Definition of a Beam
A beam is a bar subject to forces or couples that lie in a plane containing the
longitudinal section 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.
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
(M = 0 and Fv = 0) can be applied, thus the beam is indeterminate to the first degree
(3 - 2 = 1).
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 Diagrams
Shear and Moment Diagrams
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 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.
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
Solution 403
From the load diagram:
MB=0MB=0
5RD+1(30)=3(50)5RD+1(30)=3(50)
RD=24kNRD=24kN
MD=0MD=0
5RB=2(50)+6(30)5RB=2(50)+6(30)
RB=56kNRB=56kN
Segment AB:
VAB=30kNVAB=30kN
MAB=30xkNmMAB=30xkNm
Segment BC:
VBC=30+56VBC=30+56
VBC=26kNVBC=26kN
MBC=30x+56(x1)MBC=30x+56(x1)
MBC=26x56kNmMBC=26x56kNm
Segment CD:
VCD=30+5650VCD=30+5650
VCD=24kNVCD=24kN
MCD=30x+56(x1)50(x4)MCD=30x+56(x1)50(x4)
MCD=30x+56x5650x+200MCD=30x+56x5650x+200
MCD=24x+144kNmMCD=24x+144kNm
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
kNm.
2. MBC = 26x - 56 is also linear. At x = 1
m, MBC = -30 kNm; at x = 4 m, MBC =
48 kNm. 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 kNm; at x = 6 m, MCD = 0.
Problem 404
Beam loaded as shown in Fig. P-404.
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Solution 404
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MA=0MA=0
12RD+4800=3(2000)12RD+4800=3(2000)
RD=100lbRD=100lb
MD=0MD=0
12RA=9(2000)+480012RA=9(2000)+4800
RA=1900lbRA=1900lb
Segment AB:
VAB=1900lbVAB=1900lb
MAB=1900xlbftMAB=1900xlbft
Segment BC:
VBC=19002000VBC=19002000
VBC=100lbVBC=100lb
MBC=1900x2000(x3)MBC=1900x2000(x3)
MBC=1900x2000x+6000MBC=1900x2000x+6000
MBC=100x+6000lbftMBC=100x+6000lbft
Segment CD:
VCD=19002000VCD=19002000
VCD=100lbVCD=100lb
MCD=1900x2000(x3)4800MCD=1900x2000(x3)4800
MCD=1900x2000x+60004800MCD=1900x2000x+60004800
MCD=100x+1200lbftMCD=100x+1200lbft
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 lbft.
2. For segment BC, MBC = -100x + 6000 is
linear; at x = 3 ft, MBC = 5700 lbft; at x
= 9 ft, MBC = 5100 lbft.
3. MCD = -100x + 1200 is again linear; at x
= 9 ft, MCD = 300 lbft; at x = 12 ft,
MCD= 0.
Problem 405
Beam loaded as shown in Fig. P-405.
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Solution 405
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MA=0MA=0
10RC=2(80)+5[10(10)]10RC=2(80)+5[10(10)]
RC=66kNRC=66kN
MC=0MC=0
10RA=8(80)+5[10(10)]10RA=8(80)+5[10(10)]
RA=114kNRA=114kN
Segment AB:
VAB=11410xkNVAB=11410xkN
MAB=114x10x(x/2)MAB=114x10x(x/2)
MAB=114x5x2kNmMAB=114x5x2kNm
Segment BC:
VBC=1148010xVBC=1148010x
VBC=3410xkNVBC=3410xkN
MBC=114x80(x2)10x(x/2)MBC=114x80(x2)10x(x/2)
MBC=160+34x5x2kNmMBC=160+34x5x2kNm
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 - 5x2 is a second degree curve for segment AB; at x = 0, MAB = 0; at x = 2 m,
MAB = 208 kNm.
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 kNm; 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 kNm.
Problem 406
Beam loaded as shown in Fig. P-406.
