Shear and Moment

Chapter 04 Shear & Moment in Beams DEFINITION OF A BEAM 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.

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

P

M

Simple Beam

w (N/m)

Overhanging Beam

P

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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

w (N/m)

Propped Beam

M

R1

R2

w2 (N/m)

Fixed or Restrained Beam

w1 (N/m)

w2 (N/m)

P1

Continuous Beam

M

P2

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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. P1

w (N/m)

P2

Uniform Load

Concentrated Loads

w (N/m)

Uniformly Varying Load

M

Applied Couple

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 x w (N/m) the beam is cut at point C a distance B of x from he left A C L support and the R1 portion of the R2 beam to the right x of C be removed. w (N/m) The portion removed must A M then be replaced C V by vertical R 1 shearing force V

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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. 30 kN

50 kN C

B

A

1m

3m

2m

Figure 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 30 kN

Segment AB: VAB = –30 kN MAB = –30x kN⋅m

D

A x

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Segment BC: VBC = –30 + 56 VBC = 26 kN MBC = –30x + 56(x – 1) MBC = 26x – 56 kN⋅m x 30 kN

C

1m

3m

RB = 56 kN 30 kN

Load Diagram

A

1m RB = 56 kN

C

B

3m

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.

D

2m RB = 24 kN

26 kN Shear Diagram

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.

–24 kN

–30 kN 56 kN⋅m 1.154 m Moment Diagram

–30 kN⋅m

Problem 404.

B

A

50 kN

1m RB = 56 kN

x

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

50 kN

B

A

30 kN

Beam loaded as shown in Fig. P-404. 2000 lb Figure P-404 A

RA

M = 4800 lb⋅ft C

B

3 ft

6 ft

3 ft

D

RD

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∑MA = 0 12RD + 4800 = 3(2000) RD = 100 lb

Solution 404.

Segment AB: VAB = 1900 lb MAB = 1900x lb⋅ft

∑MD = 0 12RA = 9(2000) + 4800 RA = 1900 lb A x

RA = 1900 lb

Segment BC: VBC = 1900 – 2000 VBC = –100 lb MBC = 1900x – 2000(x – 3) MBC = 1900x – 2000x + 6000 MBC = –100x + 6000

6 ft B

A

C M = 4800 lb⋅ft x

RA = 1900 lb

A

2000 lb B 3 ft

M = 4800 lb⋅ft C 6 ft

D

3 ft

RA = 1900 lb

B

A x RA = 1900 lb

Segment CD: VCD = 1900 – 2000 VCD = –100 lb MCD = 1900x – 2000(x – 3) – 4800 MCD = 1900x – 2000x + 6000 – 4800 MCD = –100x + 1200

2000 lb 3 ft

2000 lb

3 ft

Load Diagram

RD = 100 lb

1900 lb

Shear Diagram –100 lb 5700 lb⋅ft 5100 lb⋅ft 300 lb⋅ft

Moment Diagram

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.

Chapter 04 Shear and Moment in Beams Problem 405.

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Beam loaded as shown in Fig. P-405. 80 kN Figure P-405

10 kN/m A

C B RA

2m

8m

∑MA = 0 10RC = 2(80) + 5[10(10)] RC = 66 kN

Solution 405.

10 kN/m

RC

∑MC = 0 10RA = 8(80) + 5[10(10)] 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 A MBC = 114x – 80(x – 2) – 10x(x/2) MBC = 160 + 34x – 5x2

A x RA = 114 kN

2m

80 kN B 10 kN/m

x

RA = 114 kN

80 kN 10 kN/m A B

2m

C

Load Diagram

8m

RA = 114 kN

RC = 66 kN

114 kN 94 kN 14 kN

Shear Diagram

1.4 m 217.8 kN⋅m

–66 kN

208 kN⋅m Moment Diagram

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 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.

Shear and Moment Equations and Diagrams; Relation between Load, Shear, and Moment; Moving Loads

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www.mathalino.com Problem 406.

Beam loaded as shown in Fig. P-406. 900 lb

400 lb

Figure P-406

60 lb/ft A

D B RA

4 ft

C 8 ft

6 ft

RC

∑MA = 0 12RC = 4(900) + 18(400) + 9[(60)(18)] RC = 1710 lb

Solution 406.

∑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

900 lb 4 ft

x RA = 670 lb

x 900 lb 60 lb/ft A C

B RA = 670 lb

A x RA = 670 lb

Segment BC: VBC = 670 – 900 – 60x VBC = –230 – 60x lb MBC = 670x – 900(x – 4) – 60x(x/2) MBC = 3600 – 230x – 30x2 lb⋅ft

60 lb/ft

A

4 ft

60 lb/ft

8 ft RC = 1710 lb

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

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.

Chapter 04 Shear and Moment in Beams 900 lb

173 www.mathalino.com 400 lb

60 lb/ft Load Diagram

A B

C

D

8’

4’ RA = 670 lb

6’ RC = 1710 lb 760 lb

670 lb 430 lb

400 lb Shear Diagram

–470 lb –950 lb

2200 lb⋅ft

Moment Diagram 3.772 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.

–3480 lb⋅ft

Problem 407.

Beam loaded as shown in Fig. P-407. 30 kN/m

Figure P-407 A

D B 3m

C 2m

1m

RA

Solution 407.

RD

∑MA = 0 6RD = 4[2(30)] RD = 40 kN Segment AB: VAB = 20 kN MAB = 20x kN⋅m

∑MD = 0 6RA = 2[2(30)] RA = 20 kN A x RA = 20 kN

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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

30 kN/m 3m

A

B x

RA = 20 kN 30 kN/m

Segment CD: VCD = 20 – 30(2) VCD = –40 kN MCD = 20x – 30(2)(x – 4) MCD = 20x – 60(x – 4)

B

A

C 3m

2m

RA = 20 kN x 30 kN/m D

A B 3m

C 2m

RA = 20 kN

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.

Load Diagram

1m RD = 40 kN

20 kN Shear Diagram

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.

0.67 m –40 kN

60 kN⋅m

66.67 kN⋅m 40 kN⋅m

Problem 408.

Moment Diagram

Beam loaded as shown in Fig. P-408. 50 kN/m

20 kN/m

Figure P-408 A

D B 2m

C 2m

RA

Solution 408.

