Shear And Moment In Beams Ch No 4 5y1h2l
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Chapter No. 4 Shear and 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 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 s 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 (ΣM = 0 and ΣFv = 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 • Relationship Between Load, Shear, and Moment • Moving Loads
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 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.
The vertical shear at C in the figure shown inprevious section (also shown to the right) is taken as
VC=( Fv)L=R1–wx where R1 = R2 = wL/2
Vc=2wL−wx
The moment at C is
MC=( MC)=2wLx−wx x2 MC=2wLx−2wx2 If we differentiate M with respect to x:
dxdM=2wL dxdx−2w 2x dxdx dxdM=2wL−wx=shear thus,
dxdM=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.
Differentiate V with respect to x gives
dxdV=0−w thus,
dxdV=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 Convention The customary sign conventions for shearing force and bending moment are represented by the figures below. A force that tends to bend the beam 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.
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 403 Beam loaded as shown in Fig. P-403. See the instruction.
Solution 403 From the load diagram:
MB=0 5RD+1(30)=3(50) RD=24kN MD=0 5RB=2(50)+6(30) RB=56kN Segment AB:
VAB=–30kN MAB=–30xkN m Segment BC:
VBC=–30+56 VBC=26kN MBC=–30x+56(x–1) MBC=26x–56kN m Segment CD:
VCD=–30+56–50 VCD=–24kN MCD=–30x+56(x–1)–50(x–4) MCD=–30x+56x–56–50x+200 MCD=–24x+144kN m
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.
Problem 404 Beam loaded as shown in Fig. P-404. See the instruction.
Solution 404
MA=0 12RD+4800=3(2000) RD=100lb MD=0 12RA=9(2000)+4800 RA=1900lb Segment AB:
VAB=1900lb MAB=1900xlb ft
Segment BC:
VBC=1900–2000 VBC=–100lb MBC=1900x–2000(x–3) MBC=1900x–2000x+6000 MBC=–100x+6000lb ft Segment CD:
VCD=1900–2000 VCD=–100lb MCD=1900x–2000(x–3)–4800 MCD=1900x–2000x+6000–4800 MCD=–100x+1200lb 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.
Problem 405 Beam loaded as shown in Fig. P-405. See the instruction.
Solution 405
MA=0 10RC=2(80)+5[10(10)] RC=66kN MC=0 10RA=8(80)+5[10(10)] RA=114kN Segment AB:
VAB=114–10xkN MAB=114x–10x(x 2) MAB=114x–5x2kN m
Segment BC:
VBC=114 –80–10x VBC=34– 10xkN MBC=114x–80(x–2)–10x(x 2) MBC=160+34x–5x2kN 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. 3.
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 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.
Problem 406 Beam loaded as shown in Fig. P-406. See the instruction.
Solution 406
MA=0 12RC=4(900)+18(400)+9[(60)(18)] RC=1710lb MC=0 12RA+6(400)=8(900)+3[60(18)] RA=670lb Segment AB:
VAB=670–60xlb MAB=670x–60x(x 2) MAB=670x–30x2lb ft
Segment BC:
VBC=670–900–60x VBC=–230–60xlb MBC=670x–900(x–4)–60x(x 2) MBC=3600–230x–30x2lb ft Segment CD:
VCD=670+1710–900–60x VCD=1480–60xlb MCD=670x+1710(x–12)–900(x–4)–60x(x 2) MCD=–16920+1480x–30x2lb ft
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 lb·ft. 2. For BC, MBC = 3600 – 230x – 30x2, 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.
Problem 407 Beam loaded as shown in Fig. P-407. See the instruction.
Solution 407
MA=0 6RD=4[2(30)] RD=40kN MD=0 6RA=2[2(30)] RA=20kN Segment AB:
VAB=20kN MAB=20xkN m Segment BC:
VBC=20–30(x–3) VBC=110–30xkN MBC=20x–30(x–3)(x–3) 2 MBC=20x–15(x–3)2kN m Segment CD:
VCD=20–30(2) VCD=–40kN MCD=20x–30(2)(x–4) MCD=20x–60(x–4)kN m
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.
Problem 408 Beam loaded as shown in Fig. P-408. See the instruction.