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Solution 406
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MA=0MA=0
12RC=4(900)+18(400)+9[(60)(18)]12RC=4(900)+18(400)+9[(60)(18)]
RC=1710lbRC=1710lb
MC=0MC=0
12RA+6(400)=8(900)+3[60(18)]12RA+6(400)=8(900)+3[60(18)]
RA=670lbRA=670lb
Segment AB:
VAB=67060xlbVAB=67060xlb
MAB=670x60x(x/2)MAB=670x60x(x/2)
MAB=670x30x2lbftMAB=670x30x2lbft
Segment BC:
VBC=67090060xVBC=67090060x
VBC=23060xlbVBC=23060xlb
MBC=670x900(x4)60x(x/2)MBC=670x900(x4)60x(x/2)
MBC=3600230x30x2lbftMBC=3600230x30x2lbft
Segment CD:
VCD=670+171090060xVCD=670+171090060x
VCD=148060xlbVCD=148060xlb
MCD=670x+1710(x12)900(x4)60x(x/2)MCD=670x+1710(x12)900(x4)60x(x/2)
MCD=16920+1480x30x2lbftMCD=16920+1480x30x2lbft
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 - 30x2 for segment AB is a second degree curve; at x = 0, MAB = 0; at x = 4 ft,
MAB = 2200 lbft.
2. For BC, MBC = 3600 - 230x - 30x2, is a second degree curve; at x = 4 ft, MBC = 2200 lbft, at
x = 12 ft, MBC = -3480 lbft; 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 lbft; at x = 18 ft, MCD = 0.
Problem 407
Beam loaded as shown in Fig. P-407.
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Solution 407
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MA=0MA=0
6RD=4[2(30)]6RD=4[2(30)]
RD=40kNRD=40kN
MD=0MD=0
6RA=2[2(30)]6RA=2[2(30)]
RA=20kNRA=20kN
Segment AB:
VAB=20kNVAB=20kN
MAB=20xkNmMAB=20xkNm
Segment BC:
VBC=2030(x3)VBC=2030(x3)
VBC=11030xkNVBC=11030xkN
MBC=20x30(x3)(x3)/2MBC=20x30(x3)(x3)/2
MBC=20x15(x3)2kNmMBC=20x15(x3)2kNm
Segment CD:
VCD=2030(2)VCD=2030(2)
VCD=40kNVCD=40kN
MCD=20x30(2)(x4)MCD=20x30(2)(x4)
MCD=20x60(x4)kNmMCD=20x60(x4)kNm
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 kNm.
2. MBC = 20x - 15(x - 3)2 for segment BC is second degree curve; at x = 3 m, MBC = 60 kNm;
at x = 5 m, MBC = 40 kNm. Note that maximum moment occurred at zero shear; at x = 3.67
m, MBC = 66.67 kNm.
3. MCD = 20x - 60(x - 4) for segment BC is linear; at x = 5 m, MCD = 40 kNm; at x = 6 m,
MCD = 0.
Problem 408
Beam loaded as shown in Fig. P-408.
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Solution 408
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MA=0MA=0
6RD=1[2(50)]+5[2(20)]6RD=1[2(50)]+5[2(20)]
RD=50kNRD=50kN
MD=0MD=0
6RA=5[2(50)]+1[2(20)]6RA=5[2(50)]+1[2(20)]
RA=90kNRA=90kN
Segment AB:
VAB=9050xkNVAB=9050xkN
MAB=90x50x(x/2)MAB=90x50x(x/2)
MAB=90x25x2kNmMAB=90x25x2kNm
Segment BC:
VBC=9050(2)VBC=9050(2)
VBC=10kNVBC=10kN
MBC=90x2(50)(x1)MBC=90x2(50)(x1)
MBC=10x+100kNmMBC=10x+100kNm
Segment CD:
VCD=902(50)20(x4)VCD=902(50)20(x4)
VCD=20x+70kNVCD=20x+70kN
MCD=90x2(50)(x1)20(x4)(x4)/2MCD=90x2(50)(x1)20(x4)(x4)/2
MCD=90x100(x1)10(x4)2MCD=90x100(x1)10(x4)2
MCD=10x2+70x60kNmMCD=10x2+70x60kNm
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 - 25x2 is second degree; at x = 0, MAB = 0; at x = 1.8 m, MAB = 81 kNm; at x =
2 m, MAB = 80 kNm.
2. MBC = -10x + 100 is linear; at x = 2 m, MBC = 80 kNm; at x = 4 m, MBC = 60 kNm.
3. MCD = -10x2 + 70x - 60; at x = 4 m, MCD = 60 kNm; at x = 6 m, MCD = 0.
Problem 409
Cantilever beam loaded as shown in Fig. P-409.