∑MA = 0 6RD = 1[2(50)] + 5[2(20)] RD = 50 kN

2m RD

∑MD = 0 6RA = 5[2(50)] + 1[2(20)] RA = 90 kN

Chapter 04 Shear and Moment in Beams Segment AB: VAB = 90 – 50x kN MAB = 90x – 50x(x/2) MAB = 90x – 25x2

50 kN/m A 2m

Segment BC: VBC = 90 – 50(2) VBC = –10 kN MBC = 90x – 2(50)(x – 1) MBC = –10x + 100 kN⋅m

B

RA = 90 kN x

50 kN/m

B

C

2m RA = 90 kN

2m x

50 kN/m

B

C 2m

–10 kN –50 kN

80 kN⋅m

60 kN⋅m

Problem 409.

x RA = 90 kN

To draw the Shear Diagram: (1) VAB = 90 – 50x is linear; at x = 0, VBC D = 90 kN; at x = 2 m, VBC = –10 kN. Load When VAB = 0, x = 1.8 m. Diagram 2m (2) VBC = –10 kN along segment BC. RD = 50 kN (3) VCD = –20x + 70 is linear; at x = 4 m, VCD = –10 kN; at x = 6 m, VCD = – 50 kN.

90 kN

81 kN⋅m

A

20 kN/m

A

1.8 m

50 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

20 kN/m

A

2m RA = 90 kN

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To draw the Moment Diagram: Shear 2 Diagram (1) MAB = 90x – 25x 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 = Moment 60 kN⋅m. Diagram (3) MCD = –10x2 + 70x – 60; at x = 4 m, MCD = 60 kN⋅m; at x = 6 m, MCD = 0.

Cantilever beam loaded as shown in Fig. P-409. wo Figure P-409 A B L/2

C L/2

Shear and Moment Equations and Diagrams; Relation between Load, Shear, and Moment; Moving Loads

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www.mathalino.com Solution 409.

Segment AB: VAB = –wox MAB = –wox(x/2) MAB = – 21 wox2

B L/2 x wo

A B

C

L/2

Load Diagram

L/2

x

1 2

To draw the Shear Diagram: (1) VAB = –wox for segment AB is linear; at x = 0, VAB = 0; at x = L/2, VAB = –

1 2

woL.

(2) At BC, the shear is uniformly distributed by – Shear Diagram

–

A

Segment BC: VBC = –wo(L/2) VBC = – 21 woL MBC = –wo(L/2)(x – L/4) MBC = – 21 woLx + 81 woL2

wo A

wo

woL

1 2

woL.

To draw the Moment Diagram: (1) MAB = –

1 2

wox2 is a second degree curve; at x =

0, MAB = 0; at x = L/2, MAB = –

–

1 8

Moment Diagram

woL2 –

Problem 410.

3 8

(2) MBC = –

1 2

woLx +

x = L/2, MBC = –

woL2

3 8

1 8

1 8

woL2.

woL2 is a second degree; at

1 8

woL2; at x = L, MBC = –

woL2.

Cantilever beam carrying the uniformly varying load shown in Fig. P-410. wo

Figure P-410

L

Solution 410.

y w = o L x wo y= x L Fx = 21 xy Fx =

1  wo  x x 2  L 

Fx

1 3

x wo

y L–x x

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wo 2 x 2L

Fx =

Shear equation: w V = – o x2 2L Moment equation:

1  wo 2  x x  3  2L  w M = – o x3 6L

M = – 13 xFx = – wo Load Diagram

L

To draw the Shear Diagram: V=–

wo 2L

2

is a second degree curve;

at x = 0, V = 0; at x = L, V = –

Shear Diagram

2nd degree

x

1 2

woL.

To draw the Moment Diagram: –

1 2

woL

M=–

wo

x

3

is a third degree curve; at

6L x = 0, M = 0; at x = L, M =– 3rd degree –

Problem 411.

1 6

woL2.

woL2

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. Figure P-411

wo

L

Solution 411.

1 6

Moment Diagram

y w = o L−x L wo y= (L − x ) L

Shear and Moment Equations and Diagrams; Relation between Load, Shear, and Moment; Moving Loads

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F1

F2

x 1 2

x

wo

y

wo

L

Shear equation:

2nd degree Shear Diagram

–

1 2

woL

–

1 3

woL2

wo 2 w x – o ( Lx − x 2 ) 2L L wo 2 wo 2 V=– x – wo x + x L 2L w V = o x 2 – wo x 2L V = –F1 – F2 = –

3rd degree

Moment Diagram

Moment equation: M = – 23 xF1 – 21 xF2

To draw the Shear Diagram:

wo

2

x – w o x is a concave 2L upward second degree curve; at x = 0, V = 0; at x = L, V = –

1 2

1 w 1  wo 2   x x  – x  o ( Lx − x 2 ) 3  2L  2  L  wo 3 wo 2 wo 3 M=– x – x + x 3L 2 2L w w M = – o x2 + o x3 2 6L M=–

woL.

To draw the Moment diagram:

wo 3 2 x + x is in third 2 6L degree; at x = 0, M = 0; at x = L,

wo

1 3

w 1   x w o − o ( L − x ) 2  L 

w  F2 = xy = x  o (L − x )  L  w F2 = o ( Lx − x 2 ) L

Load Diagram

M=–

F1 =

w  1  x wo − wo + o x  2  L  wo 2 F1 = x 2L

x

M=–

1 2

F1 =

L–x

V =

x( w o − y )

F1 =

woL2.

Problem 412.

Beam loaded as shown in Fig. P-412. 800 lb/ft

Figure P-406 A

D B RA

2 ft

C 4 ft

RC

2 ft

Chapter 04 Shear and Moment in Beams ∑MA = 0 6RC = 5[6(800)] RC = 4000 lb

Solution 412.

A

RA = 800 lb

Segment BC: VBC = 800 – 800(x – 2) VBC = 2400 – 800x MBC = 800x – 800(x – 2)(x – 2)/2 MBC = 800x – 400(x – 2)2

800 lb/ft A

B x

RA = 800 lb

800 lb/ft B

A

∑MC = 0 6RA = 1[6(800)] RA = 800 lb

Segment AB: VAB = 800 lb MAB = 800x

x

2 ft

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C 4 ft RC = 4000 lb

2 ft RA = 800 lb x

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

800 lb/ft Load D Diagram To draw the Shear Diagram: (1) 800 lb of shear force is uniformly 4 ft 2 ft distributed along segment AB. RC = 4000 lb (2) VBC = 2400 – 800x is linear; at x = 2 ft, VBC = 800 lb; at x = 6 ft, VBC = –2400 1600 lb 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 = Shear 6 ft, VCD = 1600 lb; at x = 8 ft, VBC = 0. Diagram

A B

C

2 ft RA = 800 lb 800 lb

1 ft

2000 lb⋅ft 1600 lb⋅ft

–1600 lb⋅ft

To draw the Moment Diagram: (1) MAB = 800x is linear; at x = 0, MAB = 0; –2400 lb 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. 2 Moment (3) MCD = 800x + 4000(x – 6) – 400(x – 2) is also a second degree curve; at x = 6 Diagram ft, MCD = –1600 lb⋅ft; at x = 8 ft, MCD = 0.

