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Chapter 04 Shear & Moment in
Beams
DEFINITION OF A BEAM A beam is a bar subject to forces or couples that lie in
a plane containing the longitudinal of the bar. According to determinacy, a beam may be determinate or indeterminate.
STATICALLY DETERMINATE BEAMS
Statically determinate beams are those beams in which the reactions of the supports may be determined by the use of the equations of static equilibrium. The beams shown below are examples of statically determinate beams.
P
Load
Cantilever Beam
Simple Beam
P M
Overhanging Beam
P w (N/m)
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166 Shear and Moment Equations and Diagrams; Relation between Load, Shear, and Moment; Moving Loads
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STATICALLY INDETERMINATE BEAMS
If the number of reactions exerted upon a beam exceeds the number of equations in static equilibrium, the beam is said to be statically indeterminate. In order to solve the reactions of the beam, the static equations must be supplemented by equations based upon the elastic deformations of the beam.
The degree of indeterminacy is taken as the difference
between the umber of reactions to the number of equations in static equilibrium that can be applied. In the case of the propped beam shown, there are three
reactions R1, R2, and M and only two equations (ΣM =
0 and ΣFv = 0) can be applied, thus the beam is indeterminate to the first degree (3 – 2 = 1).
P
Propped Beam
Continuous Beam
P1
w (N/m)
R1 R2
M
Fixed or Restrained Beam
w2 (N/m)
P2
M
w1 (N/m) w2 (N/m)
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TYPES OF LOADING
Loads applied to the beam may consist of a concentrated load (load applied at a point), uniform load, uniformly varying load, or an applied couple or moment. These loads are shown in the following figures.
SHEAR AND MOMENT DIAGRAM Consider a simple beam shown of length L that
carries a uniform load of w (N/m) throughout its length and is held in equilibrium by reactions R1 and R2. Assume that the beam is cut at point C a distance of x from he left support and the portion of the beam to the right of C be removed. The portion removed must then be replaced by vertical shearing force V
P1 P2
Concentrated Loads
w (N/m)
Uniform Load
w (N/m)
Uniformly Varying Load Applied Couple
M
w (N/m)
L C
x
R1 R2
A B
w (N/m)
C
x
R1
V
M A
168 Shear and Moment Equations and Diagrams; Relation between Load, Shear, and Moment; Moving Loads
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together with a couple M to hold the left portion of the bar in equilibrium under the action of R1 and wx.
The couple M is called the resisting moment or
moment and the force V is called the resisting shear or shear. The sign of V and M are taken to be positive if they have the senses indicated above.
SOLVED PROBLEMS INSTRUCTION: Write shear and moment equations for the beams in
the following problems. In each problem, let x be the distance measured from left end of the beam. Also, draw shear and moment diagrams, specifying values at all change of loading positions and at points of zero shear. Neglect the mass of the beam in each problem.
Problem 403. Beam loaded as shown in Fig. P-403. Solution 403. From the load diagram:
∑MB = 0 5RD + 1(30) = 3(50) RD = 24 kN
∑MD = 0 5RB = 2(50) + 6(30) RB = 56 kN Segment AB: VAB = –30 kN
MAB = –30x kN⋅m
30 kN 50 kN
1 m 3 m 2 m
A B C D
Figure P-403
30 kN
A
x
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Segment BC: VBC = –30 + 56 VBC = 26 kN
MBC = –30x + 56(x – 1)
MBC = 26x – 56 kN⋅m Segment CD: VCD = –30 + 56 – 50 VCD = –24 kN MCD = –30x + 56(x – 1) – 50(x – 4) MCD = –30x + 56x – 56 – 50x + 200 MCD = –24x + 144 Problem 404. Beam loaded as shown in Fig. P-404.
30 kN
1 m
A B
RB = 56 kN
x
30 kN 50 kN
1 m 3 m
A B C
RB = 56 kN
x
2000 lb
M = 4800 lb⋅ft
3 ft 6 ft 3 ft
A B C
D
Figure P-404
RA RD
To draw the Shear Diagram: (1) In segment AB, the shear is
uniformly distributed over the
segment at a magnitude of –30
kN.
(2) In segment BC, the shear is
uniformly distributed at a
magnitude of 26 kN.
(3) In segment CD, the shear is
uniformly distributed at a
magnitude of –24 kN.
To draw the Moment Diagram: (1) The equation MAB = –30x is
linear, at x = 0, MAB = 0 and at
x = 1 m, MAB = –30 kN⋅m.
(2) MBC = 26x – 56 is also linear.
At x = 1 m, MBC = –30 kN⋅m; at
x = 4 m, MBC = 48 kN⋅m. When
MBC = 0, x = 2.154 m, thus the
moment is zero at 1.154 m
from B.
(3) MCD = –24x + 144 is again
linear. At x = 4 m, MCD = 48
kN⋅m; at x = 6 m, MCD = 0.
30 kN 50 kN
1 m 3 m 2 m
A B C D
LoadDiagram
RB = 56 kN RB = 24 kN
26 kN
–30 kN –24 kN
ShearDiagram
MomentDiagram
–30 kN⋅m
56 kN⋅m
1.154 m
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Solution 404. ∑MA = 0 ∑MD = 0 12RD + 4800 = 3(2000) 12RA = 9(2000) + 4800 RD = 100 lb RA = 1900 lb Segment AB: VAB = 1900 lb
MAB = 1900x lb⋅ft Segment BC: VBC = 1900 – 2000 VBC = –100 lb MBC = 1900x – 2000(x – 3) MBC = 1900x – 2000x + 6000 MBC = –100x + 6000 Segment CD: VCD = 1900 – 2000 VCD = –100 lb MCD = 1900x – 2000(x – 3) – 4800 MCD = 1900x – 2000x + 6000 – 4800 MCD = –100x + 1200
2000 lb
B A
RA = 1900 lb
x
3 ft
6 ft C
2000 lb
B A
RA = 1900 lb x
3 ft
M = 4800 lb⋅ft
x
A
RA = 1900 lb
2000 lb M = 4800 lb⋅ft
3 ft 6 ft 3 ft
A B C
D
RA = 1900 lb RD = 100 lb
Load Diagram
1900 lb
–100 lb
Shear Diagram
Moment Diagram
5700 lb⋅ft
5100 lb⋅ft
300 lb⋅ft
To draw the Shear Diagram: (1) At segment AB, the shear is
uniformly distributed at 1900 lb.
(2) A shear of –100 lb is uniformly
distributed over segments BC
and CD.
To draw the Moment Diagram: (1) MAB = 1900x is linear; at x = 0,
MAB = 0; at x = 3 ft, MAB = 5700
lb⋅ft. (2) For segment BC, MBC = –100x +
6000 is linear; at x = 3 ft, MBC =
5700 lb⋅ft; at x = 9 ft, MBC =
5100 lb⋅ft. (3) MCD = –100x + 1200 is again
linear; at x = 9 ft, MCD = 300
lb⋅ft; at x = 12 ft, MCD = 0.
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Problem 405. Beam loaded as shown in Fig. P-405.
Solution 405. ∑MA = 0 ∑MC = 0 10RC = 2(80) + 5[10(10)] 10RA = 8(80) + 5[10(10)] RC = 66 kN RA = 114 kN Segment AB: VAB = 114 – 10x kN MAB = 114x – 10x(x/2)
MAB = 114x – 5x2 kN⋅m Segment BC: VBC = 114 – 80 – 10x VBC = 34 – 10x kN MBC = 114x – 80(x – 2) – 10x(x/2) MBC = 160 + 34x – 5x2
80 kN
10 kN/m
2 m 8 m
A
B
C
Figure P-405
RA RC
10 kN/m
x
A
RA = 114 kN
80 kN
B 10 kN/m
A
RA = 114 kN
x
2 m
To draw the Shear Diagram: (1) For segment AB, VAB = 114 – 10x
is linear; at x = 0, VAB = 14 kN; at
x = 2 m, VAB = 94 kN.
(2) VBC = 34 – 10x for segment BC is
linear; at x = 2 m, VBC = 14 kN; at
x = 10 m, VBC = –66 kN. When VBC
= 0, x = 3.4 m thus VBC = 0 at 1.4
m from B.
To draw the Moment Diagram: (1) MAB = 114x – 5x
2 is a second
degree curve for segment AB; at x
= 0, MAB = 0; at x = 2 m, MAB =
208 kN⋅m.
(2) The moment diagram is also a
second degree curve for segment
BC given by MBC = 160 + 34x –
5x2; at x = 2 m, MBC = 208 kN⋅m;
at x = 10 m, MBC = 0.
(3) Note that the maximum moment
occurs at point of zero shear.
Thus, at x = 3.4 m, MBC = 217.8 kN⋅m.
80 kN
10 kN/m
2 m 8 m
A
B C
RA = 114 kN RC = 66 kN
Load Diagram
1.4 m
114 kN
94 kN
14 kN
–66 kN
Shear Diagram
208 kN⋅m
217.8 kN⋅m
Moment Diagram
172 Shear and Moment Equations and Diagrams; Relation between Load, Shear, and Moment; Moving Loads
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Problem 406. Beam loaded as shown in Fig. P-406.
Solution 406. ∑MA = 0 12RC = 4(900) + 18(400) + 9[(60)(18)] RC = 1710 lb
∑MC = 0 12RA + 6(400) = 8(900) + 3[60(18)] RA = 670 lb Segment AB: VAB = 670 – 60x lb MAB = 670x – 60x(x/2)
MAB = 670x – 30x2 lb⋅ft Segment BC: VBC = 670 – 900 – 60x VBC = –230 – 60x lb MBC = 670x – 900(x – 4) – 60x(x/2)
MBC = 3600 – 230x – 30x2 lb⋅ft Segment CD: VCD = 670 + 1710 – 900 – 60x VCD = 1480 – 60x lb MCD = 670x + 1710(x – 12) – 900(x – 4) – 60x(x/2)
MCD = –16920 + 1480x – 30x2 lb⋅ft
Figure P-406
900 lb
60 lb/ft
4 ft 8 ft
A
B
D
RA 6 ft
C
RC
400 lb
x
A
RA = 670 lb
60 lb/ft
900 lb
x
A
RA = 670 lb
4 ft 60 lb/ft
B C
RC = 1710 lb
900 lb
A
RA = 670 lb
4 ft
x
8 ft
60 lb/ft
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Problem 407. Beam loaded as shown in Fig. P-407.
Solution 407. ∑MA = 0 ∑MD = 0 6RD = 4[2(30)] 6RA = 2[2(30)] RD = 40 kN RA = 20 kN Segment AB: VAB = 20 kN
MAB = 20x kN⋅m
Figure P-407 30 kN/m
3 m
A
B
D
RA
2 m C
1 m
RD
x
A
RA = 20 kN
To draw the Shear Diagram: (1) VAB = 670 – 60x for segment AB is
linear; at x = 0, VAB= 670 lb; at x
= 4 ft, VAB = 430 lb.
(2) For segment BC, VBC = –230 – 60x
is also linear; at x= 4 ft, VBC = –
470 lb, at x = 12 ft, VBC = –950 lb.
(3) VCD = 1480 – 60x for segment CD
is again linear; at x = 12, VCD =
760 lb; at x = 18 ft, VCD = 400 lb.
To draw the Moment Diagram: (1) MAB = 670x – 30x
2 for segment AB
is a second degree curve; at x =
0, MAB = 0; at x = 4 ft, MAB = 2200
lb⋅ft. (2) For BC, MBC = 3600 – 230x – 30x
2,
is a second degree curve; at x = 4
ft, MBC = 2200 lb⋅ft, at x = 12 ft, MBC = –3480 lb⋅ft; When MBC = 0,
3600 – 230x – 30x2 = 0, x = –
15.439 ft and 7.772 ft. Take x =
7.772 ft, thus, the moment is zero
at 3.772 ft from B.
(3) For segment CD, MCD = –16920 +
1480x – 30x2 is a second degree
curve; at x = 12 ft, MCD = –3480
lb⋅ft; at x = 18 ft, MCD = 0.
2200 lb⋅ft
3.772 ft
–3480 lb⋅ft
Moment Diagram
900 lb
60 lb/ft
4’ 8’
A
B D
RA = 670 lb
6’ C
RC = 1710 lb
400 lb
Load Diagram
670 lb 430 lb
–470 lb
–950 lb
760 lb
Shear Diagram
400 lb
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Segment BC: VBC = 20 – 30(x – 3) VBC = 110 – 30x kN MBC = 20x – 30(x – 3)(x – 3)/2 MBC = 20x – 15(x – 3)2 Segment CD: VCD = 20 – 30(2) VCD = –40 kN MCD = 20x – 30(2)(x – 4) MCD = 20x – 60(x – 4) Problem 408. Beam loaded as shown in Fig. P-408.
Solution 408. ∑MA = 0 ∑MD = 0 6RD = 1[2(50)] + 5[2(20)] 6RA = 5[2(50)] + 1[2(20)] RD = 50 kN RA = 90 kN
x
30 kN/m
A B
RA = 20 kN
2 m C
3 m
A B
RA = 20 kN
x
3 m
30 kN/m
Figure P-408 20 kN/m
2 m
A
B
D
RA RD
2 m 2 m C
50 kN/m
30 kN/m
3 m
A
B
D
RA = 20 kN
2 m C
1 m
RD = 40 kN
Load Diagram
20 kN
0.67 m –40 kN
Shear Diagram
60 kN⋅m 66.67 kN⋅m
40 kN⋅m Moment Diagram
To draw the Shear Diagram: (1) For segment AB, the shear is
uniformly distributed at 20 kN.
(2) VBC = 110 – 30x for segment BC;
at x = 3 m, VBC = 20 kN; at x = 5
m, VBC = –40 kN. For VBC = 0, x =
3.67 m or 0.67 m from B.
(3) The shear for segment CD is
uniformly distributed at –40 kN.
To draw the Moment Diagram: (1) For AB, MAB = 20x; at x = 0, MAB =
0; at x = 3 m, MAB = 60 kN⋅m.
(2) MBC = 20x – 15(x – 3)2 for
segment BC is second degree
curve; at x = 3 m, MBC = 60 kN⋅m;
at x = 5 m, MBC = 40 kN⋅m. Note
that maximum moment occurred
at zero shear; at x = 3.67 m, MBC
= 66.67 kN⋅m.