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Solution 409
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Segment AB:
VAB=woxVAB=wox
MAB=wox(x/2)MAB=wox(x/2)
MAB=12wox2MAB=12wox2
Segment BC:
VBC=wo(L/2)VBC=wo(L/2)
VBC=12woLVBC=12woL
MBC=wo(L/2)(xL/4)MBC=wo(L/2)(xL/4)
MBC=12woLx+18woL2MBC=12woLx+18woL2
To draw the Shear Diagram:
1. VAB = -wox for segment AB is linear; at x = 0, VAB = 0; at x = L/2, VAB = -woL.
2. At BC, the shear is uniformly distributed by -woL.
To draw the Moment Diagram:
1. MAB = -wox2 is a second degree curve; at x = 0, MAB = 0; at x = L/2, MAB = -1/8 woL2.
2. MBC = -woLx + 1/8 woL2 is a second degree; at x = L/2, MBC = -1/8 woL2; at x = L, MBC =
-3/8 woL2.
Problem 410
Cantilever beam carrying the uniformly varying load shown in Fig. P-410.
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Solution 410
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yx=woLyx=woL
y=woLxy=woLx
Fx=12xyFx=12xy
Fx=12x(woLx)Fx=12x(woLx)
Fx=wo2Lx2Fx=wo2Lx2
Shear equation:
V=wo2Lx2V=wo2Lx2
Moment equation:
M=13xFx=13x(wo2Lx2)M=13xFx=13x(wo2Lx2)
M=wo6Lx3M=wo6Lx3
To draw the Shear Diagram:
1. V = - wo x2 / 2L is a second degree curve; at x =
0, V = 0; at x = L, V = - woL.
To draw the Moment Diagram:
1. M = - wo x3 / 6L is a third degree curve; at x =
0, M = 0; at x = L, M = - 1/6 woL2.
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.
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Solution 411
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yLx=woLyLx=woL
y=woL(Lx)y=woL(Lx)
F1=12x(woy)F1=12x(woy)
F1=12x[wowoL(Lx)]F1=12x[wowoL(Lx)]
F1=12x[wowo+woLx]F1=12x[wowo+woLx]
F1=wo2Lx2F1=wo2Lx2
F2=xy=x[woL(Lx)]F2=xy=x[woL(Lx)]
F2=woL(Lxx2)F2=woL(Lxx2)
Shear equation:
V=F1F2=wo2Lx2woL(Lxx2)V=F1F2=wo2Lx2woL(Lxx2)
V=wo2Lx2wox+woLx2V=wo2Lx2wox+woLx2
V=wo2Lx2woxV=wo2Lx2wox
Moment equation:
M=23xF112xF2M=23xF112xF2
M=23x(wo2Lx2)12x[woL(Lxx2)]M=23x(wo2Lx2)12x[woL(Lxx2)]
M=wo3Lx3wo2x2+wo2Lx3M=wo3Lx3wo2x2+wo2Lx3
M=wo2x2+wo6Lx3M=wo2x2+wo6Lx3
To draw the Shear Diagram:
1. V = wox2/2L - wox is a concave upward second
degree curve; at x = 0, V = 0; at x = L, V = -1/2 woL.
To draw the Moment diagram:
1. M = -wox2/2 + wox3/6L is in third degree; at x = 0, M
= 0; at x = L, M = -1/3 woL2.
/ol>
Problem 412
Beam loaded as shown in Fig. P-412.
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Solution 412
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MA=0MA=0
6RC=5[6(800)]6RC=5[6(800)]
RC=4000lbRC=4000lb
MC=0MC=0
6RA=1[6(800)]6RA=1[6(800)]
RA=800lbRA=800lb
Segment AB:
VAB=800lbVAB=800lb
MAB=800xlbftMAB=800xlbft
Segment BC:
VBC=800800(x2)VBC=800800(x2)
VBC=2400800xlbVBC=2400800xlb
MBC=800x800(x2)(x2)/2MBC=800x800(x2)(x2)/2
MBC=800x400(x2)2lbftMBC=800x400(x2)2lbft
Segment CD:
VCD=800+4000800(x2)VCD=800+4000800(x2)
VCD=4800800x+1600VCD=4800800x+1600
VCD=6400800xlbVCD=6400800xlb
MCD=800x+4000(x6)800(x2)(x2)/2MCD=800x+4000(x6)800(x2)(x2)/2
MCD=800x+4000(x6)400(x2)2lbftMCD=800x+4000(x6)400(x2)2lbft
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 lbft.
