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Solution 447
To draw the Load Diagram
- A 2400 lb upward force is acting at point A. No load in segment AB.
- A point force of 2400 - 400 = 2000 lb is acting downward at point B. No load in segment BC.
- Another downward force of magnitude 400 + 4000 = 4400 lb at point C. No load in segment CD.
- Upward point force of 4000 + 1000 = 5000 lb is acting at D. No load in segment DE.
- A downward force of 1000 lb is concentrated at point E.
To draw the Moment Diagram
- MA = 0
- MB = MA + Area in shear diagram MB = 0 + 2400(2) = 4800 lb·ft MAB is linear and upward
- MC = MB + Area in shear diagram MC = 4800 + 400(3) = 6000 lb·ft MBC is linear and upward
- MD = MC + Area in shear diagram MD = 6000 - 4000(2) = -2000 lb·ft MCD is linear and downward
- ME = MD + Area in shear diagram ME = -2000 + 1000(2) = 0 MDE is linear and upward
Problem 448
Solution 448
To draw the Load Diagram
- A uniformly distributed load in AB is acting downward at a magnitude of 40/2 = 20 kN/m.
- Upward concentrated force of 40 + 36 = 76 kN acts at B. No load in segment BC.
- A downward point force acts at C at a magnitude of 36 - 16 = 20 kN.
- 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.
To draw the Load Diagram
- Downward 4000 lb force is concentrated at A and no load in segment AB.
- The shear in BC is uniformly increasing, thus a uniform upward force is acting at a magnitude of (3700 + 4000)/ = 3850 lb/ft. No load in segment CD.
- Another point force acting downward with 3700 - 1700 = 1200 lb at D and no load in segment DE.
- The shear in EF is uniformly decreasing, thus a uniform downward force is acting with magnitude of (1700 + 3100)/ = 600 lb/ft.
- Upward force of 3100 lb is concentrated at end of span F.
To draw the Moment Diagram
- The locations of zero shear (points G and H) can be easily determined by ratio and proportion of triangle.
- MA = 0; MB = MA + Area in shear diagram MB = 0 - 4000(3) = -12, lb·ft
- MG = MB + Area in shear diagram MG = -12,000 - ½ (80/77)(4000) MG = -14,077.92 lb·ft
- MC = MG + Area in shear diagram MC = -14,077.92 + ½ (74/77)(3700) MC = -12,300 lb·ft
- MD = MC + Area in shear diagram MD = -12,300 + 3700(3) = -1200 lb·ft
- ME = MD + Area in shear diagram ME = -1200 + 1700(4) = 5600 lb·ft
- MH = ME + Area in shear diagram MH = 5600 + ½ (17/6)(1700) MH = 8,008.33 lb·ft
- MF = MH + Area in shear diagram MF = 8,008.33 - ½ (31/6)(3100) = 0
Problem 450
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Solution 450
To draw the Load Diagram
- 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.
Solution 451
To draw the Load Diagram
- Upward concentrated load at A is 10 kN.
- 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.
- A downward point force is acting at C in a magnitude of 8 - 2 = 6 kN.
- 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
- To find the location of zero shear, F: x^2 /10 = 3^2 /(10 + 2) x = 2.74 m
- MA = 0
- MF = MA + Area in shear diagram MF = 0 + 2/3 (2.74)(10) = 18. kN·m
- 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
- MC = MB + Area in shear diagram MC = 18 - 2(1) = 16 kN·m
- MD = MC + Area in shear diagram MD = 16 - 8(1) = 8 kN·m
- ME = MD + Area in shear diagram ME = 8 - ½ (2)(8) = 0
- 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.
SHEAR AND MOMENT DIAGRAM PROBLEMS
Problem 415 Cantilever beam loaded as shown in Fig. P-415.
Solution 415
HideClick here to show or hide the solution Segment AB: VAB=−20xkNVAB=−20xkN MAB=−20x(x/2)MAB=−20x(x/2) MAB=−10x2kN⋅mMAB=−10x2kN⋅m
Segment BC: VBC=−20(3)VBC=−20(3) VBC=−60kNVBC=−60kN
Problem 416 Beam carrying uniformly varying load shown in Fig. P-416.