Solution 408
MA=0 6RD=1[2(50)]+5[2(20)] RD=50kN MD=0 6RA=5[2(50)]+1[2(20)] RA=90kN Segment AB:
VAB=90–50xkN MAB=90x–50x(x 2) MAB=90x–25x2kN m
Segment BC:
VBC=90–50(2) VBC=–10kN MBC=90x–2(50)(x–1) MBC=–10x+100kN m Segment CD:
VCD=90–2(50)–20(x–4) VCD=–20x+70kN MCD=90x–2(50)(x–1)–20(x–4)(x–4) 2 MCD=90x–100(x–1)–10(x–4)2 MCD=–10x2+70x–60kN m
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 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.
Problem 409 Cantilever beam loaded as shown in Fig. P-409. See the instruction.
Solution 409 Segment AB:
VAB=–wox MAB=–wox(x 2) MAB=–21wox2 Segment BC:
VBC=–wo(L 2) VBC=–21woL MBC=–wo(L 2)(x–L 4) MBC=–21woLx+81woL2
To draw the Shear Diagram: 1. VAB = –wox for segment AB is linear; at x = 0, VAB = 0; at x = L/2, VAB = –½woL. 2. At BC, the shear is uniformly distributed by – ½woL.
To draw the Moment Diagram: 1. MAB = –½wox2 is a second degree curve; at x = 0, MAB = 0; at x = L/2, MAB = –1/8 woL2. 2. MBC = –½woLx + 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. See the instruction.
Solution 410
xy=Lwo y=Lwox Fx=21xy Fx=21x Lwox Fx=2Lwox2 Shear equation:
V=−2Lwox2 Moment equation:
M=–31xFx=−31x 2Lwox2 M=−6Lwox3
To draw the Shear Diagram: 1. V = – wo x2 / 2L is a second degree curve; at x = 0, V = 0; at x = L, V = –½ 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. See the instruction.
Solution 411
yL−x=Lwo y=Lwo(L−x) F121x(wo−y) F1=21x wo−Lwo(L−x) F1=21x wo−woL−Lwox F1=2Lwox2
F2=xy=x Lwo(L−x) F2=Lwo(Lx−x2) Shear equation:
V=–F1–F2=–2Lwox2−Lwo(Lx−x2) V=–2Lwox2−wox+Lwox2 V=2Lwox2−wox Moment equation:
M=–32xF1–21xF2 M=–32x 2Lwox2 –21x Lwo(Lx−x2)
M=–3Lwox3–2wox2+2Lwox3 M=–2wox2+6Lwox3
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.
Problem 412 Beam loaded as shown in Fig. P-412. See the instruction.
Solution 412
MA=0 6RC=5[6(800)] RC=4000lb MC=0 6RA=1[6(800)] RA=800lb Segment AB:
VAB=800lb MAB=800xlb ft Segment BC:
VBC=800–800(x–2) VBC=2400–800xlb MBC=800x–800(x–2)(x–2) 2 MBC=800x–400(x–2)2lb ft Segment CD:
VCD=800+4000–800(x–2) VCD=4800–800x+1600 VCD=6400–800xlb MCD=800x+4000(x–6)–800(x–2)(x–2) 2 MCD=800x+4000(x–6)–400(x–2)2lb 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.
Problem 413 Beam loaded as shown in Fig. P-413. See the instruction.
Solution 413
MB=0 6RE=1200+1[6(100)] RE=300lb ME=0 6RB+1200=5[6(100)] RB=300lb Segment AB:
VAB=–100xlb MAB=–100x(x 2) MAB=–50x2lb ft Segment BC:
VBC=–100x+300lb MBC=–100x(x 2)+300(x–2) MBC=–50x2+300x–600lb ft
Segment CD:
VCD=–100(6)+300 VCD=–300lb MCD=–100(6)(x–3)+300(x–2) MCD=–600x+1800+300x–600 MCD=–300x+1200lb ft
Segment DE:
VDE=–100(6)+300 VDE=–300lb MDE=–100(6)(x–3)+1200+300(x–2) MDE=–600x+1800+1200+300x–600 MDE=–300x+2400lb ft
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 = –50x2 is a second degree curve; at x= 0, MAB = 0; at x = ft, MAB = –200 lb·ft. 2. MBC = –50x2 + 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 lb·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.
Problem 414 Cantilever beam carrying the load shown in Fig. P-414. See the instruction.
