTICE HALLby R.C. HIBBELER, PRENTICE HALL

LECTURE NO. 33 to 36LECTURE NO. 33 to 36SHEAR AND MOMENT EQUATIONS AND DIAGRAMSSHEAR AND MOMENT EQUATIONS AND DIAGRAMS

Objectives:Objectives: To show how to form the equations of shear force To show how to form the equations of shear force

((VV) and bending moment () and bending moment (MM) and how to plot the ) and how to plot the shear force and bending moment diagramsshear force and bending moment diagrams

SHEAR AND MOMENT EQUATIONS AND DIAGRAMS SHEAR AND MOMENT EQUATIONS AND DIAGRAMS ProcedureProcedure

The equations of shear force, V, and bending moment, M, in terms of the distance, x, along the axis of a member can be obtained by usingthe method of sections, as follows: Identify the points of discontinuity (e.g., points where a

distributed load changes or where concentrated forces or couplemoments are applied) along the member.

The V and M equations must be obtained separately for eachsegment of the member located between any two discontinuities ofloadings.

For example, for the beam shown in figure below, sections locatedat distances x1, x2, and x3 will have to be used to describe thevariation of V and M throughout the length of the beam

SHEAR AND MOMENT EQUATIONS AND DIAGRAMS SHEAR AND MOMENT EQUATIONS AND DIAGRAMS ProcedureProcedure

Draw the F.B.D. at the sections located at x1, x2, and x3 and obtain the equations for V and M separately for each segment by applying the equilibrium conditions

These equations for V and M will be valid only within the regionsfrom O to a for x1, from a to b for x2, and from b to L for x3

Using the equations of V and M, their ordinates for different values of x may be calculated and V and M diagrams may be plotted.

V and M diagrams showing the variations of V and M along the axis of a member are required for the design of the member.

SHEAR AND MOMENT EQUATIONS AND DIAGRAMS SHEAR AND MOMENT EQUATIONS AND DIAGRAMS Sign Convention for Shear Force and Bending MomentSign Convention for Shear Force and Bending Moment

As shown in the above figure: A shear force, which causes clockwise rotation of the member on

which it acts, is considered to be with positive sign.

A bending moment, which causes compression or pushing on theupper part of the member on which it acts, is considered to be withpositive sign. Also, a positive bending moment tends to bend themember in concave upward manner.

PROBLEM SOLVING: PROBLEM SOLVING: Example # 1Example # 1

For the beam shown below, draw the SFD and BMD.


Support Reactions:

Reactions at fixed support A can be determined by applying equilibriumconditions to the free-body diagram of the entire beam, as follows:

A x

5 ft 5 ft

BC

MA 100 lb800 lb-ft

5 ft 5 ft

BC

MA 100 lb800 lb-ft

A y

A x

5 ft 5 ft

BC

MA 100 lb800 lb-ft

5 ft 5 ft

BC

MA 100 lb800 lb-ft

A y5 ft 5 ft

BC

MA 100 lb800 lb-ft

5 ft 5 ft

BC

MA 100 lb800 lb-ft

A y

Fx = 0 Ax = 0 Fy = 0 Ay 100 = 0 Ay = 100 lb.

M about A = 0 MA + 800 + 100 10 = 0 MA = 1800 lb-ft. Shear and Moment Equations:

Segment AB (0 x < 5) ft

The equations for V and M for segment AB can be obtained by applying equilibrium conditions tothe free-body diagram of the segment AB, as follows:

x

M

100 lb

1800 lb-ft

V

x

M

100 lb

1800 lb-ft

V

Fy = 0 100 V = 0 V = 100 lb M about cut section = 0 1800 + 100 x M = 0 M = (100 x 1800) lb-ft


A x

5 ft 5 ft

BC

MA 100 lb800 lb-ft

5 ft 5 ft

BC

MA 100 lb800 lb-ft

A y

A x

5 ft 5 ft

BC

MA 100 lb800 lb-ft

5 ft 5 ft

BC

MA 100 lb800 lb-ft

A y5 ft 5 ft

BC

MA 100 lb800 lb-ft

5 ft 5 ft

BC

MA 100 lb800 lb-ft

A y

Segment BC (5 < x 10) ft

The equations for V and M for segment BC can be obtained by applyingequilibrium conditions to the free-body diagram of the segment BC, as follows:

B

800 lb-ft

x

M

100 lb

1800 lb-ft

V

B800 lb-ft

x

M

100 lb

1800 lb-ft

V

x

M

100 lb

1800 lb-ft

V

Fy = 0 100 V = 0 V = 100 lb M about cut section = 0 1800 + 100 x + 800 M = 0 M = (100 x 1000) lb-ft

Shear and Moment Values for Plotting the V and M Diagrams: Segment AB (0 x < 5) ft

The V and M equations obtained for segment AB are as follows:

V = 100 lb and M = (100 x 1800) lb-ft V at any point between A and B = 100 lb M at A = 100 0 1800 = 1800 lb-ft M at B = 100 5 1800 = 1300 lb-ft

Segment BC (5 < x 10) ft

The V and M equations obtained for segment BC are as follows:

V = 100 lb and M = (100 x 1000) lb-ft

V at any point between B and C = 100 lb M at B = 100 5 1000 = 500 lb-ft M at C = 100 10 1000 = 0

PROBLEM SOLVING: PROBLEM SOLVING: Example # 1Example # 1V at any point between A and B = 100 lb M at A = 100 0 1800 = 1800 lb-ft M at B = 100 5 1800 = 1300 lb-ft

V at any point between B and C = 100 lb M at B = 100 5 1000 = 500 lb-ft M at C = 100 10 1000 = 0

V and M Diagrams:

Following are the V and M diagrams plotted using the calculated values of Vand M:

BC

V

100 lb

800 lb -ft

500 lb-ft1800 lb -ft

M

100 lb

800 lb -ft

100 lb

x

x

V - diagram

M - diagram

BC

V

100 lb

800 lb -ft


M

100 lb

800 lb -ft

100 lb

x

x

V - diagram

M - diagram1300 lb

BC

V

100 lb

800 lb -ft


M

100 lb

800 lb -ft

100 lb

x

x

V - diagram

M - diagram

BC

V

100 lb

800 lb -ft


M

100 lb

800 lb -ft

100 lb

x

x

V - diagram

M - diagram

BC

V

100 lb

800 lb -ft


M

100 lb

800 lb -ft

100 lb

x

x

V - diagram

M - diagram

BC

V

100 lb

800 lb -ft


M

100 lb

800 lb -ft

100 lb

x

x

V - diagram

M - diagram1300 lb

Note: As can be seen from the M-diagram, the value of M is changing from 1300 lb-ft to 500 lb-ft at point B due to the coupleof 800 lb-ft applied at point B.


For the beam shown below, draw the AFD, SFD and BMD.

A

B

C

50N4

3A

B

C

50N4

3

3m 3m3m 3m3m 3m

A

B

C

50N4

3A

B

C

50N4

3

3m 3m3m 3m3m 3m


A

B

C

50N4

3A

B

C

50N4

3

3m 3m3m 3m3m 3m

A

B

C

50N4

3A

B

C

50N4

3

3m 3m3m 3m3m 3m

Support Reactions:

Reactions at supports A and C can be determined by applying equilibriumconditions to the free-body diagram ofthe entire beam, as follows:

30 N

40 N

B C

RAy RCy3m 3m

RCxA 30 N

40 N

B C

RAy RCy3m 3m

RCx30 N

40 N

B C

RAy RCy3m 3m

RCxA 30 N

40 N

B C

RAy RCy3m 3m

RCx

Fx = 0 RCx + 30 = 0 RCx = 30 N = 30 N ()

M about C = 0 RAy 6 40 3 = 0 RAy = 20 N Fy = 0 RAy 40 + RCy = 0 RCy = 40 RAy = 40 20 = 20 N Normal, Shear and Moment Equations:

Segment AB (0 x < 3) m

The equations for N, V and M for segment AB can be obtained by applying equilibrium conditions to the free-body diagram of the segment AB, as follows:

A

xM

N

V20N

A

xM

N

V20N

Fy = 0 20 V = 0 V = 20 N Fx = 0 N = 0

M about cut section = 0 20 x M = 0 M = (20 x) N-m

PROBLEM SOLVING: PROBLEM SOLVING: Example # 2Example # 230 N

40 N

B C

RAy RCy3m 3m

RCxA 30 N

40 N

B C

RAy RCy3m 3m

RCx30 N

40 N

B C

RAy RCy3m 3m

RCxA 30 N

40 N

B C

RAy RCy3m 3m

RCx

Segment BC (3 < x 6) m

The equations for N, V and M for segment BC can be obtained by applyingequilibrium conditions to the free-body diagram of the segment BC, as follows:

A

x

M

N

V20N

B

40N30N

A

x

M

N

V20N

B

40N30N

Fx = 0 30 + N = 0 N = 30 N

M about cut section = 0 20 x 40 (x3) M = 0 M = (120 20 x) N-m

Fy = 0 20 40 V = 0 V = 20 N

Normal, Shear and Moment Values forPlotting the N, V and M Diagrams: Segment AB (0 x < 3) m

The N, V and M equations obtained for segment AB are as follows:

N = 0, V = 20 N and M = (20 x) N-m N at any point between A and B = 0 V at any point between A and B = 20 N M at A = 20 0 = 0 M at B = 20 3 = 60 N-m Segment BC (3 < x 6) m

The N, V and M equations obtained for segment BC are as follows:

N = 30 N, V = 20 N and M = (120 20 x) N-m N at any point between B and C = 30 N V at any point between B and C = 20 N M at C = 120 20 6 = 0


N at any point between A and B = 0 V at any point between A and B = 20 N M at A = 20 0 = 0 M at B = 20 3 = 60 N-m

N at any point between B and C = 30 N V at any point between B and C = 20 N M at C = 120 20 6 = 0

N, V and M Diagrams:

Following are the N, V and M diagrams plotted using the calculated values of N, V and M:

A

B

C

50N4

3

A 30 N

40 N

B C

RAy

x

RCy3m 3m

RCx

-

+

-

+

N

-30+20

-30+20

-20-20

+60

V (N)

M = (N-m)

A

B

C

50N4

3

A 30 N

40 N

B C

RAy

x

RCy3m 3m

RCxA 30 N

40 N

B C

RAy

x

RCy3m 3m

RCx

--

+

-

+

-

++

N (N)

-30+20

-30+20

-20-20

+60

V (N)

M = (N-m)

N-Diagram

V-Diagram

M-Diagram

A

B

C

50N4

3

A 30 N

40 N

B C

RAy

x

RCy3m 3m

RCx

-

+

-

+

N

-30+20

-30+20

-20-20

+60

V (N)

M = (N-m)

A

B

C

50N4

3

A 30 N

40 N

B C

RAy

x

RCy3m 3m

RCxA 30 N

40 N

B C

RAy

x

RCy3m 3m

RCx

--

+

-

+

-

++

N (N)

-30+20

-30+20

-20-20

+60

V (N)

M = (N-m)

A

B

C

50N4

3

A 30 N

40 N

B C

RAy

x

RCy3m 3m

RCx

-

+

-

+

N

-30+20

-30+20

-20-20

+60

V (N)

M = (N-m)

A

B

C

50N4

3

A 30 N

40 N

B C

RAy

x

RCy3m 3m

RCxA 30 N

40 N

B C

RAy

x

RCy3m 3m

RCx

--

+

-

+

-

++

N (N)

-30+20

-30+20

-20-20

+60

V (N)

M = (N-m)

N-Diagram

V-Diagram

M-Diagram


For the beam shown below, draw theSFD and BMD.

A B

20N/m

A B

6 m

A B

20N/m

A B

6 m


A B

20N/m

A B

6 m

A B

20N/m

A B

6 m

A B

RAy R6m

RA B

RAy RBy6m

RBx

20 6 = 120 N

3 mA B

RAy R6m

RA B

RAy RBy6m

RBx

20 6 = 120 N

3 m

Support Reactions:

Reactions at supports A and B can be determined by applying equilibriumconditions to the free-body diagram ofthe entire beam, as follows:

Fx = 0 RBx = 0 M about B = 0 6 RAy 120 3 = 0 RAy = 60 N Fy = 0 RAy 120 + RBy = 0 RBy = 120 RAy = 120 60 = 60 N

Shear and Moment Equations:

Since there is no any point of discontinuity, the equations for V and Mcan be obtained for entire beam byapplying equilibrium conditions to thefree-body diagram of the section cut at adistance x from support A, as follows:

x

A

M

N

V

60N

20N/m

xx

A

M

N

V

20N/m

x

A

M

N

V

60N

20N/m

xx

A

M

N

V

20N/m

Fy = 0 60 20 x V = 0 V = (60 20 x) N M about cut section = 0 60 x 20 x 2

x M = 0

M = (60 x 10 x2) N-m

PROBLEM SOLVING: PROBLEM SOLVING: Example # 3Example # 3Shear and Moment Values for Plottingthe V and M Diagrams: The V and M equations obtained for theentire beam AB are as follows:

V = (60 20 x) N M = (60 x 10 x2) N-m V at A = 60 20 0 = 60 N V at B = 60 20 6 = 60 N M at A = 60 x 10 x2= 60 0 10 02 = 0M at B = 60 x 10 x2= 60 6 10 62 = 0

Mmax: For M to be Mmax, 0dMdx = 60 20 x = 0 x = 3 m Mmax = 60 3 10 32 = 90 N-m

V and M Diagrams:

Following are the V and M diagramsplotted using the calculated values of Vand M:


Support Reactions:

Reaction at support A can be determined by applying equilibriumconditions to the free-body diagram ofthe entire beam, as follows:

BC

50 kN-mFR=10kN

Ay5m 5m

Cy

2.5m

A B C

50 kN-mFR=10kN

Ay5m 5m

Cy

2.5m

A

M about C = 0 10 Ay 10 (10 2.5) + 50 = 0 Ay = 2.5 kN



The equations for V and M for segment AB can be obtained by applyingequilibrium conditions to the free-body diagram of the segment AB, as follows:

V

FR=2x

2.5kNx

0.5x

A

x

M

x

VFR=2x

2.5kNx

0.5x

A

x

M

x

VFR=2x

2.5kNx

0.5x

A

x

M

x

VFR=2x

2.5kNx

0.5x

A

x

M

x

Fy = 0 2.5 V 2 x = 0 V = (2.5 2 x) kN M about cut section x x = 0 2.5 x 2 x 0.5 x M = 0 M = (2.5 x x2) kN-m


BC

50 kN-mFR=10kN

Ay5m 5m

Cy

2.5m

A B C

50 kN-mFR=10kN

Ay5m 5m

Cy

2.5m

A



BV

50 kN-mFR=10kN

2.5kN x

2.5m

x

x

MB

V50 kN-m

FR=10kN

2.5kN x

2.5m

x

x

M

Shear and Moment Values for Plottingthe V and M Diagrams: The V and M equations obtained for the segments AB and BC are as follows: ( )

( ) ( )22.5 2 kN 7.5 kN

; 75 7.5 kN-m2.5 kN-m

V x VAB BC

M xM x x

= = = =

Fy = 0 2.5 10 V = 0 V = 7.5 kN M about cut section x-x = 0 2.5 x 10 (x 2.5) + 50 M = 0 M = (75 7.5 x) kN-m

V at A = 2.5 2 0 = 2.5 kN V at B = 2.5 2 5 = 7.5 kN V between B and C = 7.5 kN

M at A = 2.5 x x2= 2.5 0 02 = 0

M just before B = 2.5 x x2= 2.5 5 52 = 12.5 kN-m

M just after B = 75 7.5 x= 75 7.5 5 = 37.5 kN-m Mmax between A and B: For M to be Mmax, 0dMdx = 2.5 2 x = 0 x = 1.25 m Mmax = 2.5 1.25 1.252 = 1.562 kN-m


V at A = 2.5 kN V at B = 7.5 kN V between B and C = 7.5 kN M at A = 0

M just before B = 12.5 kN-m

M just after B = 37.5 kN-m

V and M Diagrams:


B C

50 kN-m2kN/m

5m 5m

A

7.5kN2.5kN

V

-+

-7.5kN

2.5kNx

x

+

-

37.5kN

12.5kN-m2.5m

+

1.25m1.562kN-m

M

B C

50 kN-m2kN/m

5m 5m

A

7.5kN2.5kN

V

--++

--7.5kN

2.5kNx

x

++

--

37.5kN

12.5kN-m2.5m

++

1.25m1.562kN-m

M

V-Diagram

M-Diagram

B C

50 kN-m2kN/m

5m 5m

A

7.5kN2.5kN

V

-+

-7.5kN

2.5kNx

x

+

-

37.5kN

12.5kN-m2.5m

+

1.25m1.562kN-m

M

B C

50 kN-m2kN/m

5m 5m

A

7.5kN2.5kN

V

--++

--7.5kN

2.5kNx

x

++

--

37.5kN

12.5kN-m2.5m

++

1.25m1.562kN-m

M

V-Diagram

M-Diagram


Support Reactions:

Reactions at supports A and C can be determined by applying equilibriumconditions to the free-body diagram ofthe entire beam, as follows:

B C

50 kN-mFR=10kN

5m 5m

A

CyAy

2.5m

B C

50 kN-mFR=10kN

5m 5m

A

CyAy

2.5m

M about C = 0 10 Ay 10 7.5 + 50 = 0 Ay = 2.5 kN Fy = 0 Ay + Cy 10 = 0 Cy= 10 Ay = 10 2.5 = 7.5 kN