Shear and Moment Equations and Diagrams; Relation between Load, Shear, and Moment; Moving Loads

180

www.mathalino.com Problem 413.

Beam loaded as shown in Fig. P-413. Figure P-413

100 lb/ft M = 1200 lb⋅ft A

E B 2 ft

C 4 ft

RB

D 1 ft

1 ft

RE

∑MB = 0 6RE = 1200 + 1[6(100)] RE = 300 lb

Solution 413.

∑ME = 0 6RB + 1200 = 5[6(100)] RB = 300 lb Segment AB: VAB = –100x lb MAB = –100x(x/2) MAB = –50x2 lb⋅ft

100 lb/ft

A x

Segment BC: VBC = –100x + 300 lb MBC = –100x(x/2) + 300(x – 2) MBC = –50x2 + 300x – 600 lb⋅ft

A B 2 ft RB = 300 lb x

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

100 lb/ft A B

C

2 ft

4 ft

RB = 300 lb x

100 lb/ft 1200 lb⋅ft A B 2 ft RB = 300 lb

100 lb/ft

C 4 ft x

D 1’

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

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100 lb/ft M = 1200 lb⋅ft E To draw the Shear Diagram: (1) VAB = –100x is linear; at x = 0, VAB = Load C D 1’ 1’ 0; at x = 2 ft, VAB = –200 lb. Diagram (2) VBC = 300 – 100x is also linear; at x RE = 300 lb = 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. Shear

A B 2 ft

4 ft

RB = 300 lb 100 lb

Diagram

–200 lb –300 lb

1 ft

300 lb⋅ft

–200 lb⋅ft –150 lb⋅ft –600 lb⋅ft

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 Moment lb⋅ft; at x = 3 ft, MBC = –150 l⋅ft. Diagram (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.

–900 lb⋅ft

Problem 414.

Cantilever beam carrying the load shown in Fig. P414. 4 kN/m Figure P-414

2 kN/m A

B 2m

Solution 414.

Segment AB: VAB = –2x kN MAB = –2x(x/2) MAB = –x2 kN⋅m

3m F2

F1 2 kN/m A

y 2 kN/m

B 2m

x–2 x

Segment BC: y 2 = x−2 3 y = 23 (x – 2)

C

3m

2 kN/m A x

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F1 = 2 x

4 kN/m 2 kN/m A

B 2m

3m Load Diagram

C

F2 =

1 2

(x – 2)y

F2 =

1 2

(x – 2)[ 23 (x – 2)]

F2 =

1 3

(x – 2)2

VBC = –F1 – F2 VBC = – 2x – 13 (x – 2)2 MBC = –(x/2)F1 –

st

1 degree

MBC = –(x/2)(2x) –

–4 kN

MBC = –x2 –

2nd degree Shear Diagram

2nd degree

–13 kN

1 9

(x – 2)F2 1 3

(x – 2)[ 13 (x – 2)2]

(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 – 13 (x – 2)2 is a second degree curve; at x = 2 m, VBC = –4 kN; at x = 5 m; VBC = –13 kN.

–4 kN⋅m 3rd degree –28 kN⋅m Moment Diagram

Problem 415.

1 3

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 – 91 (x – 2)3 is a third degree curve; at x = 2 m, MBC = –4 kN⋅m; at x = 5 m, MBC = –28 kN⋅m.

Cantilever beam loaded as shown in Fig. P-415. Figure P-415 20 kN/m A B 3m

2m

C 40 kN

Solution 415.

Segment AB: VAB = –20x kN MAB = –20x(x/2) MAB = –10x2 kN⋅m

20 kN/m A x

2m

D

Chapter 04 Shear and Moment in Beams

183 www.mathalino.com 20 kN/m

Segment BC: VBC = –20(3) VAB = –60 kN MBC = –20(3)(x – 1.5) MAB = –60(x – 1.5) kN⋅m

20 kN/m A B 3m

2m

A B 3m x

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)

C 40 kN

x

20 kN/m A B 3m

2m

C

Load Diagram

2m

D

40 kN

–20 kN –60 kN Shear Diagram

–90 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.

–210 kN⋅m –250 kN⋅m

Moment Diagram

wo

Problem 416.

Beam carrying uniformly varying load shown in Fig. P-416.

L R1

R2

Figure P-416

F = ½ Lwo

Solution 416.

∑MR2 = 0 LR1 =

2/3 L 1 3

1/3 L

LF

R1 =

1 1 3(2

R1 =

1 6

wo

Lwo) L

Lwo R1

R2

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∑MR1 = 0 LR2 = 23 LF R2 =

2 1 3(2

R2 =

1 3

Lwo)

Lwo

y wo = x L wo y= x L Fx = ½ xy 1/3 x

2/3 x

y

wo

Fx =

1 2

Fx =

wo 2 x 2L

xy =

1  wo  x x 2  L 

x

V = R 1 – Fx

R1

V=

wo

1 6

Lwo –

wo 2 x 2L

M = R1x – Fx( 13 x ) w M = 61 Lwox – o x 2 ( 13 x ) 2L w M = 61 Lwox – o x 3 6L

L R1

Load Diagram

R2

R1

a Shear Diagram

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 a=

–R2 Mmax

Moment Diagram

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

Chapter 04 Shear and Moment in Beams Problem 417.

185 www.mathalino.com

Beam carrying the triangular loading shown in Fig. P417. wo Figure P-417

L/2

L/2

R1

Solution 417.

R2

By symmetry: R1 = R2 = 21 ( 21 Lw o ) =

1 4

1/3 x

Lw o F

y

y w 2 wo = o ; y= x L x L /2

wo

x

1  2 wo  F = 21 xy = x  x 2  L  w F = o x2 L

wo

R1 L/2

V = R1 – F V= L/2 R1 1 4

1 4

Lw o –

wo 2 x L

L/2 R2

Load Diagram

M = R1x – F ( 13 x )

w  Lw o x –  o x 2 ( 13 x )  L  w M = 14 Lw o x – o x 3 3L M=

Lw o

Shear Diagram

− 1 12

2

L wo

Moment Diagram

1 4

Lw o

1 4

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.

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www.mathalino.com Problem 418.

Cantilever beam loaded as shown in Fig. P-418. M = 80 kN⋅m

20 kN

Figure P-418

B A 4m

Solution 418.