Solution 414 Segment AB:
VAB=–2xkN MAB=–2x(x 2) MAB=–x2kN m Segment BC:
yx−2=22 y=32(x−2) F1=2x F2=21(x–2)y F2=21(x–2)[32(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)[31(x–2)2] MBC=–x2–91(x–2)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 kN·m. 2. MBC = –x2 –1/9 (x – 2)3 is a third degree curve; at x = 2 m, MBC = –4 kN·m; at x = 5 m, MBC = –28 kN·m.
Problem 415 Cantilever beam loaded as shown in Fig. P-415. See the instruction.
Solution 415 Segment AB:
VAB=–20xkN MAB=–20x(x 2) MAB=–10x2kN m Segment BC:
VBC=–20(3) VAB=–60kN MBC=–20(3)(x–1 5) MAB=–60(x–1 5)kN m Segment CD:
VCD=–20(3)+40 VC D= – 20 kN MCD=–20(3)(x–1 5)+40(x–5) MCD=–60(x–1 5)+40(x–5)kN m
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 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.
Problem 416 Beam carrying uniformly varying load shown in Fig. P-416. See the instruction.
Solution 416
MR2=0 LR1=31LF R1=31(21Lwo) R1=61Lwo MR1=0 LR2=32LF R2=32(21Lwo) R2=31Lwo xy=Lwo y=Lwox Fx=21xy=21x Lwox Fx=2Lwox2 V=R1–Fx V=61Lwo−2Lwox2 M=R1x–Fx(31x) M=61Lwox−2Lwox2(31x) M=61Lwox−6Lwox3
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. See the instruction.
Solution 417 By symmetry:
R1=R2=21(21Lwo)=41Lwo xy=woL 2 y=L2wox
F=21xy=21x L2wo F=Lwox2 V=R1–F V=41Lwo–Lwox2 M=R1x–F(31x) M=41Lwox− Lwox2 (31x) 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. See the instruction.
Solution 418 Segment AB:
VAB=–20kN MAB=–20xkN m Segment BC:
VAB=–20kN MAB=–20x+80kN m
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 kN·m. 2. MBC = –20x + 80 is also linear; when x = 4 m, MBC = 0; when x = 6 m, MBC = –60 kN·m
Problem 419 Beam loaded as shown in Fig. P-419. See the instruction.
Solution
MC=0 9R1=5(810) R1=450lb MA=0 9R2=4(810) R2=360lb Segment AB:
xy=6270 y=45x F=21xy=21x(45x) F=22 5x2
VAB=R1–F VAB=450–22 5x2lb MAB=R1x–F(31x) MAB=450x–22 5x2(31x) MAB=450x–7 5x3lb ft Segment BC:
VBC=450–810 VBC=–360lb MBC=450x–810(x–4) MBC=450x–810x+3240
MBC=3240–360xlb ft
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.
Problem 420 A total distributed load of 30 kips ed by a uniformly distributed reaction as shown in Fig. P420. See the instruction.
Solution 420
w=30(1000) 12 w=2500lb/ft FV=0 R=W 20r=30(1000) r=1500lb/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
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 = 750x2is 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. 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 θ for the built-in arch shown inFig. P-421.
Solution 421 For θ that is less than 90° Components of Q and P:
Qx=Qsin Qy=Qcos Px=Psin(90 – ) Px=P(sin90 cos –cos90 sin ) Px=Pcos Py=Pcos(90 – ) Py=P(cos90 cos +sin90 sin ) Py=Psin Shear:
V= Fy V=Qy–Py V=Qcos –Psin Moment arms:
dQ=Rsin
answer
dP=R–Rcos dP=R(1–cos ) Moment:
M= Mcounterclockwise– Mclockwise M=Q(dQ)–P(dP) M=QRsin –PR(1–cos ) answer For θ that is greater than 90° Components of Q and P:
Qx=Qsin(180 – ) Qx=Q(sin180 cos –cos180 sin ) Qx=Qcos Qy=Qcos(180 – ) Qy=Q(cos180 cos +sin180 sin ) Qy=–Qsin Px=Psin( –90 ) Px=P(sin cos90 –cos sin90 ) Px=–Pcos Py=Pcos( –90 ) Py=P(cos cos90 +sin sin90 ) Py=Psin Shear:
V= Fy V=–Qy–Py V=–(–Qsin )–Psin V=Qsin –Psin answer Moment arms:
dQ=Rsin(180 – ) dQ=R(sin180 cos –cos180 sin ) dQ=Rsin
dP=R+Rcos(180 – ) dP=R+R(cos180 cos +sin180 sin ) dP=R–Rcos dP=R(1–cos ) Moment:
M= Mcounterclockwise– Mclockwise M=Q(dQ)–P(dP) M=QRsin –PR(1–cos ) 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
MC=0 2R(RA)=RP RA=21P For θ that is less than 90° Shear:
VAB=RAcos(90 – ) VAB=21P(cos90 cos +sin90 sin ) VAB=21Psin answer Moment arm:
d=R–Rcos d=R(1–cos ) Moment:
MAB=RA(d) MAB=21PR(1–cos ) answer
For θ that is greater than 90° Components of P and RA:
Px=Psin( –90 ) Px=P(sin cos90 –cos sin90 ) Px=–Pcos Py=Pcos( –90 ) Py=P(cos cos90 +sin sin90 ) Py=Psin RAx=RAsin( –90 ) RAx=21P(sin cos90 –cos sin90 ) RAx=–21Pcos RAy=RAcos( –90 ) RAy=21P(cos cos90 +sin sin90 ) RAy=21Psin Shear:
VBC= Fy VBC=RAy–Py VBC=21Psin –Psin VBC=–21Psin answer Moment arm:
d=Rcos(180 – ) d=R(cos180 cos +sin180 sin ) d=–Rcos Moment:
MBC= Mcounterclockwise– Mclockwise MBC=RA(R+d)–Pd MBC=21P(R–Rcos )–P(–Rcos ) MBC=21PR–21PRcos +PRcos MBC=21PR+21PRcos MBC=21PR(1+cos ) answer
Relationship Between Load, Shear, and Moment
The vertical shear at C in the figure shown inprevious section (also shown to the right) is taken as
VC=( Fv)L=R1–wx where R1 = R2 = wL/2
Vc=2wL−wx The moment at C is
MC=( MC)=2wLx−wx x2 MC=2wLx−2wx2 If we differentiate M with respect to x:
dxdM=2wL dxdx−2w 2x dxdx dxdM=2wL−wx=shear thus,
dxdM=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.
Differentiate V with respect to x gives
dxdV=0−w thus,
dxdV=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 Convention The customary sign conventions for shearing force and bending moment are represented by the figures below. A force that tends to bend the beam 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.
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.)
Statically Determinate Beams Statically determinate beams are those beams in which the reactions of the s 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 (ΣM = 0 and ΣFv = 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 • Relationship Between Load, Shear, and Moment • Moving Loads
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 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.
The vertical shear at C in the figure shown inprevious section (also shown to the right) is taken as
VC=( Fv)L=R1–wx where R1 = R2 = wL/2
Vc=2wL−wx
The moment at C is
MC=( MC)=2wLx−wx x2 MC=2wLx−2wx2 If we differentiate M with respect to x:
dxdM=2wL dxdx−2w 2x dxdx dxdM=2wL−wx=shear thus,
dxdM=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.
Differentiate V with respect to x gives
dxdV=0−w thus,
dxdV=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 Convention The customary sign conventions for shearing force and bending moment are represented by the figures below. A force that tends to bend the beam 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.
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 403 Beam loaded as shown in Fig. P-403. See the instruction.
Solution 403 From the load diagram:
MB=0 5RD+1(30)=3(50) RD=24kN MD=0 5RB=2(50)+6(30) RB=56kN Segment AB:
VAB=–30kN MAB=–30xkN m Segment BC:
VBC=–30+56 VBC=26kN MBC=–30x+56(x–1) MBC=26x–56kN m Segment CD:
VCD=–30+56–50 VCD=–24kN MCD=–30x+56(x–1)–50(x–4) MCD=–30x+56x–56–50x+200 MCD=–24x+144kN m
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.
Problem 404 Beam loaded as shown in Fig. P-404. See the instruction.
Solution 404
MA=0 12RD+4800=3(2000) RD=100lb MD=0 12RA=9(2000)+4800 RA=1900lb Segment AB:
VAB=1900lb MAB=1900xlb ft
Segment BC:
VBC=1900–2000 VBC=–100lb MBC=1900x–2000(x–3) MBC=1900x–2000x+6000 MBC=–100x+6000lb ft Segment CD:
VCD=1900–2000 VCD=–100lb MCD=1900x–2000(x–3)–4800 MCD=1900x–2000x+6000–4800 MCD=–100x+1200lb 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.
Problem 405 Beam loaded as shown in Fig. P-405. See the instruction.