(3) MCD = 20x – 60(x – 4) for segment
BC is linear; at x = 5 m, MCD = 40
kN⋅m; at x = 6 m, MCD = 0.
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Chapter 04 Shear and Moment in Beams www.mathalino.com
Segment AB: VAB = 90 – 50x kN MAB = 90x – 50x(x/2) MAB = 90x – 25x2 Segment BC: VBC = 90 – 50(2) VBC = –10 kN MBC = 90x – 2(50)(x – 1)
MBC = –10x + 100 kN⋅m Segment CD: VCD = 90 – 2(50) – 20(x – 4) VCD = –20x + 70 kN MCD = 90x – 2(50)(x – 1) – 20(x – 4)(x – 4)/2 MCD = 90x – 100(x – 1) – 10(x – 4)2
MCD = –10x2 + 70x – 60 kN⋅m Problem 409. Cantilever beam loaded as shown in Fig. P-409.
x
A
RA = 90 kN
50 kN/m
B
A
50 kN/m
x
2 m
RA = 90 kN
20 kN/m
2 m C B
A
50 kN/m
x
2 m
RA = 90 kN
Figure P-409
L/2
A
B L/2
C
wo
To draw the Shear Diagram: (1) VAB = 90 – 50x is linear; at x = 0, VBC
= 90 kN; at x = 2 m, VBC = –10 kN.
When VAB = 0, x = 1.8 m.
(2) VBC = –10 kN along segment BC.
(3) VCD = –20x + 70 is linear; at x = 4
m, VCD = –10 kN; at x = 6 m, VCD = –
50 kN.
To draw the Moment Diagram: (1) MAB = 90x – 25x
2 is second degree;
at x = 0, MAB = 0; at x = 1.8 m, MAB
= 81 kN⋅m; at x = 2 m, MAB = 80
kN⋅m.
(2) MBC = –10x + 100 is linear; at x = 2
m, MBC = 80 kN⋅m; at x = 4 m, MBC =
60 kN⋅m.
(3) MCD = –10x2 + 70x – 60; at x = 4 m,
MCD = 60 kN⋅m; at x = 6 m, MCD = 0.
20 kN/m
D
RD = 50 kN
2 m C B
A
50 kN/m
2 m
RA = 90 kN
2 m
90 kN
–10 kN
–50 kN
Load Diagram
Shear Diagram
Moment Diagram
81 kN⋅m
1.8 m
80 kN⋅m
60 kN⋅m
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Solution 409. Segment AB: VAB = –wox MAB = –wox(x/2)
MAB = – 21 wox2
Segment BC: VBC = –wo(L/2)
VBC = – 21 woL
MBC = –wo(L/2)(x – L/4)
MBC = – 21 woLx + 8
1 woL2
Problem 410. Cantilever beam carrying the uniformly varying load
shown in Fig. P-410.
Solution 410. x
y =
L
wo
y = xL
wo
Fx = 21 xy
Fx =
x
L
wx o
2
1
x
A
wo
B A
wo
x
L/2
Figure P-410
L
wo
x
y
3
1 x Fx
wo
L – x
L/2
A
B L/2
C
wo
Load Diagram
–2
1woL
Shear Diagram
Moment Diagram
–8
1 woL2
–8
3 woL2
To draw the Shear Diagram: (1) VAB = –wox for segment AB is linear; at x = 0, VAB
= 0; at x = L/2, VAB = –2
1woL.
(2) At BC, the shear is uniformly distributed by –
2
1woL.
To draw the Moment Diagram:
(1) MAB = –2
1wox
2 is a second degree curve; at x =
0, MAB = 0; at x = L/2, MAB = – 8
1 woL2.
(2) MBC = –2
1woLx + 8
1 woL2 is a second degree; at
x = L/2, MBC = – 8
1 woL2; at x = L, MBC = –
8
3 woL2.
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Chapter 04 Shear and Moment in Beams www.mathalino.com
Fx = 2
2x
L
wo
Shear equation:
V = – 2
2x
L
wo
Moment equation:
M = – 31 xFx = –
2
23
1x
L
wx o
M = – 3
6x
L
wo
Problem 411. Cantilever beam carrying a distributed load with
intensity varying from wo at the free end to zero at the wall, as shown in Fig. P-411.
Solution 411. L
w
xL
y o=−
y = )( xLL
wo −
Figure P-411
L
wo
L
wo
Load Diagram
Shear Diagram
–2
1woL
–6
1 woL2
Moment Diagram
2nd degree
3rd degree
To draw the Shear Diagram:
V = –2o
x
2L
w is a second degree curve;
at x = 0, V = 0; at x = L, V = –2
1woL.
To draw the Moment Diagram:
M = –3o
x
6L
w is a third degree curve; at
x = 0, M = 0; at x = L, M =–6
1 woL2.
178 Shear and Moment Equations and Diagrams; Relation between Load, Shear, and Moment; Moving Loads
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F1 = )(21 ywx o −
F1 =
−− )(2
1xL
L
wwx o
o
F1 =
+− xL
wwwx o
oo2
1
F1 = 2
2x
L
wo
F2 = xy =
− )( xLL
wx o
F2 = )( 2xLxL
wo −
Shear equation:
V = –F1 – F2 = – 2
2x
L
wo – )( 2xLxL
wo −
V = – 2
2x
L
wo – xwo + 2xL
wo
V = 2
2x
L
wo – xwo
Moment equation:
M = – 32 xF1 – 2
1 xF2
M = –
2
23
1x
L
wx o –
− )(2
1 2xLxL
wx o
M = – 3
3x
L
wo – 2
2x
wo + 3
2x
L
wo
M = – 2
2x
wo + 3
6x
L
wo
Problem 412. Beam loaded as shown in Fig. P-412.
x
wo
y
L – x
2
1x
3
2 x
F1 F2
Figure P-406 800 lb/ft
2 ft 4 ft
A
B
D
RA 2 ft
C
RC
To draw the Shear Diagram:
V = 2o
x
2L
w – xw
o is a concave
upward second degree curve; at x
= 0, V = 0; at x = L, V = –2
1 woL.
To draw the Moment diagram:
M = –2o
x
2
w +
3ox
6L
w is in third
degree; at x = 0, M = 0; at x = L,
M = –3
1 woL2.
L
wo
Load Diagram
–2
1 woL Shear Diagram
2nd degree
–3
1 woL2
Moment Diagram
3rd degree
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Solution 412. ∑MA = 0 ∑MC = 0 6RC = 5[6(800)] 6RA = 1[6(800)] RC = 4000 lb RA = 800 lb Segment AB: VAB = 800 lb MAB = 800x Segment BC: VBC = 800 – 800(x – 2) VBC = 2400 – 800x MBC = 800x – 800(x – 2)(x – 2)/2 MBC = 800x – 400(x – 2)2 Segment CD: VCD = 800 + 4000 – 800(x – 2) VCD = 4800 – 800x + 1600 VCD = 6400 – 800x MCD = 800x + 4000(x – 6) – 800(x – 2)(x – 2)/2 MCD = 800x + 4000(x – 6) – 400(x – 2)2
x
A
RA = 800 lb
x
800 lb/ft
B A
RA = 800 lb
2 ft
C
RC = 4000 lb
800 lb/ft
B A
RA = 800 lb
2 ft
x
4 ft
To draw the Shear Diagram: (1) 800 lb of shear force is uniformly
distributed along segment AB.
(2) VBC = 2400 – 800x is linear; at x = 2 ft,
VBC = 800 lb; at x = 6 ft, VBC = –2400
lb. When VBC = 0, 2400 – 800x = 0,
thus x = 3 ft or VBC = 0 at 1 ft from B.
(3) VCD = 6400 – 800x is also linear; at x =
6 ft, VCD = 1600 lb; at x = 8 ft, VBC = 0.
To draw the Moment Diagram: (1) MAB = 800x is linear; at x = 0, MAB = 0;
at x = 2 ft, MAB = 1600 lb⋅ft. (2) MBC = 800x – 400(x – 2)
2 is second
degree curve; at x = 2 ft, MBC = 1600
lb⋅ft; at x = 6 ft, MBC = –1600 lb⋅ft; at x = 3 ft, MBC = 2000 lb⋅ft.
(3) MCD = 800x + 4000(x – 6) – 400(x – 2)2
is also a second degree curve; at x = 6
ft, MCD = –1600 lb⋅ft; at x = 8 ft, MCD =
0.
800 lb/ft
2 ft 4 ft
A
B D
RA = 800 lb
2 ft C
RC = 4000 lb
800 lb
–2400 lb
1600 lb
1 ft
2000 lb⋅ft
1600 lb⋅ft
–1600 lb⋅ft
Moment Diagram
Shear Diagram
Load Diagram
180 Shear and Moment Equations and Diagrams; Relation between Load, Shear, and Moment; Moving Loads
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Problem 413. Beam loaded as shown in Fig. P-413.
Solution 413. ∑MB = 0 6RE = 1200 + 1[6(100)] RE = 300 lb
∑ME = 0 6RB + 1200 = 5[6(100)] RB = 300 lb Segment AB: VAB = –100x lb MAB = –100x(x/2)
MAB = –50x2 lb⋅ft Segment BC: VBC = –100x + 300 lb MBC = –100x(x/2) + 300(x – 2)
MBC = –50x2 + 300x – 600 lb⋅ft Segment CD: VCD = –100(6) + 300 VCD = –300 lb MCD = –100(6)(x – 3) + 300(x – 2) MCD = –600x + 1800 + 300x – 600
MCD = –300x + 1200 lb⋅ft Segment DE: VDE = –100(6) + 300 VDE = –300 lb MDE = –100(6)(x – 3) + 1200 + 300(x – 2) MDE = –600x + 1800 + 1200 + 300x – 600 MDE = –300x + 2400
Figure P-413 100 lb/ft
2 ft
A
B
E
RB 4 ft
C
1 ft RE
M = 1200 lb⋅ft
1 ft
D
x
A
100 lb/ft
100 lb/ft
2 ft
A
B
RB = 300 lb
x
100 lb/ft
2 ft
A
B
RB = 300 lb
4 ft C
x
1200 lb⋅ft 100 lb/ft
2 ft
A
B
RB = 300 lb
4 ft C
1’ D
x
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Chapter 04 Shear and Moment in Beams www.mathalino.com
Problem 414. Cantilever beam carrying the load shown in Fig. P-
414. Solution 414. Segment AB: VAB = –2x kN MAB = –2x(x/2)
MAB = –x2 kN⋅m Segment BC:
3
2
2=
−x
y
y = 32 (x – 2)
2 kN/m
x
A
2 kN/m
2 m
A B
2 kN/m
y
3 m
x
F1
F2
x – 2
2 kN/m
2 m 3 m
A B C
4 kN/m Figure P-414
To draw the Shear Diagram: (1) VAB = –100x is linear; at x = 0, VAB =
0; at x = 2 ft, VAB = –200 lb.
(2) VBC = 300 – 100x is also linear; at x
= 2 ft, VBC = 100 lb; at x = 4 ft, VBC
= –300 lb. When VBC = 0, x = 3 ft,
or VBC =0 at 1 ft from B.
(3) The shear is uniformly distributed at
–300 lb along segments CD and DE.
To draw the Moment Diagram: (1) MAB = –50x
2 is a second degree
curve; at x= 0, MAB = 0; at x = ft,
MAB = –200 lb⋅ft. (2) MBC = –50x
2 + 300x – 600 is also
second degree; at x = 2 ft; MBC = –
200 lb⋅ft; at x = 6 ft, MBC = –600
lb⋅ft; at x = 3 ft, MBC = –150 l⋅ft. (3) MCD = –300x + 1200 is linear; at x =
6 ft, MCD = –600 lb⋅ft; at x = 7 ft, MCD = –900 lb⋅ft.
(4) MDE = –300x + 2400 is again linear;
at x = 7 ft, MDE = 300 lb⋅ft; at x = 8 ft, MDE = 0.
100 lb/ft
2 ft
A
B
E
RB = 300 lb
4 ft C
1’
RE = 300 lb
M = 1200 lb⋅ft
1’ D
–300 lb
300 lb⋅ft
100 lb
–200 lb
1 ft
–200 lb⋅ft –150 lb⋅ft
–600 lb⋅ft
–900 lb⋅ft
Load Diagram
Shear Diagram
Moment Diagram
182 Shear and Moment Equations and Diagrams; Relation between Load, Shear, and Moment; Moving Loads
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F1 = 2x
F2 = 21 (x – 2)y
F2 = 21 (x – 2)[ 3
2 (x – 2)]
F2 = 31 (x – 2)2
VBC = –F1 – F2
VBC = – 2x – 31 (x – 2)2
MBC = –(x/2)F1 – 31 (x – 2)F2
MBC = –(x/2)(2x) – 31 (x – 2)[ 3
1 (x – 2)2]
MBC = –x2 – 91 (x – 2)3
Problem 415. Cantilever beam loaded as shown in Fig. P-415. Solution 415. Segment AB: VAB = –20x kN MAB = –20x(x/2)
MAB = –10x2 kN⋅m
3 m 2 m
A B C
Figure P-415
2 m D
20 kN/m
40 kN
A
20 kN/m
x
To draw the Shear Diagram: (1) VAB = –2x is linear; at x = 0, VAB = 0; at x = 2 m, VAB =
–4 kN.
(2) VBC = – 2x – 31 (x – 2)2 is a second degree curve; at x
= 2 m, VBC = –4 kN; at x = 5 m; VBC = –13 kN.
To draw the Moment Diagram: (1) MAB = –x
2 is a second degree curve; at x = 0, MAB = 0;
at x = 2 m, MAB = –4 kN⋅m.
(2) MBC = –x2 – 9
1 (x – 2)3 is a third degree curve; at x = 2
m, MBC = –4 kN⋅m; at x = 5 m, MBC = –28 kN⋅m.