2. MBC = 800x - 400(x - 2)2 is second degree curve; at x = 2 ft, MBC = 1600 lbft; at x = 6 ft,
MBC = -1600 lbft; at x = 3 ft, MBC = 2000 lbft.
3. MCD = 800x + 4000(x - 6) - 400(x - 2)2 is also a second degree curve; at x = 6 ft, MCD = -
1600 lbft; at x = 8 ft, MCD = 0.
Problem 413
Beam loaded as shown in Fig. P-413. See the instruction.
Solution 413
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MB=0MB=0
6RE=1200+1[6(100)]6RE=1200+1[6(100)]
RE=300lbRE=300lb
ME=0ME=0
6RB+1200=5[6(100)]6RB+1200=5[6(100)]
RB=300lbRB=300lb
Segment AB:
VAB=100xlbVAB=100xlb
MAB=100x(x/2)MAB=100x(x/2)
MAB=50x2lbftMAB=50x2lbft
Segment BC:
VBC=100x+300lbVBC=100x+300lb
MBC=100x(x/2)+300(x2)
Problem 414
Cantilever beam carrying the load shown in Fig. P-414.
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Solution 414
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Segment AB:
VAB=2xkNVAB=2xkN
MAB=2x(x/2)MAB=2x(x/2)
MAB=x2kNmMAB=x2kNm
Segment BC:
yx2=23yx2=23
y=23(x2)y=23(x2)
F1=2xF1=2x
F2=12(x2)yF2=12(x2)y
F2=12(x2)[23(x2)]F2=12(x2)[23(x2)]
F2=13(x2)2F2=13(x2)2
VBC=F1F2VBC=F1F2
VBC=2x13(x2)2VBC=2x13(x2)2
MBC=(x/2)F113(x2)F2MBC=(x/2)F113(x2)F2
MBC=(x/2)(2x)13(x2)[13(x2)2]MBC=(x/2)(2x)13(x2)[13(x2)2]
MBC=x219(x2)3MBC=x219(x2)3
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 - 1/3 (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 = -x2 is a second degree curve; at x = 0,
MAB = 0; at x = 2 m, MAB = -4 kNm.
2. MBC = -x2 -1/9 (x - 2)3 is a third degree curve; at
x = 2 m, MBC = -4 kNm; at x = 5 m, MBC = -28
kNm.
Problem 415
Cantilever beam loaded as shown in Fig. P-415.
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Solution 415
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Segment AB:
VAB=20xkNVAB=20xkN
MAB=20x(x/2)MAB=20x(x/2)
MAB=10x2kNmMAB=10x2kNm
Segment BC:
VBC=20(3)VBC=20(3)
VBC=60kNVBC=60kN
MBC=20(3)(x1.5)MBC=20(3)(x1.5)
MBC=60(x1.5)kNmMBC=60(x1.5)kNm
Segment CD:
VCD=20(3)+40VCD=20(3)+40
VCD=20kNVCD=20kN
MCD=20(3)(x1.5)+40(x5)MCD=20(3)(x1.5)+40(x5)
MCD=60(x1.5)+40(x5)kNmMCD=60(x1.5)+40(x5)kNm
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 = -10x2 for segment AB is
second degree curve; at x = 0, MAB =
0; at x = 3 m, MAB = -90 kNm.
2. MBC = -60(x - 1.5) for segment BC is
linear; at x = 3 m, MBC = -90 kNm;
at x = 5 m, MBC = -210 kNm.
3. MCD = -60(x - 1.5) + 40(x - 5) for
segment CD is also linear; at x = 5 m,
MCD = -210 kNm, at x = 7 m, MCD = -250 kNm.
Problem 416
Beam carrying uniformly varying load shown in Fig. P-416.
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Solution 416
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MR2=0MR2=0
LR1=13LFLR1=13LF
R1=13(12Lwo)R1=13(12Lwo)
R1=16LwoR1=16Lwo
MR1=0MR1=0
LR2=23LFLR2=23LF
R2=23(12Lwo)R2=23(12Lwo)
R2=13LwoR2=13Lwo
yx=woLyx=woL
y=woLxy=woLx
Fx=12xy=12x(woLx)Fx=12xy=12x(woLx)
Fx=wo2Lx2Fx=wo2Lx2
V=R1FxV=R1Fx
V=16Lwowo2Lx2V=16Lwowo2Lx2
M=R1xFx(13x)M=R1xFx(13x)
M=16Lwoxwo2Lx2(13x)M=16Lwoxwo2Lx2(13x)
M=16Lwoxwo6Lx3M=16Lwoxwo6Lx3
To draw the Shear Diagram:
V = 1/6 Lwo - wox2/2L 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
To draw the Moment Diagram:
M = 1/6 Lwox - wox3/6L 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 - 0.0321L2wo
Mmax = 0.0641L2wo
Problem 417
Beam carrying the triangular loading shown in Fig. P-417.