Solution 405
MA=0 10RC=2(80)+5[10(10)] RC=66kN MC=0 10RA=8(80)+5[10(10)] RA=114kN Segment AB:
VAB=114–10xkN MAB=114x–10x(x 2) MAB=114x–5x2kN m
Segment BC:
VBC=114 –80–10x VBC=34– 10xkN MBC=114x–80(x–2)–10x(x 2) MBC=160+34x–5x2kN 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. 3.
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 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.
Problem 406 Beam loaded as shown in Fig. P-406. See the instruction.
Solution 406
MA=0 12RC=4(900)+18(400)+9[(60)(18)] RC=1710lb MC=0 12RA+6(400)=8(900)+3[60(18)] RA=670lb Segment AB:
VAB=670–60xlb MAB=670x–60x(x 2) MAB=670x–30x2lb ft
Segment BC:
VBC=670–900–60x VBC=–230–60xlb MBC=670x–900(x–4)–60x(x 2) MBC=3600–230x–30x2lb ft Segment CD:
VCD=670+1710–900–60x VCD=1480–60xlb MCD=670x+1710(x–12)–900(x–4)–60x(x 2) MCD=–16920+1480x–30x2lb ft
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 lb·ft. 2. For BC, MBC = 3600 – 230x – 30x2, 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.
Problem 407 Beam loaded as shown in Fig. P-407. See the instruction.
Solution 407
MA=0 6RD=4[2(30)] RD=40kN MD=0 6RA=2[2(30)] RA=20kN Segment AB:
VAB=20kN MAB=20xkN m Segment BC:
VBC=20–30(x–3) VBC=110–30xkN MBC=20x–30(x–3)(x–3) 2 MBC=20x–15(x–3)2kN m Segment CD:
VCD=20–30(2) VCD=–40kN MCD=20x–30(2)(x–4) MCD=20x–60(x–4)kN m
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.
Problem 408 Beam loaded as shown in Fig. P-408. See the instruction.
Solution 408
MA=0 6RD=1[2(50)]+5[2(20)] RD=50kN MD=0 6RA=5[2(50)]+1[2(20)] RA=90kN Segment AB:
VAB=90–50xkN MAB=90x–50x(x 2) MAB=90x–25x2kN m
Segment BC:
VBC=90–50(2) VBC=–10kN MBC=90x–2(50)(x–1) MBC=–10x+100kN m Segment CD:
VCD=90–2(50)–20(x–4) VCD=–20x+70kN MCD=90x–2(50)(x–1)–20(x–4)(x–4) 2 MCD=90x–100(x–1)–10(x–4)2 MCD=–10x2+70x–60kN m
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 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.
Problem 409 Cantilever beam loaded as shown in Fig. P-409. See the instruction.
Solution 409 Segment AB:
VAB=–wox MAB=–wox(x 2) MAB=–21wox2 Segment BC:
VBC=–wo(L 2) VBC=–21woL MBC=–wo(L 2)(x–L 4) MBC=–21woLx+81woL2
To draw the Shear Diagram: 1. VAB = –wox for segment AB is linear; at x = 0, VAB = 0; at x = L/2, VAB = –½woL. 2. At BC, the shear is uniformly distributed by – ½woL.
To draw the Moment Diagram: 1. MAB = –½wox2 is a second degree curve; at x = 0, MAB = 0; at x = L/2, MAB = –1/8 woL2. 2. MBC = –½woLx + 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. See the instruction.
Solution 410
xy=Lwo y=Lwox Fx=21xy Fx=21x Lwox Fx=2Lwox2 Shear equation:
V=−2Lwox2 Moment equation:
M=–31xFx=−31x 2Lwox2 M=−6Lwox3
To draw the Shear Diagram: 1. V = – wo x2 / 2L is a second degree curve; at x = 0, V = 0; at x = L, V = –½ 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. See the instruction.
Solution 411
yL−x=Lwo y=Lwo(L−x) F121x(wo−y) F1=21x wo−Lwo(L−x) F1=21x wo−woL−Lwox F1=2Lwox2
F2=xy=x Lwo(L−x) F2=Lwo(Lx−x2) Shear equation:
V=–F1–F2=–2Lwox2−Lwo(Lx−x2) V=–2Lwox2−wox+Lwox2 V=2Lwox2−wox Moment equation:
M=–32xF1–21xF2 M=–32x 2Lwox2 –21x Lwo(Lx−x2)
M=–3Lwox3–2wox2+2Lwox3 M=–2wox2+6Lwox3
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.