2 kN/m
2 m 3 m
A B C
4 kN/m
1st degree
2nd degree
2nd degree
3rd degree
Moment Diagram
–4 kN
Shear Diagram
Load Diagram
–13 kN
–4 kN⋅m
–28 kN⋅m
183
Chapter 04 Shear and Moment in Beams www.mathalino.com
Segment BC: VBC = –20(3) VAB = –60 kN MBC = –20(3)(x – 1.5)
MAB = –60(x – 1.5) kN⋅m Segment CD: VCD = –20(3) + 40 VCD = –20 kN MCD = –20(3)(x – 1.5) + 40(x – 5) MCD = –60(x – 1.5) + 40(x – 5) Problem 416. Beam carrying
uniformly varying load shown in Fig. P-416.
Solution 416. ∑MR2 = 0
LR1 = 31 LF
R1 = 21
31 ( Lwo)
R1 = 61 Lwo
Figure P-416
L
R1
wo
R2
L
R1
wo
R2
F = ½ Lwo
2/3 L 1/3 L
3 m
A B
20 kN/m
x
3 m 2 m
A B C
20 kN/m
40 kN x
3 m 2 m
A B C
2 m D
20 kN/m
40 kN
–20 kN
–60 kN
–90 kN⋅m
–210 kN⋅m –250 kN⋅m
Load Diagram
Shear Diagram
Moment Diagram
To draw the Shear Diagram
(1) VAB = –20x for segment AB is linear; at
x = 0, V = 0; at x = 3 m, V = –60 kN.
(2) VBC = –60 kN is uniformly distributed
along segment BC.
(3) Shear is uniform along segment CD at
–20 kN.
To draw the Moment Diagram (1) MAB = –10x
2 for segment AB is second
degree curve; at x = 0, MAB = 0; at x =
3 m, MAB = –90 kN⋅m.
(2) MBC = –60(x – 1.5) for segment BC is
linear; at x = 3 m, MBC = –90 kN⋅m; at
x = 5 m, MBC = –210 kN⋅m.
(3) MCD = –60(x – 1.5) + 40(x – 5) for
segment CD is also linear; at x = 5 m,
MCD = –210 kN⋅m, at x = 7 m, MCD = –
250 kN⋅m.
184 Shear and Moment Equations and Diagrams; Relation between Load, Shear, and Moment; Moving Loads
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∑MR1 = 0
LR2 = 32 LF
R2 = 21
32 ( Lwo)
R2 = 31 Lwo
L
w
x
y o=
y = xL
wo
Fx = 21 xy =
x
L
wx o
2
1
Fx = 2
2x
L
wo
V = R1 – Fx
V = 61 Lwo – 2
2x
L
wo
M = R1x – Fx( x31 )
M = 61 Lwox – 2
2x
L
wo ( x31 )
M = 61 Lwox – 3
6x
L
wo
x
R1
wo
Fx = ½ xy
y
2/3 x 1/3 x
To draw the Shear Diagram: V = 1/6 Lwo – wox
2/2L is a second degree curve; at x =
0, V = 1/6 Lwo = R1; at x = L, V = –1/3 Lwo = –R2; If
a is the location of zero shear from left end, 0 = 1/6 Lwo
– wox2/2L, x = 0.5774L = a; to check, use the squared
property of parabola:
a2/R1 = L2/(R1 + R2)
a2/(1/6 Lwo) = L2/(1/6 Lwo + 1/3 Lwo)
a2 = (1/6 L3wo)/(1/2 Lwo) = 1/3 L2
a = 0.5774L a =
To draw the Moment Diagram M = 1/6 Lwox – wox
3/6L is a third degree curve; at x =
0, M = 0; at x = L, M = 0; at x = a = 0.5774L, M =
Mmax
Mmax = 1/6 Lwo(0.5774L) – wo(0.5774L)3/6L
Mmax = 0.0962L2wo – 0.0321L
2wo
Mmax = 0.0641L2wo
L
R1
wo
R2
–R2
Mmax
Load Diagram
Shear Diagram
Moment Diagram
a
R1
185
Chapter 04 Shear and Moment in Beams www.mathalino.com
Problem 417. Beam carrying the triangular loading shown in Fig. P-417.
Solution 417. By symmetry:
R1 = R2 = )( 21
21
oLw = oLw41
2/L
w
x
y o= ; y = xL
wo2
F = xy21 =
x
L
wx o2
2
1
F = 2xL
wo
V = R1 – F
V = oLw41 – 2x
L
wo
M = R1x – F )( 31 x
M = xLwo41 – )( 3
12 xxL
wo
M = xLwo41 – 3
3x
L
wo
L/2 L/2
R1 R2
wo
Figure P-417
R1
L/2
wo
x
y
F
1/3 x
L/2 L/2
R1 R2
wo
Load Diagram
Shear Diagram
o4
1Lw−
o4
1Lw
o
2
12
1wL
Moment Diagram
To draw the Shear Diagram: V = Lwo/4 – wox
2/L is a second degree curve;
at x = 0, V = Lwo/4; at x = L/2, V = 0. The
other half of the diagram can be drawn by the
concept of symmetry.
To draw the Moment Diagram M = Lwox/4 – wox
3/3L is a third degree curve;
at x = 0, M = 0; at x = L/2, M = L2wo/12. The
other half of the diagram can be drawn by the concept of symmetry.
186 Shear and Moment Equations and Diagrams; Relation between Load, Shear, and Moment; Moving Loads
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Problem 418. Cantilever beam loaded as shown in Fig. P-418. Solution 418. Segment AB: VAB = –20 kN
MAB = –20x kN⋅m Segment BC: VAB = –20 kN
MAB = –20x + 80 kN⋅m Problem 419. Beam loaded as shown in Fig. P-419.
20 kN
4 m
A
B
Figure P-418
2 m C
M = 80 kN⋅m
20 kN
x
A 20 kN
4 m
A
B
x
M = 80 kN⋅m
–80 kN⋅m
–40 kN⋅m
20 kN
4 m
A
B
2 m C
M = 80 kN⋅m
–20 kN
Load Diagram
To draw the Shear Diagram: VAB and VBC are equal and constant
at –20 kN.
To draw the Moment Diagram: (1) MAB = –20x is linear; when x = 0,
MAB = 0; when x = 4 m, MAB = –
80 kN⋅m.
(2) MBC = –20x + 80 is also linear;
when x = 4 m, MBC = 0; when x = 6 m, MBC = –60 kN⋅m
Shear Diagram
Moment Diagram
6 ft 3 ft
R1 R2
Figure P-419
A B
270 lb/ft
C
187
Chapter 04 Shear and Moment in Beams www.mathalino.com
Solution 419.
[ ∑MC = 0 ] 9R1 = 5(810) R1 = 450 lb
[ ∑MA = 0 ] 9R2 = 4(810) R2 = 360 lb Segment AB:
6
270=x
y
y = 45x
F = xy21 = )45(2
1 xx
F = 22.5x2 VAB = R1 – F VAB = 450 – 22.5x2 lb
MAB = R1x – F( x31 )
MAB = 450x – 22.5x2( x31 )
MAB = 450x – 7.5x3 lb⋅ft Segment BC: VBC = 450 – 810 VBC = –360 lb MBC = 450x – 810(x – 4) MBC = 450x – 810x + 3240
MBC = 3240 – 360x lb⋅ft
5 ft
6 ft 3 ft
R1 R2
A B
270 lb/ft
4 ft
810 lb
C
6 ft
R1
A
270 lb/ft y
x
1/3 x F
6 ft
x
R1 = 450 lb
A B
270 lb/ft
4 ft 810 lb
188 Shear and Moment Equations and Diagrams; Relation between Load, Shear, and Moment; Moving Loads
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Problem 420. A total distributed load of 30 kips supported by a
uniformly distributed reaction as shown in Fig. P-420. Solution 420.
1080 lb⋅ft
450 lb
6 ft 3 ft
R1 = 450 lb R2 = 360 lb
A B
270 lb/ft
C
Load Diagram
a =
–360 lb
Shear Diagram
Moment Diagram
a = √20
1341.64 lb⋅ft
3rd degree
linear
To draw the Shear Diagram: (1) VAB = 450 – 22.5x2 is a second degree
curve; at x = 0, VAB = 450 lb; at x = 6
ft, VAB = –360 lb.
(2) At x = a, VAB = 0,
450 – 22.5x2 = 0
22.5x2 = 450
x2 = 20
x = √20
To check, use the squared property of
parabola.
a2/450 = 62/(450 + 360)
a2 = 20
a = √20
(3) VBC = –360 lb is constant.
To draw the Moment Diagram: (1) MAB = 450x – 7.5x3 for segment AB is
third degree curve; at x = 0, MAB = 0;
at x = √20, MAB = 1341.64 lb⋅ft; at x =
6 ft, MAB = 1080 lb⋅ft. (2) MBC = 3240 – 360x for segment BC is
linear; at x = 6 ft, MBC = 1080 lb⋅ft; at
x = 9 ft, MBC = 0.
4 ft 12 ft 4 ft
W = 30 kips
Figure P-420
4 ft 12 ft 4 ft
W = 30 kips
r lb/ft
w lb/ft
189
Chapter 04 Shear and Moment in Beams www.mathalino.com
w = 30(1000)/12 w = 2500 lb/ft
∑FV = 0 R = W 20r = 30(1000) r = 1500 lb/ft First segment (from 0 to 4 ft from left): V1 = 1500x M1 = 1500x(x/2) M1 = 750x2 Second segment (from 4 ft to mid-span): V2 = 1500x – 2500(x – 4) V2 = 10000 – 1000x M2 = 1500x(x/2) – 2500(x – 4)(x – 4)/2 M2 = 750x2 – 1250(x – 4)2
r = 1500 lb/ft
x
4 ft
2500 lb/ft
r = 1500 lb/ft
x
12,000 lb⋅ft
30,000 lb⋅ft
4 ft 12 ft 4 ft
25 lb/ft
1500 lb/ft
Load Diagram
–6000 lb
6000 lb
Shear Diagram
6 ft
Moment Diagram
12,000
To draw the Shear Diagram: (1) For the first segment, V1 = 1500x is
linear; at x = 0, V1 = 0; at x = 4 ft, V1 =
6000 lb.
(2) For the second segment, V2 = 10000 –
1000x is also linear; at x = 4 ft, V1 =
6000 lb; at mid-span, x = 10 ft, V1 = 0.
(3) For the next half of the beam, the shear
diagram can be accomplished by the
concept of symmetry.
To draw the Moment Diagram: (1) For the first segment, M1 = 750x2 is a
second degree curve, an open upward
parabola; at x = 0, M1 = 0; at x = 4 ft,
M1 = 12000 lb⋅ft. (2) For the second segment, M2 = 750x2 –
1250(x – 4)2 is a second degree curve,
an downward parabola; at x = 4 ft, M2 =
12000 lb⋅ft; at mid-span, x = 10 ft, M2 =
30000 lb⋅ft. (2) The next half of the diagram, from x =
10 ft to x = 20 ft, can be drawn by using
the concept of symmetry.
190 Shear and Moment Equations and Diagrams; Relation between Load, Shear, and Moment; Moving Loads
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Problem 421. Write the shear and moment equations as functions of
the angle θ for the built-in arch shown in Fig. P-421.
Solution 421. For θ that is less than 90° Components of Q and P:
Qx = Q sin θ
Qy = Q cos θ
Px = P sin (90° – θ)
Px = P (sin 90° cos θ – cos 90° sin θ)
Px = P cos θ
Py = P cos (90° – θ)
Py = P (cos 90° cos θ + sin 90° sin θ)
Py = P sin θ Shear:
V = ∑Fy V = Qy – Py
V = Q cos θθθθ – P sin θθθθ Moment arms:
dQ = R sin θ
dP = R – R cos θ
dP = R (1 – cos θ) Moment:
M = ∑Mcounterclockwise – ∑Mclockwise M = Q(dQ) – P(dP)
M = QR sin θθθθ – PR(1 – cos θθθθ)
θ
R
Figure P-421 P
Q
B A
θ
P
Q
V
θ
90° – θ
dQ R
dP
191
Chapter 04 Shear and Moment in Beams www.mathalino.com
For θ that is greater than 90° Components of Q and P:
Qx = Q sin (180° – θ)
Qx = Q (sin 180° cos θ – cos 180° sin θ)
Qx = Q cos θ
Qy = Q cos (180° – θ)
Qy = Q (cos 180° cos θ + sin 180° sin θ)
Qy = –Q sin θ
Px = P sin (θ – 90°)
Px = P (sin θ cos 90° – cos θ sin 90°)
Px = –P cos θ
Py = P cos (θ – 90°)
Py = P (cos θ cos 90° + sin θ sin 90°)
Py = P sin θ Shear:
V = ∑Fy V = –Qy – Py
V = –(–Q sin θ) – P sin θ
V = Q sin θθθθ – P sin θθθθ Moment arms:
dQ = R sin (180° – θ)
dQ = R (sin 180° cos θ – cos 180° sin θ)
dQ = R sin θ
dP = R + R cos (180° – θ)
dP = R + R (cos 180° cos θ + sin 180° sin θ)
dP = R – R cos θ
dP = R(1 – cos θ) Moment:
M = ∑Mcounterclockwise – ∑Mclockwise M = Q(dQ) – P(dP)
M = QR sin θθθθ – PR(1 – cos θθθθ)
θ
P
Q
V
dQ R
dP
180° – θ
180° – θ
θ – 90°
192 Shear and Moment Equations and Diagrams; Relation between Load, Shear, and Moment; Moving Loads
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Problem 422. Write the shear and moment equations for the semicircular arch as shown in Fig. P-422 if (a) the load P is vertical as shown, and (b) the load is applied horizontally to the left at the top of the arch.