Segment AB: VAB = –20 kN MAB = –20x kN⋅m

M = 80 kN⋅m

20 kN

2m

C

20 kN A x

B

Segment BC: VAB = –20 kN MAB = –20x + 80 kN⋅m

A 4m x

M = 80 kN⋅m

20 kN

B A 4m

2m

C

Load Diagram

Shear Diagram –20 kN

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

Moment Diagram –40 kN⋅m –80 kN⋅m

Problem 419.

Beam loaded as shown in Fig. P-419. 270 lb/ft

A 6 ft

B

R1

C 3 ft R2

Figure P-419

Chapter 04 Shear and Moment in Beams

187 www.mathalino.com

Solution 419.

810 lb 4 ft

5 ft 270 lb/ft

A

C

B

6 ft

3 ft

R1

R2

[ ∑MC = 0 ]

9R1 = 5(810) R1 = 450 lb

[ ∑MA = 0 ]

9R2 = 4(810) R2 = 360 lb Segment AB: y 270 = x 6 y = 45x

1/3 x F 270 lb/ft

y A

F=

R1

F=

x

1 2

xy =

1 2

x( 45x )

22.5x2

6 ft

VAB = R1 – F VAB = 450 – 22.5x2 lb MAB = R1x – F( 13 x ) MAB = 450x – 22.5x2( 13 x ) MAB = 450x – 7.5x3 lb⋅ft 810 lb

4 ft

270 lb/ft

A

B

6 ft R1 = 450 lb x

Segment BC: VBC = 450 – 810 VBC = –360 lb MBC = 450x – 810(x – 4) MBC = 450x – 810x + 3240 MBC = 3240 – 360x lb⋅ft

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270 lb/ft

A

C

B 6 ft

3 ft R2 = 360 lb

R1 = 450 lb Load Diagram 450 lb

a= a= √20 –360 lb

Shear Diagram

1341.64 lb⋅ft 1080 lb⋅ft

3rd degree

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.

linear

Moment Diagram

Problem 420.

A total distributed load of 30 kips supported by a uniformly distributed reaction as shown in Fig. P-420. W = 30 kips Figure P-420

4 ft

12 ft

4 ft

Solution 420. W = 30 kips w lb/ft

r lb/ft 4 ft

12 ft

4 ft

Chapter 04 Shear and Moment in Beams

189 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

2500 lb/ft

x

r = 1500 lb/ft

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 4 ft x

25 lb/ft

1500 lb/ft 4 ft

12 ft

4 ft

Load Diagram 6000 lb

6 ft Shear Diagram –6000 lb 30,000 lb⋅ft

12,000

12,000 lb⋅ft

Moment Diagram

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.

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www.mathalino.com Problem 421.

Write the shear and moment equations as functions of the angle θ for the built-in arch shown in Fig. P-421.

R

A

θ

B

Q P

For θ that is less than 90° Components of Q and P: Qx = Q sin θ Qy = Q cos θ

Solution 421.

V dQ

R θ

θ

Q

dP P

90° – θ

Figure P-421

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 θ – P sin θ 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 θ – PR(1 – cos θ)

Chapter 04 Shear and Moment in Beams

dQ

R θ

Q θ – 90°

180° – θ dP P

191 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) V Qy = Q (cos 180° cos θ + sin 180° sin θ) Qy = –Q sin θ

180° – 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 θ – P sin θ 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 θ – PR(1 – cos θ)

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www.mathalino.com 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. P B

R θ

A

C O

Figure P-422

∑MC = 0 2R(RA) = RP RA = 21 P

Solution 422. P B

For θ that is less than 90° R θ

A

C O

Shear: VAB = RA cos (90° – θ) VAB = 21 P (cos 90° cos θ + sin 90° sin θ) VAB =

1 2

P sin θ

RA

Moment arm: d = R – R cos θ d = R(1 – cos θ)

V

Moment: MAB = Ra (d) MAB = 21 PR(1 – cos θ)

R A

θ O 90° – θ

RA d

Chapter 04 Shear and Moment in Beams

P

For θ that is greater than 90°

θ – 90°

B

V R θ

A R θ – 90°

RA

180° – θ O d

193 www.mathalino.com

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 =

1 2

P sin θ

Shear: VBC = ∑Fy VBC = RAy – Py VBC = 21 P sin θ – P sin θ VBC = – 21 P sin θ 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 =

1 2

PR –

1 2

PR cos θ + PR cos θ

MBC =

1 2

PR +

1 2

PR cos θ

MBC =

1 2

PR(1 + cos θ)

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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 =

wL – wx 2

The moment at C is

wL x x – wx  2 2 wLx wx 2 MC = – 2 2 MC = (ΣMC) =

If we differentiate M with respect to x:

wL dx dM w dx = – 2x 2 dx dx 2 dx wL dM = – wx = shear dx 2 thus,

dM =V dx 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.

Chapter 04 Shear and Moment in Beams

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Differentiate V with respect to x gives

dV = 0 – w = load dx dV = Load dx 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

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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.

Positive Bending

Negative Bending

Positive Shear

Negative Shear

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. 60 kN

Figure P-425

2m R1

30 kN

4m

1m R2

Chapter 04 Shear and Moment in Beams

197 www.mathalino.com ∑MA = 0 6R2 = 2(60) + 7(30) R2 = 55 kN

Solution 425. 60 kN

30 kN

B D A

C 2m

4m

R1 = 35 kN

1m R2 = 55 kN

∑MC = 0 6R1 + 1(30) = 4(60) R1 = 35 kN

Load Diagram

35 kN

30 kN

–25 kN

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

Shear Diagram 70 kN⋅m

Moment Diagram

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

–30 kN⋅m

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www.mathalino.com Problem 426.

Cantilever beam acted upon by a uniformly distributed load and a couple as shown in Fig. P-426. 5 kN/m Figure P-426

M = 60 kN⋅m

2m

2m

1m

Solution 426. 5 kN/m M = 60 kN⋅m D C

B A 2m

2m Load Diagram

1m

–10 kN Shear Diagram 30 kN⋅m 20 kN⋅m

2nd deg

–10 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

1st deg Moment Diagram

Problem 427.

–30 kN⋅m

Beam loaded as shown in Fig. P-427. 800 lb 100 lb/ft Figure P-427

9 ft R1

3 ft R2

Chapter 04 Shear and Moment in Beams

∑MC = 0 12R1 = 100(12)(6) + 800(3) R1 = 800 lb

Solution 427. 800 lb 100 lb/ft

A

∑MA = 0 12R2 = 100(12)(6) + 800(9) R2 = 1200 lb

C

B 9 ft

3 ft

R1 = 800 lb

199 www.mathalino.com

R2 = 1200 lb Load Diagram

800 lb

x = 8 ft

–100 lb –900 lb

Shear Diagram

–1200 lb

A’

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.