Problem 412 Beam loaded as shown in Fig. P-412. See the instruction.
Solution 412
MA=0 6RC=5[6(800)] RC=4000lb MC=0 6RA=1[6(800)] RA=800lb Segment AB:
VAB=800lb MAB=800xlb ft Segment BC:
VBC=800–800(x–2) VBC=2400–800xlb MBC=800x–800(x–2)(x–2) 2 MBC=800x–400(x–2)2lb ft Segment CD:
VCD=800+4000–800(x–2) VCD=4800–800x+1600 VCD=6400–800xlb MCD=800x+4000(x–6)–800(x–2)(x–2) 2 MCD=800x+4000(x–6)–400(x–2)2lb 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.
Problem 413 Beam loaded as shown in Fig. P-413. See the instruction.
Solution 413
MB=0 6RE=1200+1[6(100)] RE=300lb ME=0 6RB+1200=5[6(100)] RB=300lb Segment AB:
VAB=–100xlb MAB=–100x(x 2) MAB=–50x2lb ft Segment BC:
VBC=–100x+300lb MBC=–100x(x 2)+300(x–2) MBC=–50x2+300x–600lb ft
Segment CD:
VCD=–100(6)+300 VCD=–300lb MCD=–100(6)(x–3)+300(x–2) MCD=–600x+1800+300x–600 MCD=–300x+1200lb ft
Segment DE:
VDE=–100(6)+300 VDE=–300lb MDE=–100(6)(x–3)+1200+300(x–2) MDE=–600x+1800+1200+300x–600 MDE=–300x+2400lb ft
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 = –50x2 is a second degree curve; at x= 0, MAB = 0; at x = ft, MAB = –200 lb·ft. 2. MBC = –50x2 + 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 lb·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.
Problem 414 Cantilever beam carrying the load shown in Fig. P-414. See the instruction.
Solution 414 Segment AB:
VAB=–2xkN MAB=–2x(x 2) MAB=–x2kN m Segment BC:
yx−2=22 y=32(x−2) F1=2x F2=21(x–2)y F2=21(x–2)[32(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)[31(x–2)2] MBC=–x2–91(x–2)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 kN·m. 2. MBC = –x2 –1/9 (x – 2)3 is a third degree curve; at x = 2 m, MBC = –4 kN·m; at x = 5 m, MBC = –28 kN·m.
Problem 415 Cantilever beam loaded as shown in Fig. P-415. See the instruction.
Solution 415 Segment AB:
VAB=–20xkN MAB=–20x(x 2) MAB=–10x2kN m Segment BC:
VBC=–20(3) VAB=–60kN MBC=–20(3)(x–1 5) MAB=–60(x–1 5)kN m Segment CD:
VCD=–20(3)+40 VC D= – 20 kN MCD=–20(3)(x–1 5)+40(x–5) MCD=–60(x–1 5)+40(x–5)kN m
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 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.
Problem 416 Beam carrying uniformly varying load shown in Fig. P-416. See the instruction.
Solution 416
MR2=0 LR1=31LF R1=31(21Lwo) R1=61Lwo MR1=0 LR2=32LF R2=32(21Lwo) R2=31Lwo xy=Lwo y=Lwox Fx=21xy=21x Lwox Fx=2Lwox2 V=R1–Fx V=61Lwo−2Lwox2 M=R1x–Fx(31x) M=61Lwox−2Lwox2(31x) M=61Lwox−6Lwox3
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. See the instruction.
Solution 417 By symmetry:
R1=R2=21(21Lwo)=41Lwo xy=woL 2 y=L2wox
F=21xy=21x L2wo F=Lwox2 V=R1–F V=41Lwo–Lwox2 M=R1x–F(31x) M=41Lwox− Lwox2 (31x) 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. See the instruction.
Solution 418 Segment AB:
VAB=–20kN MAB=–20xkN m Segment BC:
VAB=–20kN MAB=–20x+80kN m
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 kN·m. 2. MBC = –20x + 80 is also linear; when x = 4 m, MBC = 0; when x = 6 m, MBC = –60 kN·m
Problem 419 Beam loaded as shown in Fig. P-419. See the instruction.