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Solution 417
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By symmetry:
R1=R2=12(12Lwo)=14LwoR1=R2=12(12Lwo)=14Lwo
yx=woL/2yx=woL/2
y=2woLxy=2woLx
F=12xy=12x(2woL)F=12xy=12x(2woL)
F=woLx2F=woLx2
V=R1FV=R1F
V=14LwowoLx2V=14LwowoLx2
M=R1xF(13x)M=R1xF(13x)
M=14Lwox(woLx2)(13x)M=14Lwox(woLx2)(13x)
To draw the Shear Diagram:
V = Lwo/4 - wox2/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 - wox3/3L 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.
Problem 418
Cantilever beam loaded as shown in Fig. P-418.
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Solution 418
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Segment AB:
VAB=20kNVAB=20kN
MAB=20xkNmMAB=20xkNm
Segment BC:
VAB=20kNVAB=20kN
MAB=20x+80kNmMAB=20x+80kNm
To draw the Shear Diagram:
1. 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 kNm.
2. MBC = -20x + 80 is also linear; when x = 4
m, MBC = 0; when x = 6 m, MBC = -60
kNm
Problem 419
Beam loaded as shown in Fig. P-419.
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Solution 419
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MC=0MC=0
9R1=5(810)9R1=5(810)
R1=450lbR1=450lb
MA=0MA=0
9R2=4(810)9R2=4(810)
R2=360lbR2=360lb
Segment AB:
yx=2706yx=2706
y=45xy=45x
F=12xy=12x(45x)F=12xy=12x(45x)
F=22.5x2F=22.5x2
VAB=R1FVAB=R1F
VAB=45022.5x2lbVAB=45022.5x2lb
MAB=R1xF(13x)MAB=R1xF(13x)
MAB=450x22.5x2(13x)MAB=450x22.5x2(13x)
MAB=450x7.5x3lbftMAB=450x7.5x3lbft
Segment BC:
VBC=450810VBC=450810
VBC=360lbVBC=360lb
MBC=450x810(x4)MBC=450x810(x4)
MBC=450x810x+3240MBC=450x810x+3240
MBC=3240360xlbftMBC=3240360xlbft
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 lbft; at x = 6 ft, MAB = 1080 lbft.
2. MBC = 3240 - 360x for segment BC is linear; at x = 6 ft, MBC = 1080 lbft; at x = 9 ft, MBC =
0.
Problem 420
A total distributed load of 30 kips supported by a uniformly distributed reaction as
shown in Fig. P-420.
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Solution 420
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w=30(1000)/12w=30(1000)/12
w=2500lb/ftw=2500lb/ft
FV=0FV=0
R=WR=W
20r=30(1000)20r=30(1000)
r=1500lb/ftr=1500lb/ft
First segment (from 0 to 4 ft from left):
V1=1500xV1=1500x
M1=1500x(x/2)M1=1500x(x/2)
M1=750x2M1=750x2
Second segment (from 4 ft to mid-span):
V2=1500x2500(x4)V2=1500x2500(x4)
V2=100001000xV2=100001000x
M2=1500x(x/2)2500(x4)(x4)/2M2=1500x(x/2)2500(x4)(x4)/2
M2=750x21250(x4)2M2=750x21250(x4)2
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 lbft.
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
lbft; at mid-span, x = 10 ft, M2 =
30000 lbft.
3. The next half of the diagram, from x
= 10 ft to x = 20 ft, can be drawn by using the concept of symmetry.
Problem 421
Write the shear and moment equations as functions of the angle for the built-in arch
shown in Fig. P-421.