3200 lb⋅ft 3150 lb⋅ft

Moment Diagram

Problem 428.

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

Beam loaded as shown in Fig. P-428. 10 kN/m

Figure P-428

25 kN⋅m

1m R1

1m

3m

2m R2

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∑MD = 0 5R1 = 50(0.5) + 25 R1 = 10 kN

Solution 428. 50 kN

10 kN/m 25 kN⋅m A B

C

1m 1m R1 = 10 kN

E

D

0.5 m 3m

2m R2 = 40 kN

Load Diagram

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=1m

20 kN 10 kN

x=1m –20 kN Shear Diagram

10 kN⋅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

–5 kN⋅m

–15 kN⋅m –20 kN⋅m Moment Diagram

Problem 429.

∑MA = 0 5R2 + 25 = 50(4.5) R2 = 40 kN

Beam loaded as shown in Fig. P-429. 100 lb 120 lb/ft

2 ft R1

120 lb/ft

2 ft

2 ft R2

Figure P-429

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Chapter 04 Shear and Moment in Beams

201 www.mathalino.com ∑MC = 0 4R1 + 120(2)(1) = 100(2) + 120(2)(3) R1 = 170 lb

Solution 429. 100 lb 120 lb/ft

A

120 lb/ft

B

2 ft R1 = 170 lb

C

∑MA = 0 4R2 = 120(2)(1) + 100(2) + 120(2)(5) R2 = 410 lb

D

2 ft

2 ft R2 = 410 lb

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

Load Diagram 240 lb 170 lb

–70 lb x = 1.42 ft

–170 lb Shear Diagram

120.42 lb⋅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

100 lb⋅ft

–240 lb⋅ft Moment Diagram

Problem 430.

Beam loaded as shown in P-430. 1000 lb

2000 lb 200 lb/ft

400 lb/ft

5 ft

10 ft R1

10 ft R2

Figure P-430

Shear and Moment Equations and Diagrams; Relation between Load, Shear, and Moment; Moving Loads

202

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∑MD = 0 20R1 = 1000(25) + 400(5)(22.5) + 2000(10) + 200(10)(5) R1 = 5000 lb

Solution 430. 1000 lb

2000 lb 200 lb/ft

400 lb/ft B

C

∑MB = 0 20R2 + 1000(5) + 400(5)(2.5) = 2000(10) + 200(10)(15) R2 = 2000 lb

A 10 ft

5 ft

10 ft

D

R2 = 2000 lb

R1 = 5000 lb Load 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

2000 lb

A’

–1000 lb

–2000 lb –3000 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.

Shear Diagram

10000 lb⋅ft 5 ft A’

–10000 lb⋅ft Moment Diagram

Problem 431.

Beam loaded as shown in Fig. P-431. 50 kN

20 kN/m

40 kN

Figure P-431 2m

1m

10 kN/m

7m R1

3m R2

Chapter 04 Shear and Moment in Beams

203 www.mathalino.com ∑MD = 0 7R1 + 40(3) = 5(50) + 10(10)(2) + 20(4)(2) R1 = 70 kN

Solution 431. 4m

50 kN 2m

20 kN/m

1m C

B

40 kN

10 kN/m

A

D 2m

5m R1 = 70 kN

∑MA = 0 7R2 = 50(2) + 10(10)(5) + 20(4)(5) + 40(10) R2 = 200 lb

E 3m

R2 = 200 lb Load Diagram

70 kN

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

40 kN

70 kN 50 kN

y

–10 kN

a = 1/3 m –130 kN Shear Diagram

A’ 120 kN⋅m B’

115 kN⋅m

C’

x

–165 kN⋅m Moment Diagram (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

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.

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.

Shear and Moment Equations and Diagrams; Relation between Load, Shear, and Moment; Moving Loads

204

www.mathalino.com Problem 432.

Beam loaded as shown in Fig. P-432. 60 kN

40 kN/m M = 120 kN⋅m

Figure P-432

1m

3m R1

40 kN/m M = 120 kN⋅m D

A

B

E

C

1m R1 = 132 kN

3m

1m 1m R2 = 48 kN

Load Diagram

72 kN

x = 1.8 m –60 kN

1m R2

∑ME = 0 5R1 + 120 = 6(60) + 40(3)(3.5) R1 = 132 kN

Solution 432. 60 kN

1m

–48 kN

Shear Diagram

48 kN⋅m

4.8 kN⋅m –24 kN⋅m –60 kN⋅m –72 kN⋅m Moment Diagram

∑MB = 0 5R2 + 60(1) = 40(3)(1.5) + 120 R2 = 48 kN 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.

Chapter 04 Shear and Moment in Beams Problem 433.

205 www.mathalino.com

Overhang beam loaded by a force and a couple as shown in Fig. P-433. 750 lb Figure P-433

3000 lb⋅ft

2 ft R1

3 ft

2 ft R2

∑MC = 0 5R1 + 2(750) = 3000 R1 = 300 lb

Solution 433.

750 lb 3000 lb⋅ft A

B

C

2 ft R1 = 300 lb

3 ft

D

Load Diagram

2 ft R2 = 450 lb 750 lb

300 lb Shear Diagram

600 lb⋅ft Moment Diagram

–1500 lb⋅ft

–2400 lb⋅ft

∑MA = 0 5R2 + 3000 = 7(750) R2 = 450 lb 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

Shear and Moment Equations and Diagrams; Relation between Load, Shear, and Moment; Moving Loads

206

www.mathalino.com Problem 434.

Beam loaded as shown in Fig. P-434. 60 kN 20 kN/m Figure P-434

M = 120 kN⋅m

2m

2m R1

60 kN 20 kN/m M = 120 kN⋅m B

C

D

2m 2m R1 = 100 kN

2m

60 kN 20 kN

–40 kN

Shear Diagram

3–x 80 kN⋅m 1m 40 kN⋅m

C’ 50 kN⋅m

x

1m –40 kN⋅m

E

2m R2 = 40 kN

Load Diagram

–40 kN

2m R2

∑ME = 0 6R1 + 120 = 20(4)(6) + 60(4) R1 = 100 kN

Solution 434.

A

2m

–40 kN⋅m

Moment Diagram

∑MB = 0 6R2 = 20(4)(0) + 60(2) + 120 R2 = 40 kN 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

Chapter 04 Shear and Moment in Beams Problem 435.