Solution 422. ∑MC = 0 2R(RA) = RP
RA = 21 P
For θ that is less than 90° Shear:
VAB = RA cos (90° – θ)
VAB = 21 P (cos 90° cos θ + sin 90° sin θ)
VAB = 21 P sin θθθθ
Moment arm:
d = R – R cos θ
d = R(1 – cos θ) Moment: MAB = Ra (d)
MAB = 21 PR(1 – cos θθθθ)
θ
R
Figure P-422
P
C
O
A
B
θ
P
C
O
A
B
RA
R
θ
O
A
RA
R
V
d
90° – θ
193
Chapter 04 Shear and Moment in Beams www.mathalino.com
For θ that is greater than 90° Components of P and RA:
Px = P sin (θ – 90°)
Px = P (sin θ cos 90° – cos θ sin 90°)
Px = –P cos θ
Py = P cos (θ – 90°)
Py = P (cos θ cos 90° + sin θ sin 90°)
Py = P sin θ
RAx = RA sin (θ – 90°)
RAx = 21 P (sin θ cos 90° – cos θ sin 90°)
RAx = – 21 P cos θ
RAy = RA cos (θ – 90°)
RAy = 21 P (cos θ cos 90° + sin θ sin 90°)
RAy = 21 P sin θ
Shear:
VBC = ∑Fy VBC = RAy – Py
VBC = 21 P sin θ – P sin θ
VBC = – 21 P sin θθθθ
Moment arm:
d = R cos (180° – θ)
d = R (cos 180° cos θ + sin 180° sin θ)
d = –R cos θ Moment:
MBC = ∑Mcounterclockwise – ∑Mclockwise
MBC = RA(R + d) – Pd
MBC = 21 P(R – R cos θ) – P(–R cos θ)
MBC = 21 PR – 2
1 PR cos θ + PR cos θ
MBC = 21 PR + 2
1 PR cos θ
MBC = 21 PR(1 + cos θθθθ)
θ
P
O
A
B
RA
R
180° – θ
V
d
θ – 90°
θ – 90°
R
194 Shear and Moment Equations and Diagrams; Relation between Load, Shear, and Moment; Moving Loads
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RELATIONSHIP BETWEEN LOAD, SHEAR, AND MOMENT The vertical shear at C in the figure shown in
previous section (Shear and Moment Diagram) is taken as
VC = (ΣFv)L = R1 – wx where R1 = R2 = wL/2
VC = 2
wL – wx
The moment at C is
MC = (ΣMC) = xwL
2 –
2
xwx
MC = 2
wLx –
2
2wx
If we differentiate M with respect to x:
dx
dM =
2
wL
dx
dx –
dx
dxx
w2
2
dx
dM =
2
wL – wx = shear
thus,
dx
dM = V
Thus, the rate of change of the bending moment with
respect to x is equal to the shearing force, or the slope of the moment diagram at the given point is the shear at that point.
195
Chapter 04 Shear and Moment in Beams www.mathalino.com
Differentiate V with respect to x gives
dx
dV = 0 – w = load
dx
dV = Load
Thus, the rate of change of the shearing force with
respect to x is equal to the load or the slope of the shear diagram at a given point equals the load at that point.
PROPERTIES OF SHEAR AND MOMENT DIAGRAMS
The following are some important properties of shear and moment diagrams:
1. The area of the shear diagram to the left or to the
right of the section is equal to the moment at that section.
2. The slope of the moment diagram at a given point is the shear at that point.
3. The slope of the shear diagram at a given point equals the load at that point.
4. The maximum moment occurs at the point of zero shears. This is in reference to property number 2, that when the shear (also the slope of the moment diagram) is zero, the tangent drawn to the moment diagram is horizontal.
5. When the shear diagram is increasing, the moment diagram is concave upward.
6. When the shear diagram is decreasing, the moment diagram is concave downward.
SIGN CONVENTIONS
The customary sign conventions for shearing force and bending moment are represented by the figures below. A force that tends to bend the beam
196 Shear and Moment Equations and Diagrams; Relation between Load, Shear, and Moment; Moving Loads
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downward is said to produce a positive bending moment. A force that tends to shear the left portion of the beam upward with respect to the right portion is said to produce a positive shearing force.
An easier way of determining the sign of the bending
moment at any section is that upward forces always cause positive bending moments regardless of whether they act to the left or to the right of the exploratory section.
SOLVED PROBLEMS INSTRUCTION: Without writing shear and moment equations, draw
the shear and moment diagrams for the beams specified in the following problems. Give numerical values at all change of loading positions and at all points of zero shear. (Note to instructor: Problems 403 to 420 may also be assigned for solution by semi graphical method describes in this article.)
Problem 425. Beam loaded as shown in Fig. P-425.
Positive Bending Negative Bending
Positive Shear Negative Shear
R1 R2
60 kN 30 kN
2 m 4 m 1 m
Figure P-425
197
Chapter 04 Shear and Moment in Beams www.mathalino.com
Solution 425. ∑MA = 0 6R2 = 2(60) + 7(30) R2 = 55 kN
∑MC = 0 6R1 + 1(30) = 4(60) R1 = 35 kN
35 kN
–25 kN
30 kN
Shear Diagram
70 kN⋅m
–30 kN⋅m
Moment Diagram
R1 = 35 kN R2 = 55 kN
60 kN 30 kN
2 m 4 m 1 m
Load Diagram
A
B
C
D
To draw the Shear Diagram: (1) VA = R1 = 35 kN
(2) VB = VA + Area in load diagram – 60 kN
VB = 35 + 0 – 60 = –25 kN
(3) VC = VB + area in load diagram + R2
VC = –25 + 0 + 55 = 30 kN
(4) VD = VC + Area in load diagram – 30 kN
VD = 30 + 0 – 30 = 0
To draw the Moment Diagram: (1) MA = 0
(2) MB = MA + Area in shear diagram
MB = 0 + 35(2) = 70 kN⋅m
(3) MC = MB + Area in shear diagram
MC = 70 – 25(4) = –30 kN⋅m
(4) MD = MC + Area in shear diagram MD = –30 + 30(1) = 0
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198 Shear and Moment Equations and Diagrams; Relation between Load, Shear, and Moment; Moving Loads
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Problem 426. Cantilever beam acted upon by a uniformly distributed load and a couple as shown in Fig. P-426.
Solution 426.
Problem 427. Beam loaded as shown in Fig. P-427.
1 m
Figure P-426
2 m 2 m
5 kN/m
M = 60 kN⋅m
To draw the Shear Diagram
(1) VA = 0
(2) VB = VA + Area in load diagram
VB = 0 – 5(2)
VB = –10 kN
(3) VC = VB + Area in load diagram
VC = –10 + 0
VC = –10 kN
(4) VD = VC + Area in load diagram
VD = –10 + 0
VD = –10 kN
To draw the Moment Diagram
(1) MA = 0
(2) MB = MA + Area in shear diagram
MB = 0 – ½ (2)(10)
MB = –10 kN⋅m
(3) MC = MB + Area in shear diagram
MC = –10 – 10(2)
MC = –30 kN⋅m
MC2 = –30 + M = –30 + 60 = 30 kN⋅m
(4) MD = MC2 + Area in shear diagram
MD = 30 – 10(1)
MD = 20 kN⋅m –10 kN⋅m
–30 kN⋅m
30 kN⋅m
20 kN⋅m
Moment Diagram
2nd deg
1st deg
Shear Diagram
–10 kN
1 m 2 m 2 m
5 kN/m
M = 60 kN⋅m
Load Diagram
A
B C D
Figure P-427
9 ft
100 lb/ft
3 ft
800 lb
R1 R2
199
Chapter 04 Shear and Moment in Beams www.mathalino.com
Solution 427. ∑MC = 0 12R1 = 100(12)(6) + 800(3) R1 = 800 lb
∑MA = 0 12R2 = 100(12)(6) + 800(9) R2 = 1200 lb Problem 428. Beam loaded as shown in Fig. P-428.
Figure P-428
10 kN/m
1 m 1 m 3 m 2 m
R1 R2
25 kN⋅m
To draw the Shear Diagram
(1) VA = R1 = 800 lb
(2) VB = VA + Area in load diagram
VB = 800 – 100(9)
VB = –100 lb
VB2 = –100 – 800 = –900 lb
(3) VC = VB2 + Area in load diagram
VC = –900 – 100(3)
VC = –1200 lb
(4) Solving for x:
x / 800 = (9 – x) / 100
100x = 7200 – 800x
x = 8 ft
To draw the Moment Diagram
(1) MA = 0
(2) Mx = MA + Area in shear diagram
Mx = 0 + ½ (8)(800) = 3200 lb⋅ft (3) MB = Mx + Area in shear diagram
MB = 3200 – ½ (1)(100) = 3150 lb⋅ft (4) MC = MB + Area in shear diagram
MC = 3150 – ½ (900 + 1200)(3) = 0
(5) The moment curve BC is downward
parabola with vertex at A’. A’ is the
location of zero shear for segment BC.
9 ft
100 lb/ft
3 ft
800 lb
R1 = 800 lb R2 = 1200 lb
A B C
Load Diagram
x = 8 ft
800 lb
–100 lb
–900 lb
–1200 lb Shear Diagram
3200 lb⋅ft 3150 lb⋅ft
Moment Diagram
A’
200 Shear and Moment Equations and Diagrams; Relation between Load, Shear, and Moment; Moving Loads
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Solution 428. ∑MD = 0 5R1 = 50(0.5) + 25 R1 = 10 kN
∑MA = 0 5R2 + 25 = 50(4.5) R2 = 40 kN Problem 429. Beam loaded as shown in Fig. P-429.
To draw the Shear Diagram
(1) VA = R1 = 10 kN
(2) VB = VA + Area in load diagram
VB = 10 + 0 = 10 kN
(3) VC = VB + Area in load diagram
VC = 10 + 0 = 10 kN
(4) VD = VC + Area in load diagram
VD = 10 – 10(3) = –20 kN
VD2 = –20 + R2 = 20 kN
(5) VE = VD2 + Area in load diagram
VE = 20 – 10(2) = 0
(6) Solving for x:
x / 10 = (3 – x) / 20
20x = 30 – 10x
x = 1 m
To draw the Moment Diagram
(1) MA = 0
(2) MB = MA + Area in shear diagram
MB = 0 + 1(10) = 10 kN⋅m
MB2 = 10 – 25 = –15 kN⋅m
(3) MC = MB2 + Area in shear diagram
MC = –15 + 1(10) = –5 kN⋅m
(4) Mx = MC + Area in shear diagram
Mx = –5 + ½ (1)(10) = 0
(5) MD = Mx + Area in shear diagram
MD = 0 – ½ (2)(20) = –20 kN⋅m
(6) ME = MD + Area in shear diagram
ME = –20 + ½ (2)(20) = 0
1 m 1 m
B
10 kN/m
3 m 2 m
R1 = 10 kN R2 = 40 kN
A
C D E
50 kN
0.5 m
25 kN⋅m
Load Diagram
–20 kN⋅m
–5 kN⋅m
Moment Diagram
–15 kN⋅m
10 kN⋅m
Shear Diagram
10 kN
–20 kN
20 kN
x = 1 m
100 lb
2 ft 2 ft 2 ft
R1 R2
120 lb/ft 120 lb/ft
Figure P-429
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Solution 429. ∑MC = 0 4R1 + 120(2)(1) = 100(2) + 120(2)(3) R1 = 170 lb
∑MA = 0 4R2 = 120(2)(1) + 100(2) + 120(2)(5) R2 = 410 lb Problem 430. Beam loaded as shown in P-430.
1000 lb
5 ft R1 R2
400 lb/ft 200 lb/ft
Figure P-430
2000 lb
10 ft 10 ft
To draw the Shear Diagram
(1) VA = R1 = 170 lb
(2) VB = VA + Area in load diagram
VB = 170 – 120(2) = –70 lb
VB2 = –70 – 100 = –170 lb
(3) VC = VB2 + Area in load diagram
VC = –170 + 0 = –170 lb
VC2 = –170 + R2
VC2 = –170 + 410 = 240 lb
(4) VD = VC2 + Area in load diagram
VD = 240 – 120(2) = 0
(5) Solving for x:
x / 170 = (2 – x) / 70
70x = 340 – 170x
x = 17 / 12 ft = 1.42 ft
To draw the Moment Diagram
(1) MA = 0
(2) Mx = MA + Area in shear diagram
Mx = 0 + ½ (17/12)(170)
Mx = 1445/12 = 120.42 lb⋅ft (3) MB = Mx + Area in shear diagram
MB = 1445/12 – ½ (2 – 17/12)(70)
MB = 100 lb⋅ft (4) MC = MB + Area in shear diagram
MC = 100 – 170(2) = –240 lb⋅ft (5) MD = MC + Area in shear diagram
MD = –240 + ½ (2)(240) = 0
–240 lb⋅ft
100 lb⋅ft
120.42 lb⋅ft
100 lb
2 ft 2 ft 2 ft
R1 = 170 lb R2 = 410 lb
120 lb/ft 120 lb/ft
A B C D
Load Diagram
–70 lb
170 lb
–170 lb
240 lb
Shear Diagram
x = 1.42 ft
Moment Diagram
202 Shear and Moment Equations and Diagrams; Relation between Load, Shear, and Moment; Moving Loads
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Solution 430. ∑MD = 0 20R1 = 1000(25) + 400(5)(22.5) + 2000(10) + 200(10)(5) R1 = 5000 lb
∑MB = 0 20R2 + 1000(5) + 400(5)(2.5) = 2000(10) + 200(10)(15) R2 = 2000 lb Problem 431. Beam loaded as shown in Fig. P-431.
1000 lb
5 ft R1 = 5000 lb R2 = 2000 lb
400 lb/ft 200 lb/ft
2000 lb
10 ft 10 ft
A
B C
D
Load Diagram
–10000 lb⋅ft
10000 lb⋅ft
5 ft
Moment Diagram
To draw the Shear Diagram (1) VA = –1000 lb
(2) VB = VA + Area in load diagram
VB = –1000 – 400(5) = –3000 lb
VB2 = –3000 + R1 = 2000 lb
(3) VC = VB2 + Area in load diagram
VC = 2000 + 0 = 2000 lb
VC2 = 2000 – 2000 = 0
(4) VD = VC2 + Area in load diagram
VD = 0 + 200(10) = 2000 lb
To draw the Moment Diagram
(1) MA = 0
(2) MB = MA + Area in shear diagram
MB = 0 – ½ (1000 + 3000)(5)
MB = –10000 lb⋅ft (3) MC = MB + Area in shear diagram
MC = –10000 + 2000(10) = 10000 lb⋅ft (4) MD = MC + Area in shear diagram
MD = 10000 – ½ (10)(2000) = 0
(5) For segment BC, the location of zero
moment can be accomplished by
symmetry and that is 5 ft from B.