Solution 421
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For that is less than 90
Components of Q and P:
Qx=QsinQx=Qsin
Qy=QcosQy=Qcos
Px=Psin(90)Px=Psin(90)
Px=P(sin90coscos90sin)Px=P(sin90coscos90sin)
Px=PcosPx=Pcos
Py=Pcos(90)Py=Pcos(90)
Py=P(cos90cos+sin90sin)Py=P(cos90cos+sin90sin)
Py=PsinPy=Psin
Shear:
V=FyV=Fy
V=QyPyV=QyPy
V=QcosPsinV=QcosPsin answer
Moment arms:
dQ=RsindQ=Rsin
dP=RRcosdP=RRcos
dP=R(1cos)dP=R(1cos)
Moment:
M=McounterclockwiseMclockwiseM=McounterclockwiseMclockwise
M=Q(dQ)P(dP)M=Q(dQ)P(dP)
M=QRsinPR(1cos)M=QRsinPR(1cos) answer
For that is greater than 90
Components of Q and P:
Qx=Qsin(180)Qx=Qsin(180)
Qx=Q(sin180coscos180sin)Qx=Q(sin180coscos180sin)
Qx=QcosQx=Qcos
Qy=Qcos(180)Qy=Qcos(180)
Qy=Q(cos180cos+sin180sin)Qy=Q(cos180cos+sin180sin)
Qy=QsinQy=Qsin
Px=Psin(90)Px=Psin(90)
Px=P(sincos90cossin90)Px=P(sincos90cossin90)
Px=PcosPx=Pcos
Py=Pcos(90)Py=Pcos(90)
Py=P(coscos90+sinsin90)Py=P(coscos90+sinsin90)
Py=PsinPy=Psin
Shear:
V=FyV=Fy
V=QyPyV=QyPy
V=(Qsin)PsinV=(Qsin)Psin
V=QsinPsinV=QsinPsin answer
Moment arms:
dQ=Rsin(180)dQ=Rsin(180)
dQ=R(sin180coscos180sin)dQ=R(sin180coscos180sin)
dQ=RsindQ=Rsin
dP=R+Rcos(180)dP=R+Rcos(180)
dP=R+R(cos180cos+sin180sin)dP=R+R(cos180cos+sin180sin)
dP=RRcosdP=RRcos
dP=R(1cos)dP=R(1cos)
Moment:
M=McounterclockwiseMclockwiseM=McounterclockwiseMclockwise
M=Q(dQ)P(dP)M=Q(dQ)P(dP)
M=QRsinPR(1cos)M=QRsinPR(1cos) answer
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
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MC=0MC=0
2R(RA)=RP2R(RA)=RP
RA=12PRA=12P
For that is less than 90
Shear:
VAB=RAcos(90)VAB=RAcos(90)
VAB=1/2P(cos90cos+sin90sin)
VAB=1/2P(cos90cos+sin90sin)
VAB=12PsinVAB=12Psin answer
Moment arm:
d=RRcosd=RRcos
d=R(1cos)d=R(1cos)
Moment:
MAB=RA(d)MAB=RA(d)
MAB=1/2PR(1cos)MAB=12PR(1cos) answer
For that is greater than 90
Components of P and RA:
Px=Psin(90)Px=Psin(90)
Px=P(sincos90cossin90)Px=P(sincos90cossin90)
Px=PcosPx=Pcos
Py=Pcos(90)Py=Pcos(90)
Py=P(coscos90+sinsin90)Py=P(coscos90+sinsin90)
Py=PsinPy=Psin
RAx=RAsin(90)RAx=RAsin(90)
RAx=12P(sincos90cossin90)RAx=12P(sincos90cossin90)
RAx=12PcosRAx=12Pcos
RAy=RAcos(90)RAy=RAcos(90)
RAy=12P(coscos90+sinsin90)RAy=12P(coscos90+sinsin90)
RAy=12PsinRAy=12Psin
Shear:
VBC=FyVBC=Fy
VBC=RAyPyVBC=RAyPy
VBC=12PsinPsinVBC=12PsinPsin
VBC=12PsinVBC=12Psin answer
Moment arm:
d=Rcos(180)d=Rcos(180)
d=R(cos180cos+sin180sin)d=R(cos180cos+sin180sin)
d=Rcosd=Rcos
Moment:
MBC=McounterclockwiseMclockwiseMBC=McounterclockwiseMclockwise
MBC=RA(R+d)PdMBC=RA(R+d)Pd
MBC=12P(RRcos)P(Rcos)MBC=12P(RRcos)P(Rcos)
MBC=12PR12PRcos+PRcosMBC=12PR12PRcos+PRcos
MBC=12PR+12PRcosMBC=12PR+12PRcos
MBC=12PR(1+cos)MBC=12PR(1+cos) answer