V

FR=2x

2.5kNx

0.5x

A

x

M

x

VFR=2x

2.5kNx

0.5x

A

x

M

x

VFR=2x

2.5kNx

0.5x

A

x

M

x

VFR=2x

2.5kNx

0.5x

A

x

M

x

Fy = 0 2.5 V 2 x = 0 V = (2.5 2 x) kN M about cut section x x = 0 2.5 x 2 x 0.5 x M = 0 M = (2.5 x x2) kN-m


B C

50 kN-mFR=10kN

5m 5m

A

CyAy

2.5m

B C

50 kN-mFR=10kN

5m 5m

A

CyAy

2.5m

Segment CB (0 x < 5) m

The equations for V and M for segment CB can be obtained by applyingequilibrium conditions to the free-body diagram of the segment CB, as follows:

V

xx 7.5 kN

x

M

50 kN-mV

xx 7.5 kN

x

M

50 kN-m

Fy = 0 V + 7.5 = 0 V = 7.5 kN M about cut section x x = 0 7.5x + 50 + M = 0 M = (7.5 x 50) kN-m

Shear and Moment Values for Plottingthe V and M Diagrams: The V and M equations obtained for thesegments AB and CB are as follows:

Segment AB (0 x < 5) m V = (2.5 2 x) kN M = (2.5x x2) kN-m

Segment AB (0 x < 5) m V = 7.5 kN M = (7.5 x 50) kN-m V at A = 2.5 2 0 = 2.5 kN V at B = 2.5 2 5 = 7.5 kN V between C and B = 7.5 kN M at A = 2.5 x x2= 2.5 0 02 = 0 M at B = 2.5 x x2= 2.5552 = 12.5 kN-m M at C =75 7.5 x= 7.5050 = 50 kN-m Mmax between A and B: For M to be Mmax, 0dMdx = 2.52 x = 0 x = 1.25 m Mmax = 2.5 1.25 1.252 = 1.562 kN-m


V at A = 2.5 kN V at B = 7.5 kN V between C and B = 7.5 kN

M at A = 0 M at B = 12.5 kN-m M at C = 50 kN-m Mmax = 2.5 1.25 1.252 = 1.562 kN-m

V and M Diagrams:


B

50 kN-m2kN/m

5m 5m 7.5kN2.5kN

V

-+

-7.5kN

2.5kNx

x

-50kN-m

12.5kN-m2.5m

+

1.25m1.562kN-m

M

B

50 kN-m2kN/m

5m 5m 7.5kN2.5kN

V

--++

--7.5kN

2.5kNx

x

--50kN-m

12.5kN-m2.5m

++

1.25m1.562kN-m

M

V-Diagram

M-Diagram

B

50 kN-m2kN/m

5m 5m 7.5kN2.5kN

V

-+

-7.5kN

2.5kNx

x

-50kN-m

12.5kN-m2.5m

+

1.25m1.562kN-m

M

B

50 kN-m2kN/m

5m 5m 7.5kN2.5kN

V

--++

--7.5kN

2.5kNx

x

--50kN-m

12.5kN-m2.5m

++

1.25m1.562kN-m

M

V-Diagram

M-Diagram


For the beam shown below, draw theSFD and BMD.


A B C

20N

A B C

20N

6 m 4m

5 N/m

80 N-mAB C

20N

A B C

20N

6 m 4m

5 N/m

A B C

20N

A B C

20N

6 m 4m

5 N/m

80 N-m

Support Reactions:

Reactions at supports A and B can be determined by applying equilibriumconditions to the free-body diagram ofthe entire beam, as follows:

RAy

A

6m 4m

80N/m20N

A

6m 4m

80N/m20N

FR = 5 6 = 30 N

3 m

RByRAy

A

6m 4m

80N/m20N

A

6m 4m

80N/m20N

FR = 5 6 = 30 N

3 m

RBy

A

6m 4m

80N/m20N

A

6m 4m

80N/m20N

FR = 5 6 = 30 N

3 m

RBy

B

C

MA = 0 303RBy 6+2010 80 = 0 6RBy = 210 RBy = 35 N Fy = 0 RAy + RBy 30 20 = 0 RAy = 50 RBy = 50 35 = 15 N