(6) The moment curve AB is a downward
parabola with vertex at A’. A’ is the
location of zero shear for segment AB at point outside the beam.
–3000 lb –1000 lb
2000 lb
–2000 lb
Shear Diagram
A’
A’
Figure P-431
7 m 3 m
10 kN/m
40 kN 50 kN 20 kN/m
1 m 2 m
R1 R2
203
Chapter 04 Shear and Moment in Beams www.mathalino.com
Solution 431. ∑MD = 0 7R1 + 40(3) = 5(50) + 10(10)(2) + 20(4)(2) R1 = 70 kN
∑MA = 0 7R2 = 50(2) + 10(10)(5) + 20(4)(5) + 40(10) R2 = 200 lb
To draw the Shear Diagram
(1) VA = R1 = 70 kN
(2) VB = VA + Area in load diagram
VB = 70 – 10(2) = 50 kN
VB2 = 50 – 50 = 0
(3) VC = VB2 + Area in load diagram
VC = 0 – 10(1) = –10 kN
(4) VD = VC + Area in load diagram
VD = –10 – 30(4) = –130 kN
VD2 = –130 + R2
VD2 = –130 + 200 = 70 kN
(5) VE = VD2 + Area in load diagram
VE = 70 – 10(3) = 40 kN
VE2 = 40 – 40 = 0
To draw the Moment Diagram
(1) MA = 0
(2) MB = MA + Area in shear diagram
MB = 0 + ½ (70 + 50)(2) = 120 kN⋅m(3) MC = MB + Area in shear diagram
MC = 120 – ½ (1)(10) = 115 kN⋅m
(4) MD = MC + Area in shear diagram
MD = 115 – ½ (10 + 130)(4)
MD = –165 kN⋅m
(5) ME = MD + Area in shear diagram
ME = –165 + ½ (70 + 40)(3) = 0
(6) Moment curves AB, CD and DE are
downward parabolas with vertices at
A’, B’ and C’, respectively. A’, B’ and
C’ are corresponding zero shear
points of segments AB, CD and DE.
115 kN⋅m 120 kN⋅m
5 m 3 m
10 kN/m
40 kN 50 kN
20 kN/m 1 m 2 m
R1 = 70 kN R2 = 200 lb
2 m
Load Diagram
A
B C
D E
50 kN
–10 kN
70 kN
–130 kN
70 kN 40 kN
Shear Diagram
4 m
–165 kN⋅m
Moment Diagram
B’
A’
C’
x
y
a = 1/3 m
(7) Solving for point of zero moment:
a / 10 = (a + 4) / 130
130a = 10a + 40
a = 1/3 m
y / (x + a) = 130 / (4 + a)
y = 130(x + 1/3) / (4 + 1/3)
y = 30x + 10
MC = 115 kN⋅m
Mzero = MC + Area in shear
0 = 115 – ½ (10 + y)x
(10 + y)x = 230
(10 + 30x + 10)x = 230
30x2 + 20x – 230 = 0
3x2 + 2x – 23 = 0
x = 2.46 m
zero moment is at 2.46 m from C
Another way to solve the
location of zero moment
is by the squared
property of parabola (see
Problem 434). This point
is the appropriate location
for construction joint of
concrete structures.
204 Shear and Moment Equations and Diagrams; Relation between Load, Shear, and Moment; Moving Loads
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Problem 432. Beam loaded as shown in Fig. P-432.
Solution 432. ∑ME = 0 5R1 + 120 = 6(60) + 40(3)(3.5) R1 = 132 kN
∑MB = 0 5R2 + 60(1) = 40(3)(1.5) + 120 R2 = 48 kN
Figure P-432
R2
1 m 3 m 1 m 1 m
R1
60 kN 40 kN/m
M = 120 kN⋅m
R2 = 48 kN
1 m 3 m 1 m 1 m
R1 = 132 kN
60 kN 40 kN/m
M = 120 kN⋅m
Load Diagram
A B C
D
E
–60 kN
72 kN
–48 kN
Shear Diagram
x = 1.8 m
To draw the Shear Diagram
(1) VA = –60 kN
(2) VB = VA + Area in load diagram
VB = –60 + 0 = –60 kN
VB2 = VB + R1 = –60 + 132 = 72 kN
(3) VC = VB2 + Area in load diagram
VC = 72 – 3(40) = –48 kN
(4) VD = VC + Area in load diagram
VD = –48 + 0 = –48 kN
(5) VE = VD + Area in load diagram
VE = –48 + 0 = –48 kN
VE2 = VE + R2 = –48 + 48 = 0
(6) Solving for x:
x / 72 = (3 – x) / 48
48x = 216 – 72x
x = 1.8 m
To draw the Moment Diagram (1) MA = 0
(2) MB = MA + Area in shear diagram
MB = 0 – 60(1) = –60 kN⋅m
(3) Mx = MB + Area in shear diagram
MX = –60 + ½ (1.8)(72) = 4.8 kN⋅m
(4) MC = MX + Area in shear diagram
MC = 4.8 – ½ (3 – 1.8)(48) = –24 kN⋅m
(5) MD = MC + Area in shear diagram
MD = –24 – ½ (24 + 72)(1) = –72 kN⋅m
MD2 = –72 + 120 = 48 kN⋅m
(6) ME = MD2 + Area in shear diagram
ME = 48 – 48(1) = 0
(7) The location of zero moment on
segment BC can be determined using
the squared property of parabola. See the solution of Problem 434.
–60 kN⋅m
4.8 kN⋅m
–24 kN⋅m
–72 kN⋅m
48 kN⋅m
Moment Diagram
205
Chapter 04 Shear and Moment in Beams www.mathalino.com
Problem 433. Overhang beam loaded by a force and a couple as shown in Fig. P-433.
Solution 433. ∑MC = 0 5R1 + 2(750) = 3000 R1 = 300 lb
∑MA = 0 5R2 + 3000 = 7(750) R2 = 450 lb
Figure P-433
R2
2 ft
R1
750 lb
3000 lb⋅ft
3 ft 2 ft
R2 = 450 lb
2 ft
R1 = 300 lb
750 lb
3000 lb⋅ft
3 ft 2 ft
B C D A
Load Diagram
600 lb⋅ft
–2400 lb⋅ft
–1500 lb⋅ft
Moment Diagram
750 lb
300 lb
Shear Diagram
To draw the Shear Diagram
(1) VA = R1 = 300 lb
(2) VB = VA + Area in load diagram
VB = 300 + 0 = 300 lb
(3) VC = VB + Area in load diagram
VC = 300 + 0 = 300 lb
VC2 = VC + R2 = 300 + 450 = 750 lb
(5) VD = VC2 + Area in load diagram
VD = 750 + 0 = 750
VD2 = VD – 750 = 750 – 750 = 0
To draw the Moment Diagram
(1) MA = 0
(2) MB = VA + Area in shear diagram
MB = 0 + 300(2) = 600 lb⋅ft MB2 = VB – 3000
MB2 = 600 – 3000 = –2400 lb⋅ft (3) MC = MB2 + Area in shear diagram
MC = –2400 + 300(3) = –1500 lb⋅ft (4) MD = MC + Area in shear diagram
MD = –1500 + 750(2) = 0
206 Shear and Moment Equations and Diagrams; Relation between Load, Shear, and Moment; Moving Loads
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Problem 434. Beam loaded as shown in Fig. P-434.
Solution 434. ∑ME = 0 6R1 + 120 = 20(4)(6) + 60(4) R1 = 100 kN
∑MB = 0 6R2 = 20(4)(0) + 60(2) + 120 R2 = 40 kN
Figure P-434
R2
2 m
R1
60 kN
20 kN/m
M = 120 kN⋅m
2 m 2 m 2 m
To draw the Shear Diagram
(1) VA = 0
(2) VB = VA + Area in load diagram
VB = 0 – 20(2) = –40 kN
VB2 = VB + R1 = –40 + 100 = 60 kN
(3) VC = VB2 + Area in load diagram
VC = 60 – 20(2) = 20 kN
VC2 = VC – 60 = 20 – 60 = –40 kN
(4) VD = VC2 + Area in load diagram
VD = –40 + 0 = –40 kN
(5) VE = VD + Area in load diagram
VE = –40 + 0 = –40 kN
VE2 = VE + R2 = –40 + 40 = 0
To draw the Moment Diagram
(1) MA = 0
(2) MB = MA + Area in shear diagram
MB = 0 – ½ (40)(2) = –40 kN⋅m
(3) MC = MB + Area in shear diagram
MC = –40 + ½ (60 + 20)(2) = 40 kN⋅m
(4) MD = MC + Area in shear diagram
MD = 40 – 40(2) = –40 kN⋅m
MD2 = MD + M = –40 + 120 = 80 kN⋅m
(5) ME = MD2 + Area in shear diagram
ME = 80 – 40(2) = 0
(6) Moment curve BC is a downward parabola
with vertex at C’. C’ is the location of zero
shear for segment BC.
(7) Location of zero moment at segment BC:
By squared property of parabola:
(3 – x)2 / 50 = 32 / (50 + 40)
3 – x = 2.236
x = 0.764 m from B
–40 kN⋅m –40 kN⋅m
80 kN⋅m
Moment Diagram
C’
R2 = 40 kN
2 m
R1 = 100 kN
60 kN
20 kN/m
M = 120 kN⋅m
2 m 2 m 2 m
A B C D E
Load Diagram
–40 kN
60 kN
20 kN
–40 kN
Shear Diagram
1 m
50 kN⋅m 40 kN⋅m
3 – x
x
1 m
207
Chapter 04 Shear and Moment in Beams www.mathalino.com
Problem 435. Beam loaded and supported as shown in Fig. P-435.
Solution 435. ∑MB = 0 2wo (5) = 10(4)(0) + 20(2) + 40(3) wo = 16 kN/m
∑Mmidpoint of EF = 0 5R1 = 10(4)(5) + 20(3) + 40(2)
R1 = 68 kN
Figure P-435
2 m
R1
20 kN
10 kN/m 40 kN
2 m 1 m wo
1 m 2 m
64 kN⋅m
–32 kN
–20 kN
48 kN
28 kN
8 kN
2 m
R1 = 68 kN
20 kN
10 kN/m 40 kN
2 m 1 m wo = 16 kN/m
1 m 2 m
A B C D E
F
Load Diagram
Shear Diagram
–20 kN⋅m
56 kN⋅m
32 kN⋅m
C’
Moment Diagram
To draw the Shear Diagram
(1) MA = 0
(2) MB = MA + Area in load diagram
MB = 0 – 10(2) = –20 kN
MB2 + MB + R1 = –20 + 68 = 48 kN
(3) MC = MB2 + Area in load diagram
MC = 48 – 10(2) = 28 kN
MC2 = MC – 20 = 28 – 20 = 8 kN
(4) MD = MC2 + Area in load diagram
MD = 8 + 0 = 8 kN
MD2 = MD – 40 = 8 – 40 = –32 kN
(5) ME = MD2 + Area in load diagram
ME = –32 + 0 = –32 kN
(6) MF = ME + Area in load diagram
MF = –32 + wo(2)
MF = –32 + 16(2) = 0
To draw the Moment Diagram
(1) MA = 0
(2) MB = MA + Area in shear diagram
MB = 0 – ½ (20)(2) = –20 kN⋅m
(3) MC = MB + Area in shear diagram
MC = –20 + ½ (48 + 28)(2)
MC = 56 kN⋅m
(4) MD = MC + Area in shear diagram
MD = 56 + 8(1) = 64 kN⋅m
(5) ME = MD + Area in shear diagram
ME = 64 – 32(1) = 32 kN⋅m
(6) MF = ME + Area in shear diagram
MF = 32 – ½ (32)(2) = 0
(7) The location and magnitude of moment
at C’ are determined from shear
diagram. By squared property of
parabola, x = 0.44 m from B.
x
2.8 m
95.2 kN⋅m
208 Shear and Moment Equations and Diagrams; Relation between Load, Shear, and Moment; Moving Loads
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Problem 436. A distributed load is supported by two distributed reactions as shown in Fig. P-436.
Solution 436. ∑Mmidpoint of CD = 0 4w1 (11) = 440(8)(5) w1 = 400 lb/ft
∑Mmidpoint of AB = 0 2w2 (11) = 440(8)(6) w2 = 960 lb/ft
R2 = w2 lb/ft
Figure P-436
R1 = w1 lb/ft
8 ft
4 ft 2 ft
440 lb/ft
R2 = 960 lb/ft R1 = 400 lb/ft
8 ft
4 ft 2 ft
440 lb/ft
A B C
D
Load Diagram
1600 lb
–1920 lb Shear Diagram
To draw the Shear Diagram
(1) VA = 0
(2) VB = VA + Area in load diagram
VB = 0 + 400(4) = 1600 lb
(3) VC = VB + Area in load diagram
VC = 1600 – 440(8) = –1920 lb
(4) VD = VC + Area in load diagram
VD = –1920 + 960(2) = 0
(5) Location of zero shear:
x / 1600 = (8 – x) / 1920
x = 40/11 ft = 3.636 ft from B
To draw the Moment Diagram
(1) MA = 0
(2) MB = MA + Area in shear diagram
MB = 0 + ½ (1600)(4) = 3200 lb⋅ft (3) Mx = MB + Area in shear diagram
Mx = 3200 + ½ (1600)(40/11)
Mx = 6109.1 lb⋅ft (4) MC = Mx + Area in shear diagram
MC = 6109.1 – ½ (8 – 40/11)(1920)
MC = 1920 lb⋅ft (5) MD = MC + Area in shear diagram
MD = 1920 – ½ (1920)(2) = 0
x = 3.636 ft
Moment Diagram
3200 lb⋅ft
6109.1 lb⋅ft
1920 lb⋅ft
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Chapter 04 Shear and Moment in Beams www.mathalino.com
Problem 437. Cantilever beam loaded as shown in Fig. P-437 Solution 437.