207 www.mathalino.com

Beam loaded and supported as shown in Fig. P-435. 20 kN 10 kN/m

Figure P-435

2m

2m

40 kN 1m

1m

2m

wo

R1

∑MB = 0 2wo (5) = 10(4)(0) + 20(2) + 40(3) wo = 16 kN/m

Solution 435. 20 kN 10 kN/m

40 kN 1m

2m F

A

C

B

D

E

2m 2m 1m R1 = 68 kN Load Diagram

wo = 16 kN/m

∑Mmidpoint of EF = 0 5R1 = 10(4)(5) + 20(3) + 40(2) R1 = 68 kN 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

2.8 m 48 kN 28 kN 8 kN

–20 kN –32 kN Shear Diagram C’ 95.2 kN⋅m 64 kN⋅m 56 kN⋅m 32 kN⋅m

x

–20 kN⋅m Moment Diagram

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.

Shear and Moment Equations and Diagrams; Relation between Load, Shear, and Moment; Moving Loads

208

www.mathalino.com Problem 436.

A distributed load is supported by two distributed reactions as shown in Fig. P-436. 440 lb/ft Figure P-436 4 ft

2 ft 8 ft

R1 = w1 lb/ft

R2 = w2 lb/ft

∑Mmidpoint of CD = 0 4w1 (11) = 440(8)(5) w1 = 400 lb/ft

Solution 436. 440 lb/ft 4 ft

2 ft

A

D B

C

8 ft

R1 = 400 lb/ft

R2 = 960 lb/ft Load 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

1600 lb

x = 3.636 ft

Shear Diagram

–1920 lb

6109.1 lb⋅ft 3200 lb⋅ft 1920 lb⋅ft

Moment Diagram

∑Mmidpoint of AB = 0 2w2 (11) = 440(8)(6) w2 = 960 lb/ft

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

Chapter 04 Shear and Moment in Beams Problem 437.

209 www.mathalino.com

Cantilever beam loaded as shown in Fig. P-437

1000 lb

400 lb/ft

2 ft

2 ft

4 ft

500 lb Figure P-437

Solution 437.

1000 lb

A

400 lb/ft

B 2 ft

C 2 ft

D

Load Diagram

4 ft

500 lb

Shear Diagram –500 lb –1000 lb

–2100 lb

Moment Diagram

–2000 lb⋅ft –3000 lb⋅ft

–8200 lb⋅ft

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

Shear and Moment Equations and Diagrams; Relation between Load, Shear, and Moment; Moving Loads

210

www.mathalino.com 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. 200 lb/ft Figure P-438

Hinge 2 ft

4 ft

2 ft

R1 200 lb/ft

Solution 438. 200 lb/ft A

B

H

2 ft A

B

H

2 ft 4 ft R1 = 900 lb Load Diagram

C 2 ft

4 ft R1

∑MH = 0 4R1 = 200(6)(3) R1 = 900 lb

500 lb

x = 2.5 ft

–300 lb

–400 lb Shear Diagram

–700 lb

1.0 ft 225 lb⋅ft

–400 lb⋅ft Moment Diagram

–1000 lb⋅ft

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.

Chapter 04 Shear and Moment in Beams Problem 439.

211 www.mathalino.com

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. 4000 lb

400 lb/ft

Figure P-439 Hinge

R1

4 ft

4 ft

4 ft

10 ft

R2

R3

Solution 439. 4000 lb 4000 lb

400 lb/ft

Hinge

Hinge

H

A A

B

H

4 ft 4 ft R1 = 2000 lb

C

VH

D

4 ft R2 = 4800 lb

10 ft R3 = 1200 lb

4 ft

4 ft

R1

∑MH = 0 8R1 = 4000(4) R1 = 2000 lb

Load Diagram 2800 lb

2000 lb

B

VH = 2000 lb

∑MA = 0 8VH = 4000(4) VH = 2000 lb 400 lb/ft

Hinge –1200 lb –2000 lb

H

x = 7 ft

C

D

4 ft

10 ft R2

Shear Diagram 3 ft 8000 lb ft⋅ Moment Diagram 1800 lb⋅ft

R3

∑MD = 0 10R2 = 2000(14) + 400(10)(5) R2 = 4800 lb ∑MH = 0 14R3 + 4(4800) = 400(10)(9) R3 = 1200 lb

–8000 lb⋅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

Shear and Moment Equations and Diagrams; Relation between Load, Shear, and Moment; Moving Loads

212

www.mathalino.com Problem 440.

A frame ABCD, with rigid corners at B and C, supports the concentrated load as shown in Fig. P440. (Draw shear and moment diagrams for each of the three parts of the frame.)

Copyright © 2011 Mathalino.com All rights reserved.

C

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.

L/2

L

D

P Figure P-440

B

A

L/2

Solution 440. Member AB

Member BC P

P PL/2 Load Diagram

0

P

– PL/2

Load Diagram

A

Shear Diagram

L/2

Moment Diagram

B P

PL/2

C

L/2

B

PL/2 P

Shear Diagram

Member CD P C

–PL/2 Moment Diagram

PL/2

L

D

Load Diagram

Shear Diagram –P PL/2 Moment Diagram

–PL/2

Chapter 04 Shear and Moment in Beams Problem 441.

213 www.mathalino.com

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.) 10 kN 4

Figure P-441

F

3

2m 2m

2m

A

B

2m C

D

2m 14 kN E

Solution 441.

FBH = 14 kN to the right MB = 14(2) MB = 28 kN⋅m counterclockwise

10 kN 5

F

4 3

2m 2m

2m

A

B

2m C

D

2m

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

14 kN

∑MD = 0 6RA + 12 + 8(2) = 28 RA = 0

E

8 kN A

2m 14 kN

B 28 kN⋅m

RA = 0

2m 6 kN

2m C 12 kN⋅m RDV = 8 kN

D RDH = 20 kN

∑MA = 0 6RDV + 12 = 28 + 8(4) RDV = 8 kN ∑FH = 0 RDH = 14 + 6 RDH = 20 kN

Shear and Moment Equations and Diagrams; Relation between Load, Shear, and Moment; Moving Loads

214

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8 kN A

2m 14 kN

2m 6 kN

B 28 kN⋅m

2m C 12 kN⋅m

Load Diagram

RA = 0

D RDH = 20 kN 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

RDV = 8 kN

8 kN

Shear Diagram

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

–16 kN⋅m –28 kN⋅m Moment Diagram

Problem 442.

Beam carrying the uniformly varying load shown in Fig. P-442.

Figure P-442 wO

L R1

∑MR2 = 0 LR1 = 13 L ( 21 Lwo)

Solution 442. ½ Lwo 2/3 L

R1 =

1/3 L wo

R2

1 6

Lwo

∑MR1 = 0 LR2 = 23 L ( 21 Lwo) R2 =

L R1

R2

1 3

Lwo

Chapter 04 Shear and Moment in Beams

215 www.mathalino.com 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.

wo A

B

L R1 = 1/6 Lwo R2 = 1/3 Lwo Load Diagram 1/6 Lwo

C x = L / √3 Shear Diagram –1/3 Lwo

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
see figure for solving A1 MB = MC – 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.