Solution
MC=0 9R1=5(810) R1=450lb MA=0 9R2=4(810) R2=360lb Segment AB:
xy=6270 y=45x F=21xy=21x(45x) F=22 5x2
VAB=R1–F VAB=450–22 5x2lb MAB=R1x–F(31x) MAB=450x–22 5x2(31x) MAB=450x–7 5x3lb ft Segment BC:
VBC=450–810 VBC=–360lb MBC=450x–810(x–4) MBC=450x–810x+3240
MBC=3240–360xlb ft
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.
Problem 420 A total distributed load of 30 kips ed by a uniformly distributed reaction as shown in Fig. P420. See the instruction.
Solution 420
w=30(1000) 12 w=2500lb/ft FV=0 R=W 20r=30(1000) r=1500lb/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
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 = 750x2is 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. 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 θ for the built-in arch shown inFig. P-421.
Solution 421 For θ that is less than 90° Components of Q and P:
Qx=Qsin Qy=Qcos Px=Psin(90 – ) Px=P(sin90 cos –cos90 sin ) Px=Pcos Py=Pcos(90 – ) Py=P(cos90 cos +sin90 sin ) Py=Psin Shear:
V= Fy V=Qy–Py V=Qcos –Psin Moment arms:
dQ=Rsin
answer
dP=R–Rcos dP=R(1–cos ) Moment:
M= Mcounterclockwise– Mclockwise M=Q(dQ)–P(dP) M=QRsin –PR(1–cos ) answer For θ that is greater than 90° Components of Q and P:
Qx=Qsin(180 – ) Qx=Q(sin180 cos –cos180 sin ) Qx=Qcos Qy=Qcos(180 – ) Qy=Q(cos180 cos +sin180 sin ) Qy=–Qsin Px=Psin( –90 ) Px=P(sin cos90 –cos sin90 ) Px=–Pcos Py=Pcos( –90 ) Py=P(cos cos90 +sin sin90 ) Py=Psin Shear:
V= Fy V=–Qy–Py V=–(–Qsin )–Psin V=Qsin –Psin answer Moment arms:
dQ=Rsin(180 – ) dQ=R(sin180 cos –cos180 sin ) dQ=Rsin
dP=R+Rcos(180 – ) dP=R+R(cos180 cos +sin180 sin ) dP=R–Rcos dP=R(1–cos ) Moment:
M= Mcounterclockwise– Mclockwise M=Q(dQ)–P(dP) M=QRsin –PR(1–cos ) 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
MC=0 2R(RA)=RP RA=21P For θ that is less than 90° Shear:
VAB=RAcos(90 – ) VAB=21P(cos90 cos +sin90 sin ) VAB=21Psin answer Moment arm:
d=R–Rcos d=R(1–cos ) Moment:
MAB=RA(d) MAB=21PR(1–cos ) answer
For θ that is greater than 90° Components of P and RA:
Px=Psin( –90 ) Px=P(sin cos90 –cos sin90 ) Px=–Pcos Py=Pcos( –90 ) Py=P(cos cos90 +sin sin90 ) Py=Psin RAx=RAsin( –90 ) RAx=21P(sin cos90 –cos sin90 ) RAx=–21Pcos RAy=RAcos( –90 ) RAy=21P(cos cos90 +sin sin90 ) RAy=21Psin Shear:
VBC= Fy VBC=RAy–Py VBC=21Psin –Psin VBC=–21Psin answer Moment arm:
d=Rcos(180 – ) d=R(cos180 cos +sin180 sin ) d=–Rcos Moment:
MBC= Mcounterclockwise– Mclockwise MBC=RA(R+d)–Pd MBC=21P(R–Rcos )–P(–Rcos ) MBC=21PR–21PRcos +PRcos MBC=21PR+21PRcos MBC=21PR(1+cos ) answer
Relationship Between Load, Shear, and Moment
The vertical shear at C in the figure shown inprevious section (also shown to the right) is taken as
VC=( Fv)L=R1–wx where R1 = R2 = wL/2
Vc=2wL−wx The moment at C is
MC=( MC)=2wLx−wx x2 MC=2wLx−2wx2 If we differentiate M with respect to x:
dxdM=2wL dxdx−2w 2x dxdx dxdM=2wL−wx=shear thus,
dxdM=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.
Differentiate V with respect to x gives
dxdV=0−w thus,
dxdV=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 Convention The customary sign conventions for shearing force and bending moment are represented by the figures below. A force that tends to bend the beam 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.
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.)
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