4 ft 2 ft 2 ft
1000 lb
500 lb
400 lb/ft
Figure P-437
To draw the Shear Diagram (1) VA = –1000 lb
VB = VA + Area in load diagram
VB = –1000 + 0 = –1000 lb
VB2 = VB + 500 = –1000 + 500
VB2 = –500 lb
(2) VC = VB2 + Area in load diagram
VC = –500 + 0 = –500 lb
(3) VD = VC + Area in load diagram
VD = –500 – 400(4) = –2100 lb
To draw the Moment Diagram (1) MA = 0
(2) MB = MA + Area in shear diagram
MB = 0 – 1000(2) = –2000 lb⋅ft (3) MC = MB + Area in shear diagram
MC = –2000 – 500(2) = –3000 lb⋅ft (4) MD = MC + Area in shear diagram
MD = –3000 – ½ (500 + 2100)(4) MD = –8200 lb⋅ft
4 ft 2 ft 2 ft
1000 lb
500 lb
400 lb/ft
A B C D
Load Diagram
–3000 lb⋅ft
–2000 lb⋅ft
Moment Diagram
–8200 lb⋅ft
Shear Diagram
–1000 lb
–500 lb
–2100 lb
210 Shear and Moment Equations and Diagrams; Relation between Load, Shear, and Moment; Moving Loads
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Problem 438. The beam loaded as shown in Fig. P-438 consists of two segments joined by a frictionless hinge at which the bending moment is zero.
Solution 438.
∑MH = 0 4R1 = 200(6)(3) R1 = 900 lb
Figure P-438
4 ft 2 ft 2 ft
R1
200 lb/ft
Hinge
4 ft 2 ft
R1
200 lb/ft
A B H
To draw the Shear Diagram
(1) VA = 0
(2) VB = VA + Area in load diagram
VB = 0 – 200(2) = –400 lb
VB2 = VB + R1 = –400 + 900 = 500 lb
(3) VH = VB2 + Area in load diagram
VH = 500 – 200(4) = –300 lb
(4) VC = VH + Area in load diagram
VC = –300 – 200(2) = –700 lb
(5) Location of zero shear:
x / 500 = (4 – x) / 300
300x = 2000 – 500x
x = 2.5 ft
To draw the Moment Diagram
(1) MA = 0
(2) MB = MA + Area in shear diagram
MB = 0 – ½ (400)(2) = –400 lb⋅ft (3) Mx = MB + Area in load diagram
Mx = –400 + ½ (500)(2.5)
Mx = 225 lb⋅ft (4) MH = Mx + Area in load diagram
MH = 225 – ½ (300)(4 – 2.5) = 0 ok!
(5) MC = MH + Area in load diagram
MC = 0 – ½ (300 + 700)(2)
MC = –1000 lb⋅ft (6) The location of zero moment in segment
BH can easily be found by symmetry.
–400 lb
–300 lb
–700 lb
500 lb
4 ft 2 ft 2 ft
R1 = 900 lb
200 lb/ft
A B H C
Load Diagram
Moment Diagram
x = 2.5 ft
–400 lb⋅ft
225 lb⋅ft
–1000 lb⋅ft
Shear Diagram
1.0 ft
211
Chapter 04 Shear and Moment in Beams www.mathalino.com
Problem 439. A beam supported on three reactions as shown in Fig. P-439 consists of two segments joined by frictionless hinge at which the bending moment is zero.
Solution 439.
∑MH = 0 ∑MA = 0 8R1 = 4000(4) 8VH = 4000(4) R1 = 2000 lb VH = 2000 lb
∑MD = 0 10R2 = 2000(14) + 400(10)(5) R2 = 4800 lb
∑MH = 0 14R3 + 4(4800) = 400(10)(9) R3 = 1200 lb
400 lb/ft
Hinge
4000 lb
R1 R2 R3 4 ft 4 ft 4 ft 10 ft
Figure P-439
400 lb/ft
Hinge
R2 R3 4 ft 10 ft
C D H
VH = 2000 lb
Hinge
4000 lb
R1 4 ft 4 ft
VH
A B H
400 lb/ft
Hinge
4000 lb
R1 = 2000 lb R2 = 4800 lb R3 = 1200 lb
4 ft 4 ft 4 ft 10 ft
A B C D H
Load Diagram
8000 lb ft⋅
–8000 lb⋅ft
1800 lb⋅ft
3 ft
Moment Diagram
2000 lb
–2000 lb
2800 lb
–1200 lb
Shear Diagram
x = 7 ft
To draw the Shear Diagram
(1) VA = 0
(2) VB = 2000 lb
VB2 = 2000 – 4000 = –2000 lb
(3) VH = –2000 lb
(3) VC = –2000 lb
VC = –2000 + 4800 = 2800 lb (4) VD = 2800 – 400(10) = –1200 lb
(5) Location of zero shear:
x / 2800 = (10 – x) / 1200
1200x = 28000 – 2800x
x = 7 ft
To draw the Moment Diagram
(1) MA = 0
(2) MB = 2000(4) = 8000 lb⋅ft
(3) MH = 8000 – 4000(2) = 0
(4) MC = –400(2)
MC = –8000 lb⋅ft (5) Mx = –800 + ½(2800)(7)
Mx = 1800 lb⋅ft (6) MD = 1800 – ½(1200)(3)
MD = 0
(7) Zero M is 4 ft from R2
212 Shear and Moment Equations and Diagrams; Relation between Load, Shear, and Moment; Moving Loads
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Problem 440. A frame ABCD, with rigid corners at B and C, supports the concentrated load as shown in Fig. P-440. (Draw shear and moment diagrams for each of the three parts of the frame.)
Solution 440.
Member BC
B
L/2
P
PL/2
Lo
ad
Dia
gra
m
P
C PL/2
0
Sh
ea
r Dia
gra
m
Mo
me
nt D
iag
ram
– P
L/2
A
P
B
C D
L/2
L/2
L
Figure P-440
Member AB
A
P
B L/2
P
PL/2
P
–PL/2
Load Diagram
Shear Diagram
Moment Diagram
Member CD
PL/2
P
D C
Load Diagram
L
PL/2
–PL/2
Moment Diagram
Shear Diagram
–P
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Chapter 04 Shear and Moment in Beams www.mathalino.com
Problem 441. A beam ABCD is supported by a roller at A and a hinge at D. It is subjected to the loads shown in Fig. P-441, which act at the ends of the vertical members BE and CF. These vertical members are rigidly attached to the beam at B and C. (Draw shear and moment diagrams for the beam ABCD only.)
Solution 441. FBH = 14 kN to the right MB = 14(2)
MB = 28 kN⋅m counterclockwise FCH = 3/5 (10) FCH = 6 kN to the right FCV = 4/5 (10) FCV = 8 kN upward
MC = FCH (2) = 6(2)
MC = 12 kN⋅m clockwise
∑MD = 0 6RA + 12 + 8(2) = 28 RA = 0
∑MA = 0 6RDV + 12 = 28 + 8(4) RDV = 8 kN
∑FH = 0 RDH = 14 + 6 RDH = 20 kN
2 m 2 m 2 m
2 m
2 m
E 14 kN
10 kN
A B C D
F Figure P-441 3
4
2 m 2 m 2 m
2 m
2 m
E 14 kN
10 kN
A B C D
F 3
4 5
2 m 2 m 2 m
A B C D
28 kN⋅m 12 kN⋅m
14 kN 6 kN
8 kN
RA = 0 RDV = 8 kN
RDH = 20 kN
214 Shear and Moment Equations and Diagrams; Relation between Load, Shear, and Moment; Moving Loads
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Problem 442. Beam carrying the uniformly varying load shown in
Fig. P-442.
Solution 442. ∑MR2 = 0
LR1 = 31 L ( 2
1 Lwo)
R1 = 61 Lwo
∑MR1 = 0
LR2 = 32 L ( 2
1 Lwo)
R2 = 31 Lwo
2 m 2 m 2 m
A B C D
28 kN⋅m 12 kN⋅m
14 kN 6 kN
8 kN
RA = 0 RDV = 8 kN
RDH = 20 kN
Load Diagram
8 kN
Shear Diagram
Moment Diagram
–28 kN⋅m
–16 kN⋅m
To draw the Shear Diagram
(1) Shear in segments AB and BC is
zero.
(2) VC = 8
(3) VD = VC + Area in load diagram
VD = 8 + 0 = 8 kN
VD2 = VD – RDV
VD2 = 8 – 8 = 0
To draw the Moment Diagram
(1) Moment in segment AB is zero
(2) MB = –28 kN⋅m
(3) MC = MB + Area in shear diagram
MC = –28 + 0 = –28 kN⋅m
MC2 = MC + 12 = –28 + 12
MC2 = –16 kN⋅m
(4) MD = MC2 + Area in shear diagram
MD = –16 + 8(2)
MD = 0
R1 R2
L
wO wO
Figure P-442
R1 R2
L
wo
½ Lwo
2/3 L 1/3 L
215
Chapter 04 Shear and Moment in Beams www.mathalino.com
Problem 443. Beam carrying the triangular loads shown in Fig. P-
443.
To draw the Shear Diagram
(1) VA = R1 = 1/6 Lwo
(2) VB = VA + Area in load diagram
VB = 1/6 Lwo – 1/2 Lwo
VB = –1/3 Lwo
(3) Location of zero shear C:
By squared property of parabola:
x2 / (1/6 Lwo) = L2 / (1/6 Lwo + 1/3 Lwo)
6x2 = 2L2
x = L / √3
(4) The shear in AB is a parabola with vertex at A,
the starting point of uniformly varying load.
The load in AB is 0 at A to downward wo or –
wo at B, thus the slope of shear diagram is
decreasing. For decreasing slope, the
parabola is open downward.
To draw the Moment Diagram (1) MA = 0
(2) MC = MA + Area in shear diagram
MC = 0 + 2/3 (L/√3)(1/6 Lwo)
MC = 0.06415L2wo = Mmax
(3) MB = MC + Area in shear diagram
MB = MC – A1 � see figure for solving A1
For A1:
A1 = 1/3 L(1/6 Lwo + 1/3 Lwo)
– 1/3 (L/√3)(1/6 Lwo)
– 1/6 Lwo (L – L/√3)
A1 = 0.16667L2wo – 0.03208L2wo
– 0.07044L2wo
A1 = 0.06415L2wo
MB = 0.06415L2wo – 0.06415L2wo = 0
(4) The shear diagram is second degree curve,
thus the moment diagram is a third degree
curve. The maximum moment (highest point)
occurred at C, the location of zero shear. The
value of shears in AC is positive then the
moment in AC is increasing; at CB the shear is
negative, then the moment in CB is
decreasing.
R1 = 1/6 Lwo R2 = 1/3 Lwo
L
wo
1/6 Lwo
–1/3 Lwo
Load Diagram
Shear Diagram
Moment Diagram
Mmax = 0.06415L2wo
A B
C
x = L / √3
Figure P-443
R1 R2
wO
L/2 L/2
1/6 Lwo
–1/3 Lwo
L / √3 A1
Figure for solving A1
216 Shear and Moment Equations and Diagrams; Relation between Load, Shear, and Moment; Moving Loads
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Solution 443. By symmetry:
R1 = R2 = )( 21
21
oLw
R1 = R2 = oLw41
Problem 444. Beam loaded as shown in Fig. P-444. Solution 444. Total load
= 2[ 21 (L/2)(wo)]
= 21 Lwo
By symmetry
R1 = R2 = 21 × total load
R1 = R2 = 41 Lwo
To draw the Shear Diagram (1) VA = R1 = ¼ Lwo
(2) VB = VA + Area in load diagram
VB = ¼ Lwo – ½ (L/2)(wo) = 0
(3) VC = VB + Area in load diagram
VC = 0 – ½ (L/2)(wo) = –¼ Lwo
(4) Load in AB is linear, thus, VAB is second degree or
parabolic curve. The load is from 0 at A to wo (wo is
downward or –wo) at B, thus the slope of VAB is
decreasing.
(5) VBC is also parabolic since the load in BC is linear.
The magnitude of load in BC is from –wo to 0 or
increasing, thus the slope of VBC is increasing.
To draw the Moment Diagram (1) MA = 0
(2) MB = MA + Area in shear diagram
MB = 0 + 2/3 (L/2)(1/4 Lwo) = 1/12 Lwo
(3) MC = MB + Area in shear diagram
MC = 1/12 Lwo – 2/3 (L/2)(1/4 Lwo) = 0
(4) MAC is third degree because the shear diagram in AC
is second degree.
(5) The shear from A to C is decreasing, thus the slope
of moment diagram from A to C is decreasing. Moment Diagram
o
2
12
1wL
Shear Diagram o4
1Lw−
o4
1Lw
Load Diagram
R2 = ¼ Lwo R1 = ¼ Lwo
wO
L/2 L/2 A B C
Figure P-444
R1 R2
wo
L/2 L/2
wo
217
Chapter 04 Shear and Moment in Beams www.mathalino.com
Problem 445. Beam carrying the loads shown in Fig. P-445.
Solution 445. ∑MR2 = 0 5R1 = 80(3) + 90(2) R1 = 84 kN
∑MR1 = 0 5R2 = 80(2) + 90(3) R2 = 86 kN Checking R1 + R2 = F1 + F2 ok!
To draw the Shear Diagram
(1) VA = R1 = ¼ Lwo
(2) VB = VA + Area in load diagram
VB = ¼ Lwo – ½ (L/2)(wo) = 0
(3) VC = VB + Area in load diagram
VC = 0 – ½ (L/2)(wo) = –¼ Lwo
(4) The shear diagram in AB is second degree
curve. The shear in AB is from –wo
(downward wo) to zero or increasing, thus,
the slope of shear at AB is increasing (upward
parabola).