2

Mmax = 0.06415L wo

Moment Diagram 1/6 Lwo

L / √3

A1

Figure for solving A1 –1/3 Lwo

Problem 443.

Beam carrying the triangular loads shown in Fig. P443. wO Figure P-443

L/2 R1

L/2 R2

Shear and Moment Equations and Diagrams; Relation between Load, Shear, and Moment; Moving Loads

216

www.mathalino.com Solution 443.

By symmetry: R1 = R2 = 21 ( 12 Lw o )

wO

A

L/2

B

L/2

R1 = ¼ Lwo

R1 = R2 =

C

R2 = ¼ Lwo Load Diagram

1 4

Lw o

Shear Diagram 1 12

−

1 4

Lw o

2

L wo

Moment Diagram

Problem 444.

1 4

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.

Beam loaded as shown in Fig. P-444. Figure P-444

wo

wo

L/2

L/2

R1

Solution 444.

R2

Total load = 2[ 21 (L/2)(wo)] =

1 2

Lwo

By symmetry R1 = R2 = R1 = R2 =

1 2 1 4

× total load Lwo

Chapter 04 Shear and Moment in Beams

wo

wo

A

C

B

L/2

L/2

R1 = ¼ Lwo

R2 = ¼ Lwo Load Diagram

1 4

Lw o

Shear Diagram

1 24

−

1 4

Lw o

2

L wo

217 www.mathalino.com 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.

Moment Diagram

Problem 445.

Beam carrying the loads shown in Fig. P-445. 80 kN/m Figure P-445

20 kN/m

1m

3m

1m

R1

R2

∑MR2 = 0 5R1 = 80(3) + 90(2) R1 = 84 kN

Solution 445. F2 = 90 kN 2m F1 = 80 kN

1m 60 kN/m 20 kN/m

R1

1m

3m

1m

R2

∑MR1 = 0 5R2 = 80(2) + 90(3) R2 = 86 kN Checking R1 + R2 = F1 + F2 ok!

Shear and Moment Equations and Diagrams; Relation between Load, Shear, and Moment; Moving Loads

218

www.mathalino.com

80 kN/m

y 20 kN/m A

B

C 3m

1m R1 = 84 kN

D 1m R2 = 86 kN

Load Diagram 84 kN 64 kN z

E x =1.72 m Shear Diagram

–86 kN

137.5 kN⋅m 74 kN⋅m

3rd deg nd

2

deg

86 kN⋅m 1st deg

Moment Diagram

4m 9.97 73.97

64 1m

A1 1.72 m

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

1.28 m A2 86

Figure for solving A1 and A2

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

Chapter 04 Shear and Moment in Beams Problem 446.

219 www.mathalino.com

Beam loaded and supported as shown in Fig. P-446. 50 kN Figure P-446 1m

4m

50 kN

1m

∑FV = 0 4wo + 2[ 21 wo(1)] = 20(4) + 2(50) 5wo = 180 wo = 36 kN/m

Solution 446.

50 kN

4m 20 kN/m

50 kN

20 kN/m 1m

1m B

C

A

D wo = 36 kN/m Load Diagram 32 kN 18 kN

–18 kN –32 kN Shear Diagram

6 Kn⋅m

6 Kn⋅m

–26 kN⋅m Moment Diagram

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.

Shear and Moment Equations and Diagrams; Relation between Load, Shear, and Moment; Moving Loads

220

www.mathalino.com 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. V (lb) 2

3

2

2

2400

Figure P-447

1000

400

x (ft)

Solution 447. –4000 2000 lb

2 ft

1000 lb

4400 lb

3 ft

2 ft

R1 = 2400 lb

2 ft

R2 = 5000 lb Load Diagram

2400 lb 1000 lb

400 lb A

B

C

D

–4000 lb Given Shear Diagram

6000 lb⋅ft 4800 lb⋅ft

–2000 lb⋅ft Moment Diagram

E

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

Chapter 04 Shear and Moment in Beams Problem 448.

221 www.mathalino.com

Shear diagram as shown in Fig. P-448.

Figure P-448

V (kN)

2

1

1.6

2.4

2

36 16

20 x (m) –24

–40

Solution 448. 20 kN/m

20 kN 10 kN/m

2m 1m 4m 2m R1 = 76 kN R2 = 44 kN Load Diagram 1.6 m

2.4 m

36 kN 16 kN A

20 kN F

B C

D –24 kN

–40 kN Given Shear Diagram

8.8 kN⋅m –4 kN⋅m –20 kN⋅m –40 kN⋅m Moment Diagram

E

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.

Shear and Moment Equations and Diagrams; Relation between Load, Shear, and Moment; Moving Loads

222

www.mathalino.com Problem 449.

Shear diagram as shown in Fig. P-449. 3

2

V (lb)

Figure P-449

3

4

8

3700 1700 x (ft)

–4000

–3100

Solution 449. 4000 lb

A

2000 lb

B

C D

600 lb/ft

E

3850 lb/ft 3 ft

2 ft

3 ft

4 ft

Load Diagram

80/77 ft 3700 lb

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, F thus a uniform upward force is acting at a magnitude of (3700 + 4000)/2 = 3850 3100 lb lb/ft. No load in segment CD. (3) Another point force acting downward with 8 ft 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. 31/6 ft

1700 lb H

G

–4000 lb

74/77 ft

17/6 ft

–3100 lb

Given Shear Diagram

8,008.33 lb⋅ft 5,600 lb⋅ft

–1,200 lb⋅ft

–12,000 lb⋅ft –12,300 lb⋅ft –14,077.92 lb⋅ft Moment Diagram

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

Chapter 04 Shear and Moment in Beams Problem 450.

223 www.mathalino.com

Shear diagram as shown in Fig. P-450. 4

V (lb)

Figure P-450

2

4

4

4

900 480 x (ft)

–900 –1380

Solution 450.

Solution 900 lb

900 lb

A

120 lb/ft

C

B

D

225 lb/ft 4 ft

E

F

1860 lb 2 ft

4 ft

4 ft

4 ft

Load Diagram

900 lb 480 lb G

–900 lb –1380 lb Given Shear Diagram

3600 lb⋅ft 1800 lb⋅ft

–960 lb⋅ft Moment Diagram

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.

Shear and Moment Equations and Diagrams; Relation between Load, Shear, and Moment; Moving Loads

224

www.mathalino.com Problem 451.

