2- handouts lecture-33 to 36
TRANSCRIPT
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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
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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
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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
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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.
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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.
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PROBLEM SOLVING: PROBLEM SOLVING: Example # 1Example # 1
For the beam shown below, draw the SFD and BMD.
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PROBLEM SOLVING: PROBLEM SOLVING: Example # 1Example # 1
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
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PROBLEM SOLVING: PROBLEM SOLVING: Example # 1Example # 1
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
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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
500 lb-ft1800 lb -ft
M
100 lb
800 lb -ft
100 lb
x
x
V - diagram
M - diagram1300 lb
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
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
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
500 lb-ft1800 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.
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PROBLEM SOLVING: PROBLEM SOLVING: Example # 2Example # 2
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
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PROBLEM SOLVING: PROBLEM SOLVING: Example # 2Example # 2
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
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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
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PROBLEM SOLVING: PROBLEM SOLVING: Example # 2Example # 2
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
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PROBLEM SOLVING: PROBLEM SOLVING: Example # 3Example # 3
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
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PROBLEM SOLVING: PROBLEM SOLVING: Example # 3Example # 3
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
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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:
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PROBLEM SOLVING: PROBLEM SOLVING: Example # 4Example # 4
For the beam shown below, draw the SFD and BMD.
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PROBLEM SOLVING: PROBLEM SOLVING: Example # 4Example # 4
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
Shear and Moment Equations:
Segment AB (0 x < 5) m
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
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PROBLEM SOLVING: PROBLEM SOLVING: Example # 4Example # 4
BC
50 kN-mFR=10kN
Ay5m 5m
Cy
2.5m
A B C
50 kN-mFR=10kN
Ay5m 5m
Cy
2.5m
A
Segment BC (5 < x 10) m
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:
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
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PROBLEM SOLVING: PROBLEM SOLVING: Example # 4Example # 4
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:
Following are the V and M diagrams plotted using the calculated values of Vand 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
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
-
PROBLEM SOLVING: PROBLEM SOLVING: Example # 5Example # 5
For the beam shown below, draw the SFD and BMD.
-
PROBLEM SOLVING: PROBLEM SOLVING: Example # 5Example # 5
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
Shear and Moment Equations:
Segment AB (0 x < 5) m
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
-
PROBLEM SOLVING: PROBLEM SOLVING: Example # 5Example # 5
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
-
PROBLEM SOLVING: PROBLEM SOLVING: Example # 5Example # 5
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:
Following are the V and M diagrams plotted using the calculated values of Vand 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
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
-
PROBLEM SOLVING: PROBLEM SOLVING: Example # 6Example # 6
For the beam shown below, draw theSFD and BMD.
-
PROBLEM SOLVING: PROBLEM SOLVING: Example # 6Example # 6
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
Shear and Moment Equations:
Segment AB (0 x < 6) m
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:
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
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PROBLEM SOLVING: PROBLEM SOLVING: Example # 6Example # 6
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
Segment BC (6 < x 10) m
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:
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
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PROBLEM SOLVING: PROBLEM SOLVING: Example # 6Example # 6
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:
Following are the V and M diagrams plotted using the calculated values of Vand 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
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
= = = = =
= =