(5) The shear diagram in BC is second degree
curve. The shear in BC is from zero to –wo
(downward wo) or decreasing, thus, the slope
of shear at BC is decreasing (downward
parabola)
To draw the Moment Diagram (1) MA = 0
(2) MB = MA + Area in shear diagram
MB = 0 + 1/3 (L/2)(¼ Lwo) = 1/24 L2wo
(3) MC = MB + Area in shear diagram
MC = 1/24 L2wo – 1/3 (L/2)(¼ Lwo) = 0
(4) The shear diagram from A to C is decreasing,
thus, the moment diagram is a concave
downward third degree curve.
o4
1Lw
o4
1Lw−Shear Diagram
o
2
24
1wL
Moment Diagram
R1 = ¼ Lwo
wo
L/2 L/2
wo
R2 = ¼ Lwo
Load Diagram
A B
C
Figure P-445
R1 R2
1 m 3 m 1 m
80 kN/m
20 kN/m
R1 R2
60 kN/m
1 m 3 m
20 kN/m
1 m
F1 = 80 kN
F2 = 90 kN 1 m 2 m
218 Shear and Moment Equations and Diagrams; Relation between Load, Shear, and Moment; Moving Loads
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To draw the Shear Diagram
(1) VA = R1 = 84 kN
(2) VB = VA + Area in load diagram
VB = 84 – 20(1) = 64 kN
(3) VC = VB + Area in load diagram
VC = 64 – ½ (20 + 80)(3) = –86 kN
(4) VD = VC + Area in load diagram
VD = –86 + 0 = –86 kN
VD2 = VD + R2 = –86 + 86 = 0
(5) Location of zero shear:
From the load diagram:
y / (x + 1) = 80 / 4
y = 20(x + 1)
VE = VB + Area in load diagram
0 = 64 – ½ (20 + y)x
(20 + y)x = 128
[20 + 20(x + 1)]x = 128
20x2 + 40x – 128 = 0
5x2 + 10x – 32 = 0
x = 1.72 and –3.72
use x = 1.72 m from B
(5) By squared property of parabola:
z / (1 + x)2 = (z + 86) / 42
16z = 7.3984z + 636.2624
8.6016z = 254.4224
z = 73.97 kN
To draw the Moment Diagram
(1) MA = 0
(2) MB = MA + Area in shear diagram
MB = 0 + ½ (84 + 64)(1) = 74 kN⋅m
(3) ME = MB + Area in shear diagram
ME = 74 + A1 � see figure for A1 and A2
For A1:
A1 = 2/3 (1 + 1.72)(73.97) – 64(1)
– 2/3 (1)(9.97)
A1 = 63.5
ME = 74 + 63.5 = 137.5 kN⋅m
(4) MC = ME + Area in shear diagram
MC = ME – A2
For A2:
A2 = 1/3 (4)(73.97 + 86)
– 1/3 (1 + 1.72)(73.97)
– 1.28(73.97)
A2 = 51.5
MC = 137.5 – 51.5 = 86 kN⋅m
(5) MD = MC + Area in shear diagram
MD = 86 – 86(1) = 0
R1 = 84 kN
80 kN/m
1 m
20 kN/m
1 m
R2 = 86 kN
Load Diagram
3 m
y
A B C D
84 kN
Shear Diagram
64 kN
–86 kN
E z
x =1.72 m
A1
A2
73.97 64
1 m 1.72 m
1.28 m
9.97
Figure for solving A1 and A2
86
4 m
86 kN⋅m
137.5 kN⋅m
74 kN⋅m
Moment Diagram
2nd deg
3rd deg
1st deg
219
Chapter 04 Shear and Moment in Beams www.mathalino.com
Problem 446. Beam loaded and supported as shown in Fig. P-446.
Solution 446. ∑FV = 0
4wo + 2[ 21 wo(1)] = 20(4) + 2(50)
5wo = 180 wo = 36 kN/m
20 kN/m
4 m 50 kN 50 kN
1 m 1 m
Figure P-446
To draw the Shear Diagram
(1) VA = 0
(2) VB = VA + Area in load diagram
VB = 0 + ½ (36)(1) = 18 kN
VB2 = VB – 50 = 18 – 50
VB2 = –32 kN
(3) The net uniformly distributed load in
segment BC is 36 – 20 = 16 kN/m
upward.
VC = VB2 + Area in load diagram
VC = –32 + 16(4) = 32 kN
VC2 = VC – 50 = 32 – 50
VC2 = –18 kN
(4) VD = VC2 + Area in load diagram
VD = –18 + ½ (36)(1) = 0
(5) The shape of shear at AB and CD are
parabolic spandrel with vertex at A and D,
respectively.
(6) The location of zero shear is obviously at
the midspan or 2 m from B.
To draw the Moment Diagram
(1) MA = 0
(2) MB = MA + Area in shear diagram
MB = 0 + 1/3 (1)(18)
MB = 6 kN⋅m
(3) Mmidspan = MB + Area in shear diagram
Mmidspan = 6 – ½ (32)(2)
Mmidspan = –26 kN⋅m
(4) MC = Mmidspan + Area in shear diagram
MC = –26 + ½ (32)(2)
MC = 6 kN⋅m
(5) MD = MC + Area in shear diagram
MD = 6 – 1/3 (1)(18) = 0
(6) The moment diagram at AB and CD are 3rd
degree curve while at BC is 2nd degree curve.
6 Kn⋅m
–26 kN⋅m
6 Kn⋅m
Moment Diagram
18 kN
–32 kN
–18 kN
32 kN
Shear Diagram
20 kN/m
4 m 50 kN 50 kN
1 m 1 m
wo = 36 kN/m
Load Diagram
A D
B C
220 Shear and Moment Equations and Diagrams; Relation between Load, Shear, and Moment; Moving Loads
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INSTRUCTION: In the following problems, draw moment and load diagrams corresponding to the given shear diagrams. Specify values at all change of load positions and at all points of zero shear.
Problem 447. Shear diagram as shown in Fig. P-447. Solution 447.
Figure P-447 2400
400
–4000
1000
V (lb)
x (ft)
2 3 2 2
To draw the Load Diagram
(1) A 2400 lb upward force is acting at
point A. No load in segment AB.
(2) A point force of 2400 – 400 = 2000
lb is acting downward at point B.
No load in segment BC.
(3) Another downward force of
magnitude 400 + 4000 = 4400 lb
at point C. No load in segment CD.
(4) Upward point force of 4000 + 1000
= 5000 lb is acting at D. No load
in segment DE.
(5) A downward force of 1000 lb is
concentrated at point E.
To draw the Moment Diagram
(1) MA = 0
(2) MB = MA + Area in shear diagram
MB = 0 + 2400(2) = 4800 lb⋅ft MAB is linear and upward
(3) MC = MB + Area in shear diagram
MC = 4800 + 400(3) = 6000 lb⋅ft MBC is linear and upward
(4) MD = MC + Area in shear diagram
MD = 6000 – 4000(2) = –2000 lb⋅ft MCD is linear and downward
(5) ME = MD + Area in shear diagram
ME = –2000 + 1000(2) = 0 MDE is linear and upward
4800 lb⋅ft 6000 lb⋅ft
–2000 lb⋅ft Moment Diagram
R1 = 2400 lb
2000 lb 4400 lb
R2 = 5000 lb
1000 lb
Load Diagram
2 ft 3 ft 2 ft 2 ft
A B C E
2400 lb
400 lb
–4000 lb
1000 lb
Given Shear Diagram
D
221
Chapter 04 Shear and Moment in Beams www.mathalino.com
Problem 448. Shear diagram as shown in Fig. P-448. Solution 448.
2 1 1.6 2.4 2 V
(kN)
x (m)
36
16
–40
–24
20
Figure P-448
To draw the Load Diagram
(1) A uniformly distributed load in AB is acting
downward at a magnitude of 40/2 = 20
kN/m.
(2) Upward concentrated force of 40 + 36 =
76 kN acts at B. No load in segment BC.
(3) A downward point force acts at C at a
magnitude of 36 – 16 = 20 kN.
(4) Downward uniformly distributed load in CD
has a magnitude of (16 + 24)/4 = 10 kN/m
& causes zero shear at point F, 1.6 m from
C.
(5) Another upward concentrated force acts at
D at a magnitude of 20 + 24 = 44 kN.
(6) The load in segment DE is uniform and
downward at 20/2 = 10 kN/m.
To draw the Moment Diagram
(1) MA = 0
(2) MB = MA + Area in shear diagram
MB = 0 – ½ (40)(2) = –40 kN⋅m
MAB is downward parabola with vertex at A.
(3) MC = MB + Area in shear diagram
MC = –40 + 36(1) = –4 kN⋅m
MBC is linear and upward
(4) MF = MC + Area in shear diagram
MF = –4 + ½ (16)(1.6) = 8.8 kN⋅m
(5) MD = MF + Area in shear diagram
MD = 8.8 – ½ (24)(2.4) = –20 kN⋅m
MCD is downward parabola with vertex at F.
(6) ME = MD + Area in shear diagram
ME = –20 + ½ (20)(2) = 0 MDE is downward parabola with vertex at E.
–4 kN⋅m
–40 kN⋅m
8.8 kN⋅m
–20 kN⋅m
Moment Diagram
R1 = 76 kN R2 = 44 kN
20 kN/m
10 kN/m
20 kN
2 m 1 m 4 m 2 m
Load Diagram
F
36 kN
16 kN
–40 kN
–24 kN
20 kN
Given Shear Diagram
1.6 m 2.4 m
A B
C D E
222 Shear and Moment Equations and Diagrams; Relation between Load, Shear, and Moment; Moving Loads
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Problem 449. Shear diagram as shown in Fig. P-449. Solution 449.
–4000
3700
1700
–3100
V
(lb)
x (ft)
3 2 3 4 8
Figure P-449
To draw the Load Diagram
(1) Downward 4000 lb force is concentrated at
A and no load in segment AB.
(2) The shear in BC is uniformly increasing,
thus a uniform upward force is acting at a
magnitude of (3700 + 4000)/2 = 3850
lb/ft. No load in segment CD.
(3) Another point force acting downward with
3700 – 1700 = 1200 lb at D and no load in
segment DE.
(4) The shear in EF is uniformly decreasing,
thus a uniform downward force is acting
with magnitude of (1700 + 3100)/8 = 600
lb/ft.
(5) Upward force of 3100 lb is concentrated at
end of span F.
To draw the Moment Diagram
(1) The locations of zero shear (points G and
H) can be easily determined by ratio and
proportion of triangle.
(2) MA = 0
(3) MB = MA + Area in shear diagram
MB = 0 – 4000(3) = –12,000 lb⋅ft (4) MG = MB + Area in shear diagram
MG = –12,000 – ½ (80/77)(4000)
MG = –14,077.92 lb⋅ft (5) MC = MG + Area in shear diagram
MC = –14,077.92 + ½ (74/77)(3700)
MC = –12,300 lb⋅ft (6) MD = MC + Area in shear diagram
MD = –12,300 + 3700(3) = –1200 lb⋅ft (7) ME = MD + Area in shear diagram
ME = –1200 + 1700(4) = 5600 lb⋅ft (8) MH = ME + Area in shear diagram
MH = 5600 + ½ (17/6)(1700)
MH = 8,008.33 lb⋅ft (9) MF = MH + Area in shear diagram
MF = 8,008.33 – ½ (31/6)(3100) = 0
8,008.33 lb⋅ft
–14,077.92 lb⋅ft –12,300 lb⋅ft
–1,200 lb⋅ft
Moment Diagram
–12,000 lb⋅ft
5,600 lb⋅ft
3 ft 2 ft 3 ft 4 ft 8 ft
4000 lb 2000 lb
3100 lb 3850 lb/ft
600 lb/ft
Load Diagram
A B C D E F
80/77 ft
–4000 lb
3700 lb
1700 lb
–3100 lb 74/77 ft
17/6 ft
31/6 ft
Given Shear Diagram
G H
223
Chapter 04 Shear and Moment in Beams www.mathalino.com
Problem 450. Shear diagram as shown in Fig. P-450. Solution 450. Solution
V
(lb)
x (ft)
4 2 4 4 4
900
–900
–1380
480
Figure P-450
To draw the Load Diagram
(1) The shear diagram in AB is uniformly upward,
thus the load is uniformly distributed upward at
a magnitude of 900/4 = 225 lb/ft. No load in
segment BC.
(2) A downward point force acts at point C with
magnitude of 900 lb. No load in segment CD.
(3) Another concentrated force is acting downward
at D with a magnitude of 900 lb.
(4) The load in DE is uniformly distributed
downward at a magnitude of (1380 – 900)/4 =
120 lb/ft.
(5) An upward load is concentrated at E with
magnitude of 480 + 1380 = 1860 lb.
(6) 480/4 = 120 lb/ft is distributed uniformly over
the span EF.
To draw the Moment Diagram
(1) MA = 0
(2) MB = MA + Area in shear diagram
MB = 0 + ½ (4)(900) = 1800 lb⋅ft (3) MC = MB + Area in shear diagram
MC = 1800 + 900(2) = 3600 lb⋅ft (4) MD = MC + Area in shear diagram
MD = 3600 + 0 = 3600 lb⋅ft (5) ME = MD + Area in shear diagram
ME = 3600 – ½ (900 + 1380)(4)
ME = –960 lb⋅ft (6) MF = ME + Area in shear diagram
MF = –960 + ½ (480)(4) = 0
(7) The shape of moment diagram in AB is upward
parabola with vertex at A, while linear in BC and
horizontal in CD. For segment DE, the diagram
is downward parabola with vertex at G. G is the
point where the extended shear in DE intersects
the line of zero shear.
(8) The moment diagram in EF is a downward
parabola with vertex at F.
1800 lb⋅ft
4 ft 2 ft 4 ft 4 ft 4 ft
225 lb/ft
120 lb/ft 900 lb
900 lb
1860 lb
Load Diagram
900 lb
–1380 lb
–900 lb
480 lb
Given Shear Diagram
3600 lb⋅ft
–960 lb⋅ft
Moment Diagram
A B C D E F
G
224 Shear and Moment Equations and Diagrams; Relation between Load, Shear, and Moment; Moving Loads
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Problem 451. Shear diagram as shown in Fig. P-451. Solution 451.
Figure P-451
To draw the Load Diagram
(1) Upward concentrated load at A is 10 kN.
(2) The shear in AB is a 2nd-degree curve, thus
the load in AB is uniformly varying. In this
case, it is zero at A to 2(10 + 2)/3 = 8 kN at
B. No load in segment BC.
(3) A downward point force is acting at C in a
magnitude of 8 – 2 = 6 kN.