Shear diagram as shown in Fig. P-451. V (kN)

Figure P-451

10

2nd-degree curve

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.

x (m) –2 –8 3

1

1

2

Solution 451. 8 kN/m 6 kN

A

B

C

D

1m

1m

4 kN/m

10 kN 3m

2m

Load Diagram 10 kN

F –2 kN x = 2.74 m –8 kN Given Shear Diagram

18.26 kN⋅m

18 kN⋅m 16 kN⋅m 8 kN⋅m

Moment Diagram

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 E 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.

Chapter 04 Shear and Moment in Beams MOVING LOADS 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.

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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). P

Position for maximum shear

Position for maximum moment

P

L/2

L/2 L

Mmax =

PL and Vmax = P 4

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 Pb > Ps

Pb

d

L

Ps

226

Shear and Moment Equations and Diagrams; Relation between Load, Shear, and Moment; Moving Loads

www.mathalino.com

Mmax =

( PL − Ps d ) 2 4 PL

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

R 40 kN

xR = 40(5) x = 200/R x = 200/100 x=2m

60 kN 5–x

5m

x

Chapter 04 Shear and Moment in Beams

227 www.mathalino.com

For maximum moment under 40 kN wheel: ∑MR2 = 0 R = 100 kN 10R1 = 3.5(100) 60 kN R1 = 35 kN 3m 2m

40 kN

CL

R1

3.5 m

1.5 m

3.5 m

R2

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

10 m

R = 100 kN 40 kN 3m

2m

For maximum moment under 60 kN wheel: ∑MR1 = 0 60 kN 10R2 = 4(100) R2 = 40 kN

CL

R1

4m

1m

R2

4m

10 m

Thus, Mmax = 160 kN⋅⋅m R = 100 kN 60 kN

40 kN 3m

R1

8m

MTo the right of 60 kN = 4R2 MTo the right of 60 kN = 4(40) MTo the right of 60 kN = 160 kN⋅m

2m

R2

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

R 30 kN

xR = 4(30) x = 120/R x = 120/80 x = 1.5 m

4–x

4m

x

50 kN

Shear and Moment Equations and Diagrams; Relation between Load, Shear, and Moment; Moving Loads

228

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Maximum moment under 30 kN wheel: ∑MR2 = 0 8R1 = 2.75(80) R1 = 27.5 kN

R = 80 kN 30 kN

1.5 m 50 kN

2.5 m CL

R1

2.75 m

1.25 m

2.75 m

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

R2

8m

Maximum moment under 50 kN wheel: ∑MR1 = 0 8R2 = 3.25(80) R2 = 32.5 kN

R = 80 kN 30 kN

2.5 m

1.5 m 50 kN CL

R1

3.25 m

0.75 m

3.25 m

R2

8m

Thus, Mmax = 105.625 kN⋅⋅m R = 80 kN 2.5 m

30 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

1.5 m 50 kN

The maximum shear will occur when the 50 kN is over a support. ∑MR1 = 0 8R2 = 6.5(80) R2 = 65 kN

6.5 m R1

8m

R2

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

Wr = 1800 lb

R = 3000 lb x

9–x

9 ft

Wf

Chapter 04 Shear and Moment in Beams

229 www.mathalino.com

9 – x = 5.4 ft

Wr = 1800 lb

R = 3000 lb

Wf = 1200 lb 5.4 ft

3.6 ft CL

R1

5.2 ft

1.8 ft

5.2 ft

R2

14 ft

Wr = 1800 lb

7 ft 14 ft

R1

R2

Wr = 1800 lb R = 3000 lb 3.6’

R1

4.3 ft

Wf = 1200 lb

4.3 ft

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

5.4’ CL

2.7 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

R2

14 ft

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

Wr = 1800 lb R = 3000 lb 3.6’ 5.4’

R1

Wf = 1200 lb

10.4 ft 14 ft

R2

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

Shear and Moment Equations and Diagrams; Relation between Load, Shear, and Moment; Moving Loads

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www.mathalino.com 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

x P1 = 4k

P2 = 8k

P3 = 6k R

9’ 18’ xR = 9P2 + (9 + 18)P3 x(18) = 9(8) + (9 + 18)(6)
the resultant R is 13 ft from P1 x = 13 ft

Maximum moment under P1 ∑MR2 = 0 P3 = 6k 44R1 = 15.5R 44R1 = 15.5(18) R1 = 6.34091 kips R1 = 6,340.91 lbs

x = 13’ P1 = 4k

P2 = 8k 9’

18’ R = 18k CL

15.5’

6.5’

15.5’

R1

R2 44’

Maximum moment under P2 ∑MR2 = 0 44R1 = 20R 44R1 = 20(18) R1 = 8.18182 kips R1 = 8,181.82 lbs

x = 13’ P1 = 4k

P2 = 8k 9’

P3 = 6k 18’ R = 18k

CL

20’

2’

R1

20’ R2

44’

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

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

Chapter 04 Shear and Moment in Beams

231 www.mathalino.com Maximum moment under P3 ∑R1 = 0 44R2 = 15R 44R2 = 15(18) R2 = 6.13636 kips R2 = 6,136.36 lbs

x = 13’ P1 = 4k

P2 = 8k

P3 = 6k

9’

18’ R = 18k CL

15’

7’

15’

R1

R2 44’

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

x = 13’ P1 = 4k

P2 = 8k

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 R2

P3 = 6k

9’

18’ R = 18k

31’ R1 44’

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

x 20 kN

10 kN

30 kN R

3m

5m

xR = 3(20) + 8(30) x(60) = 3(20) + 8(30) x=5m

232

Shear and Moment Equations and Diagrams; Relation between Load, Shear, and Moment; Moving Loads

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Maximum moment under 10 kN x=5m 20 kN

10 kN

30 kN

3m

5m R = 60 kN CL

3.5 m

2.5 m

3.5 m

R1

R2 12 m

∑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 x=5m 20 kN

10 kN

30 kN

3m

5m R = 60 kN CL

5m

5m 1m

R1

12 m

∑MR2 = 0 12R1 = 5R 12R1 = 5(60) R1 = 25 kN

R2

Chapter 04 Shear and Moment in Beams

233 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. x=3m 30 kN

20 kN

10 kN 3m

5m R = 50 kN CL

5m R1

5m 1m

R2

12 m

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

234

Shear and Moment Equations and Diagrams; Relation between Load, Shear, and Moment; Moving Loads

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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. x=5m 20 kN

10 kN 3m

30 kN 5m R = 60 kN

9m 12 m R1

R2

∑MR1 = 0 12R2 = 9R 12R2 = 9(60) R2 = 45 kN Thus, Vmax = 45 kN

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