A

x

M

N

V15N

B

A

x

M

N

V15N

B

x

x

5 x

0.5 x

A

x

M

N

V15N

B

A

x

M

N

V15N

B

x

x

5 x

0.5 x

Fy = 0 15 V 5 x = 0 V = (15 5 x) N M about cut section x x = 0 15 x 5 x 0.5 x M = 0 M = (15 x 2.5 x2) N-m


RAy

A

6m 4m

80N/m20N

A

6m 4m

80N/m20N

FR = 5 6 = 30 N

3 m

RByRAy

A

6m 4m

80N/m20N

A

6m 4m

80N/m20N

FR = 5 6 = 30 N

3 m

RBy

A

6m 4m

80N/m20N

A

6m 4m

80N/m20N

FR = 5 6 = 30 N

3 m

RBy

B

C



A

x

M

N

V15N

B

35N6m x-6

5N/m

A

x

M

N

V15N

B

35N6m6m x-6

5N/m

Fy = 0 15 V 5 6 + 35 = 0 V = 20 N M about cut section = 0 15 x 5 6 (x3) + 35(x 6) M = 0 M = (20 x 120) N-m

Shear and Moment Values for Plottingthe V and M Diagrams: The V and M equations obtained for the segments AB and BC are as follows:

Segment AB (0 x < 6) m V = (15 5 x) N M = (15 x 2.5 x2) N-m

Segment BC (6 < x 10) m V = 20 N M = (20 x 120) N-m V at A = 15 5 0 = 15 kN V just before B = 15 5 6 = 15 N V just after B = 20 N V between B and C = 20 N M at A= 15 x 2.5 x2 = 15 0 2.5 02 = 0M at B= 15 x 2.5 x2 = 15 6 2.5 62 = 0M at C= 20 x 120 = 20 10120 = 80 N-mMmax between A and B: For M to be Mmax, 0dMdx = 15 5 x = 0 x = 3 m Mmax = 15 x2.5 x2 = 15 3 2.5 32 = 22.5 N-m


V at A = 15 kN V just before B = 15 N V just after B = 20 N V between B and C = 20 N M at A= 0 M at B= 0 M at C= 80 N-m Mmax = 22.5 N-m

V and M Diagrams:


A B

A

30N

6m

M = (N-m)

5N/m

+

22.5Nm

C80N/m

20N

4m15N35N

80N/m20N

-

++15

V(N)+

+20 +20

-15

+

80N-m

A B

A

30N

6m

M = (N-m)

5N/m

+

22.5Nm

C80N/m

20N

4m15N35N

80N/m20N

-

++15

V(N)+

+20 +20

-15

+

80N-m

V-Diagram

M-Diagram

A B

A

30N

6m

M = (N-m)

5N/m

+

22.5Nm

C80N/m

20N

4m15N35N

80N/m20N

-

++15

V(N)+

+20 +20

-15

+

80N-m

A B

A

30N

6m

M = (N-m)

5N/m

+

22.5Nm

C80N/m

20N

4m15N35N

80N/m20N

-

++15

V(N)+

+20 +20

-15

+

80N-m

V-Diagram

M-Diagram

Multiple Choice Problems

Answer the following questions related to a beam segment having moment equation

as: 3

(9 ) kN-m9xM x=

1. The type of load on the segment is (a) Point load (b) Rectangular load (c) Triangular load (d) None of these

Ans: (c)

Feedback:

Triangular load always has third degree of the moment equation.

2. The degree of the shear force equation for the segment is (a) 0 (b) 1 (c) 2 (d) 3

Ans: (c)

Feedback:

Since a triangular load always has third degree of the moment equation, thedegree of the shear force equation for the segment will be two, because degree of moment equation is always one more than the degree of shear force equation.

Multiple Choice ProblemsAnswer the following questions related to a beam segment having moment equation

as: 3

(9 ) kN-m9xM x=

3. Shear force at x = 0 will be (a) 0 (b) 9 kN (c) 81 kN (d) None of these

Ans: (b)

Feedback:

Since shear force is always taken as the slope of the moment curve, therefore,2 2 2

x = 03 09 9 ; 9 9 kN9 3 3at

dM x xV Vdx

= = = = =

4. Maximum moment for the segment will be (a) 0 (b) 5.2 kN-m (c) 15.2 kN-m (d) 31.2 kN-m

Ans: (d)

Feedback: 2 2

3

max

3For M to be maximum, 0 9 9 0 27 5.2 m9 3

5.29 5.2 31.2 kN-m9

dM x x xdx

M

= = = = =

= =

2- handouts lecture-33 to 36

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