(4) The shear in DE is uniformly increasing, thus
the load in DE is uniformly distributed and
upward. This load is spread over DE at a
magnitude of 8/2 = 4 kN/m.
To draw the Moment Diagram
(1) To find the location of zero shear, F:
x2/10 = 32/(10 + 2)
x = 2.74 m
(2) MA = 0
(3) MF = MA + Area in shear diagram
MF = 0 + 2/3 (2.74)(10) = 18.26 kN⋅m
(4) MB = MF + Area in shear diagram
MB = 18.26 – [1/3 (10 + 2)(3)
– 1/3 (2.74)(10) – 10(3 – 2.74)]
MB = 18 kN⋅m
(5) MC = MB + Area in shear diagram
MC = 18 – 2(1) = 16 kN⋅m
(6) MD = MC + Area in shear diagram
MD = 16 – 8(1) = 8 kN⋅m
(7) ME = MD + Area in shear diagram
ME = 8 – ½ (2)(8) = 0
(8) The moment diagram in AB is a second
degree curve, at BC and CD are linear and
downward. For segment DE, the moment
diagram is parabola open upward with vertex at E.
3 m 1 m 1 m 2 m
10 kN 4 kN/m
8 kN/m
6 kN
Load Diagram
A B C D
E
V (kN)
x (m)
10
–2
–8
3 1 1 2
2nd-degree curve
10 kN
–2 kN
–8 kN
Given Shear Diagram
F
x = 2.74 m
18.26 kN⋅m 18 kN⋅m 16 kN⋅m
8 kN⋅m
Moment Diagram
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Chapter 04 Shear and Moment in Beams www.mathalino.com
MOVING LOADS From the previous section, we see that the maximum moment occurs at a point of zero shears. For beams loaded with concentrated loads, the point of zero shears usually occurs under a concentrated load and so the maximum moment.
Beams and girders such as in a bridge or an overhead
crane are subject to moving concentrated loads, which are at fixed distance with each other. The problem here is to determine the moment under each load when each load is in a position to cause a maximum moment. The largest value of these moments governs the design of the beam.
SINGLE MOVING LOAD
For a single moving load, the maximum moment occurs when the load is at the midspan and the maximum shear occurs when the load is very near the support (usually assumed to lie over the support).
Mmax = 4
PL and Vmax = P
TWO MOVING LOADS
For two moving loads, the maximum shear occurs at the reaction when the larger load is over that support. The maximum moment is given by
P P Position for
maximum shear
Position for
maximum moment
L
L/2 L/2
Pb Ps d
L
Pb > Ps
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226 Shear and Moment Equations and Diagrams; Relation between Load, Shear, and Moment; Moving Loads
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Mmax = PL
dPPL s
4
)( 2−
where Ps is the smaller load, Pb is the bigger load, and
P is the total load (P = Ps + Pb). THREE OR MORE MOVING LOADS In general, the bending moment under a particular
load is a maximum when the center of the beam is midway between that load and the resultant of all the loads then on the span. With this rule, we compute the maximum moment under each load, and use the biggest of the moments for the design. Usually, the biggest of these moments occurs under the biggest load.
The maximum shear occurs at the reaction where the
resultant load is nearest. Usually, it happens if the biggest load is over that support and as many a possible of the remaining loads are still on the span.
In determining the largest moment and shear, it is
sometimes necessary to check the condition when the bigger loads are on the span and the rest of the smaller loads are outside.
SOLVED PROBLEMS Problem 453. A truck with axle loads of 40 kN and 60 kN on a
wheel base of 5 m rolls across a 10-m span. Compute the maximum bending moment and the maximum shearing force.
Solution 453. R = 40 + 60 = 100 kN xR = 40(5) x = 200/R x = 200/100 x = 2 m
5 m
40 kN 60 kN
R
x 5 – x
227
Chapter 04 Shear and Moment in Beams www.mathalino.com
For maximum moment under 40 kN wheel:
∑MR2 = 0 10R1 = 3.5(100) R1 = 35 kN MTo the left of 40 kN = 3.5R1 MTo the left of 40 kN = 3.5(35)
MTo the left of 40 kN = 122.5 kN⋅m For maximum moment under 60 kN wheel:
∑MR1 = 0 10R2 = 4(100) R2 = 40 kN MTo the right of 60 kN = 4R2 MTo the right of 60 kN = 4(40)
MTo the right of 60 kN = 160 kN⋅m
Thus, Mmax = 160 kN⋅⋅⋅⋅m
The maximum shear will occur when the 60 kN is over a support.
∑MR1 = 0 10R2 = 100(8) R2 = 80 kN Thus, Vmax = 80 kN Problem 454. Repeat Prob. 453 using axle loads of 30 kN and 50 kN
on a wheel base of 4 m crossing an 8-m span. Solution 454. R = 30 + 50 = 80 kN xR = 4(30) x = 120/R x = 120/80 x = 1.5 m
40 kN 60 kN
R = 100 kN
2 m 3 m
C L
10 m
3.5 m 1.5 m 3.5 m R1 R2
40 kN 60 kN
R = 100 kN
2 m 3 m
10 m
4 m 1 m 4 m R1 R2
C L
8 m R1 R2
40 kN 60 kN
R = 100 kN
2 m 3 m
4 m
30 kN 50 kN
R
x 4 – x
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Maximum moment under 30 kN wheel:
∑MR2 = 0 8R1 = 2.75(80) R1 = 27.5 kN MTo the left of 30 kN = 2.75R1 MTo the left of 30 kN = 2.75(27.5)
MTo the left of 30 kN = 75.625 kN⋅m Maximum moment under 50 kN wheel:
∑MR1 = 0 8R2 = 3.25(80) R2 = 32.5 kN MTo the right of 50 kN = 3.25R2 MTo the right of 50 kN = 3.25(32.5)
MTo the right of 50 kN = 105.625 kN⋅m
Thus, Mmax = 105.625 kN⋅⋅⋅⋅m
The maximum shear will occur when the 50 kN is over a support.
∑MR1 = 0 8R2 = 6.5(80) R2 = 65 kN Thus, Vmax = 65 kN Problem 455. A tractor weighing 3000 lb, with a wheel base of 9 ft,
carries 1800 lb of its load on the rear wheels. Compute the maximum moment and maximum shear when crossing a 14 ft-span.
Solution 455. R = Wr + Wf 3000 = 1800 + Wf Wf = 1200 lb Rx = 9Wf 3000x = 9(1200) x = 3.6 ft
30 kN 50 kN R = 80 kN
1.5 m 2.5 m
C L
2.75 m
8 m
1.25 m 2.75 m R1 R2
30 kN 50 kN R = 80 kN
1.5 m 2.5 m
C L
3.25 m
8 m
0.75 m 3.25 m R1 R2
30 kN 50 kN
R = 80 kN 1.5 m
2.5 m
6.5 m
8 m R1 R2
Wr = 1800 lb
Wf
R = 3000 lb
9 – x x
9 ft
229
Chapter 04 Shear and Moment in Beams www.mathalino.com
9 – x = 5.4 ft
When the midspan is midway between Wr and R, the front wheel Wf will be outside the span (see figure). In this case, only the rear wheel Wr = 1800 lb is the load. The maximum moment for this condition is when the load is at the midspan.
R1 = R2 = ½ (1800) R1 = 900 lb Maximum moment under Wr MTo the left of rear wheel = 7R1 MTo the left of rear wheel = 7(900)
MTo the left of rear wheel = 6300 lb⋅ft Maximum moment under Wf
∑MR1 = 0 14R2 = 4.3R
14R2 = 4.3(3000) R2 = 921.43 lb MTo the right of front wheel = 4.3R2 MTo the right of front wheel = 4.3(921.43)
MTo the right of front wheel = 3962.1 lb⋅ft Thus, Mmax = MTo the left of rear wheel
Thus, Mmax = 6300 lb⋅⋅⋅⋅ft
The maximum shear will occur when the
rear wheel (wheel of greater load) is directly over the support.
∑MR2 = 0 14R1 = 10.4R 14R1 = 10.4(3000) R1 = 2228.57 lb Thus, Vmax = 2228.57 lb
14 ft
R1 R2
Wr = 1800 lb Wf = 1200 lb R = 3000 lb
5.4 ft 3.6 ft
5.2 ft
C L
5.2 ft 1.8 ft
14 ft R1 R2
Wr = 1800 lb
7 ft
14 ft
R1 R2
Wr = 1800 lb Wf = 1200 lb R = 3000 lb
5.4’ 3.6’
4.3 ft
C L
4.3 ft 2.7 ft
R2
14 ft
R1
Wr = 1800 lb
Wf = 1200 lb
R = 3000 lb
5.4’ 3.6’
10.4 ft
230 Shear and Moment Equations and Diagrams; Relation between Load, Shear, and Moment; Moving Loads
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Problem 456. Three wheel loads roll as a unit across a 44-ft span. The loads are P1 = 4000 lb and P2 = 8000 lb separated by 9 ft, and P3 = 6000 lb at 18 ft from P2. Determine the maximum moment and maximum shear in the simply supported span.
Solution 456. R = P1 + P2+ P3 R = 4k + 8k + 6k R = 18 kips R = 18,000 lbs xR = 9P2 + (9 + 18)P3 x(18) = 9(8) + (9 + 18)(6) x = 13 ft � the resultant R is 13 ft from P1 Maximum moment under P1
∑MR2 = 0 44R1 = 15.5R 44R1 = 15.5(18) R1 = 6.34091 kips R1 = 6,340.91 lbs MTo the left of P1 = 15.5R1 MTo the left of P1 = 15.5(6340.91)
MTo the left of P1 = 98,284.1 lb⋅ft Maximum moment under P2
∑MR2 = 0 44R1 = 20R 44R1 = 20(18) R1 = 8.18182 kips R1 = 8,181.82 lbs MTo the left of P2 = 20R1 – 9P1 MTo the left of P2 = 20(8,181.82) – 9(4000)
MTo the left of P2 = 127,636.4 lb⋅ft
P1 = 4k P2 = 8k P3 = 6k
R = 18k
x = 13’
18’ 9’
R1 R2
C L
20’
44’
2’ 20’
P1 = 4k P2 = 8k P3 = 6k
9’ 18’
R
x
P1 = 4k P2 = 8k P3 = 6k
R = 18k
x = 13’
R1 R2
C L
15.5’
44’
6.5’ 15.5’
18’ 9’
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Chapter 04 Shear and Moment in Beams www.mathalino.com
Maximum moment under P3
∑R1 = 0 44R2 = 15R 44R2 = 15(18) R2 = 6.13636 kips R2 = 6,136.36 lbs MTo the right of P3 = 15R2
MTo the right of P3 = 15(6,136.36)
MTo the right of P3 = 92,045.4 lb⋅ft Thus, Mmax = MTo the left of P2
Thus, Mmax = 127,636.4 lb⋅⋅⋅⋅ft
The maximum shear will occur when P1 is over the support.
∑MR2 = 0 44R1 = 35R 44R1 = 31(18) R1 = 12.6818 kips R1 = 12,681.8 lbs Thus, Vmax = 12,681.8 lbs Problem 457. A truck and trailer combination crossing a 12-m span
has axle loads of 10, 20, and 30 kN separated respectively by distances of 3 and 5 m. Compute the maximum moment and maximum shear developed in the span.
Solution 457. R = 10 + 20 + 30 R = 60 kN xR = 3(20) + 8(30) x(60) = 3(20) + 8(30) x = 5 m
P1 = 4k P2 = 8k P3 = 6k
R = 18k
x = 13’
18’ 9’
R1 R2
C L
15’
44’
7’ 15’
P1 = 4k P2 = 8k P3 = 6k
R = 18k
x = 13’
18’ 9’
R1 R2
44’
31’
5 m
10 kN 20 kN 30 kN
R
x
3 m
232 Shear and Moment Equations and Diagrams; Relation between Load, Shear, and Moment; Moving Loads
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Maximum moment under 10 kN
∑MR2 = 0 12R1 = 3.5R 12R1 = 3.5(60) 12R1 = 210 R1 = 12.7 kN MTo the left of 10 kN = 3.5R1 MTo the left of 10 kN = 3.5(12.7)
MTo the left of 10 kN = 61.25 kN⋅m Maximum moment under 20 kN
∑MR2 = 0 12R1 = 5R 12R1 = 5(60) R1 = 25 kN
5 m
10 kN 20 kN 30 kN
R = 60 kN
x = 5 m
3 m
C L
R1 R2
3.5 m
12 m
3.5 m 2.5 m
5 m
10 kN 20 kN 30 kN
R = 60 kN
x = 5 m
3 m
C L
R1 R2
5 m
12 m
5 m
1 m
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Chapter 04 Shear and Moment in Beams www.mathalino.com
MTo the left of 20 kN = 5R1 – 3(10) MTo the left of 20 kN = 5(25) – 30
MTo the left of 20 kN = 95 kN⋅m When the centerline of the beam is midway between
reaction R = 60 kN and 30 kN, the 10 kN comes off the span.
R = 20 + 30 R = 50 kN xR = 5(30) x(50) = 150 x = 3 m from 20 kN
∑MR1 = 0 12R2 = 5R 12R2 = 5(50) R2 = 20.83 kN MTo the right of 30 kN = 5R2 MTo the right of 30 kN = 5(20.83)
MTo the right of 30 kN = 104.17 kN⋅m Thus, the maximum moment will occur when only
the 20 and 30 kN loads are on the span. Mmax = MTo the right of 30 kN
Mmax = 104.17 kN⋅⋅⋅⋅m
5 m
10 kN 20 kN 30 kN
R = 50 kN
x = 3 m
3 m
C L
R1 R2
5 m
12 m
5 m
1 m
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The maximum shear will occur when the three loads are on the span and the 30 kN load is directly over the support.
∑MR1 = 0 12R2 = 9R 12R2 = 9(60) R2 = 45 kN Thus, Vmax = 45 kN
5 m
10 kN 20 kN 30 kN
R = 60 kN
x = 5 m
3 m
R1 R2
12 m 9 m
Copyright © 2011 Mathalino.com. All rights reserved.
This eBook is NOT FOR SALE. Please download this eBook only from www.mathalino.com.
In doing so, you are inderictly helping the author to create more free contents. Thank you for your support.