Solution 416
ΣMR2=0ΣMR2= LR1=13LFLR1=13LF R1=13(12Lwo)R1=13(12Lwo) R1=16LwoR1=16Lwo
ΣMR1=0ΣMR1=
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=wo2Lx
V=R1−FxV=R1−Fx V=16Lwo−wo2Lx2V=16Lwo−wo2Lx
M=R1x−Fx(13x)M=R1x−Fx(13x)
M=16Lwox−wo2Lx2(13x)M=16Lwox−wo2Lx2(13x) M=16Lwox−wo6Lx3M=16Lwox−wo6Lx
To draw the Shear Diagram: V = 1/6 Lwo - wox^2 /2L is a second degree curve; at x = 0, V = 1/6 Lwo = R 1 ; at x = L, V = -1/ Lwo= -R 2 ; If a is the location of zero shear from left end, 0 = 1/6 Lwo - wox^2 /2L, x = 0.5774L = a; to check, use the squared property of parabola:
a^2 /R 1 = L^2 /(R 1 + R 2 ) a^2 /(1/6 Lwo) = L^2 /(1/6 Lwo + 1/3 Lwo) a^2 = (1/6 L^3 wo)/(1/2 Lwo) = 1/3 L 2 a = 0.5774L
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 = 0.0962L^2 wo - 0.0321L^2 wo Mmax = 0.0641L^2 wo
Problem 405 Beam loaded as shown in Fig. P-405.
To draw the Moment Diagram:
- 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.
- The moment diagram is also a second degree curve for segment BC given by MBC = 160 + 34x - 5x^2 ; at x = 2 m, MBC = 208 kN·m; at x = 10 m, MBC = 0.
- Note that the maximum moment occurs at point of zero shear. Thus, at x = 3. m, MBC = 217.8 kN·m.
Problem 407 Beam loaded as shown in Fig. P-407.
Solution 407
ΣMA=0ΣMA= 6RD=4[2(30)]6RD=4[2(30)] RD=40kNRD=40kN
ΣMD=0ΣMD=
6RA=2[2(30)]6RA=2[2(30)]
RA=20kNRA=20kN
Segment AB: VAB=20kNVAB=20kN MAB=20xkN⋅mMAB=20xkN⋅m
Segment BC: VBC=20−30(x−3)VBC=20−30(x−3) VBC=110−30xkNVBC=110−30xkN MBC=20x−30(x−3)(x−3)/2MBC=20x−30(x−3)(x− )/ MBC=20x−15(x−3)2kN⋅mMBC=20x−15(x−3)2kN⋅m
Segment CD: VCD=20−30(2)VCD=20−30(2) VCD=−40kNVCD=−40kN MCD=20x−30(2)(x−4)MCD=20x−30(2)(x −4) MCD=20x−60(x−4)kN⋅mMCD=20x−60(x −4)kN⋅m
To draw the Shear Diagram:
Solution 408
ΣMA=0ΣMA= 6RD=1[2(50)]+5[2(20)]6RD=1[2(50)]+5[2(20)] RD=50kNRD=50kN
ΣMD=0ΣMD=
6RA=5[2(50)]+1[2(20)]6RA=5[2(50)]+1[2(20)]
RA=90kNRA=90kN
Segment AB: VAB=90−50xkNVAB=90−50xkN MAB=90x−50x(x/2)MAB=90x−50x(x/2) MAB=90x−25x2kN⋅mMAB=90x−25x2kN⋅m
Segment BC: VBC=90−50(2)VBC=90−50(2) VBC=−10kNVBC=−10kN MBC=90x−2(50)(x−1)MBC=90x−2(50)(x−1) MBC=−10x+100kN⋅mMBC=−10x+100kN⋅m
Segment CD: VCD=90−2(50)−20(x−4)VCD=90−2(50)−20(x−4) VCD=−20x+70kNVCD=−20x+70kN
MCD=90x−2(50)(x− 1 )−20(x−4)(x−4)/2MCD=90x−2(50)(x−1)−20(x−4)(x−4)/ MCD=90x−100(x− 1 )−10(x−4)2MCD=90x−100(x−1)−10(x−4) MCD=−10x2+70x−60kN⋅mMCD=−10x2+70x−60kN⋅m
To draw the Shear Diagram:
- 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.
- VBC = -10 kN along segment BC.
- VCD = -20x + 70 is linear; at x = 4 m, VCD = -10 kN; at x = 6 m, VCD = -50 kN.
To draw the Moment Diagram:
VBC=−2x−13(x−2)2VBC=−2x−13(x−2)
MBC=−(x/2)F1−13(x−2)F2MBC=−(x/2)F1−13(x−2)F MBC=−(x/2)(2x)−13(x−2)[13(x−2)2]MBC=−(x/2)(2x)−13(x−2)[13(x−2)2] MBC=−x2−19(x−2)3MBC=−x2−19(x−2)
To draw the Shear Diagram:
- VAB = -2x is linear; at x = 0, VAB = 0; at x = 2 m, VAB = -4 kN.
- 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:
- MAB = -x^2 is a second degree curve; at x = 0, MAB = 0; at x = 2 m, MAB = - kN·m.
- MBC = -x^2 -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.
