Shear and Moment

Chapter 04 Shear & Moment in

Beams

DEFINITION OF A BEAM A beam is a bar subject to forces or couples that lie in

a plane containing the longitudinal of the bar. According to determinacy, a beam may be determinate or indeterminate.

STATICALLY DETERMINATE BEAMS

Statically determinate beams are those beams in which the reactions of the supports may be determined by the use of the equations of static equilibrium. The beams shown below are examples of statically determinate beams.

P

Load

Cantilever Beam

Simple Beam

P M

Overhanging Beam

P w (N/m)

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

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STATICALLY INDETERMINATE BEAMS

If the number of reactions exerted upon a beam exceeds the number of equations in static equilibrium, the beam is said to be statically indeterminate. In order to solve the reactions of the beam, the static equations must be supplemented by equations based upon the elastic deformations of the beam.

The degree of indeterminacy is taken as the difference

between the umber of reactions to the number of equations in static equilibrium that can be applied. In the case of the propped beam shown, there are three

reactions R1, R2, and M and only two equations (ΣM =

0 and ΣFv = 0) can be applied, thus the beam is indeterminate to the first degree (3 – 2 = 1).

P

Propped Beam

Continuous Beam

P1

w (N/m)

R1 R2

M

Fixed or Restrained Beam

w2 (N/m)

P2

M

w1 (N/m) w2 (N/m)

167

Chapter 04 Shear and Moment in Beams www.mathalino.com

TYPES OF LOADING

Loads applied to the beam may consist of a concentrated load (load applied at a point), uniform load, uniformly varying load, or an applied couple or moment. These loads are shown in the following figures.

SHEAR AND MOMENT DIAGRAM Consider a simple beam shown of length L that

carries a uniform load of w (N/m) throughout its length and is held in equilibrium by reactions R1 and R2. Assume that the beam is cut at point C a distance of x from he left support and the portion of the beam to the right of C be removed. The portion removed must then be replaced by vertical shearing force V

P1 P2

Concentrated Loads

w (N/m)

Uniform Load

w (N/m)

Uniformly Varying Load Applied Couple

M

w (N/m)

L C

x

R1 R2

A B

w (N/m)

C

x

R1

V

M A


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together with a couple M to hold the left portion of the bar in equilibrium under the action of R1 and wx.

The couple M is called the resisting moment or

moment and the force V is called the resisting shear or shear. The sign of V and M are taken to be positive if they have the senses indicated above.

SOLVED PROBLEMS INSTRUCTION: Write shear and moment equations for the beams in

the following problems. In each problem, let x be the distance measured from left end of the beam. Also, draw shear and moment diagrams, specifying values at all change of loading positions and at points of zero shear. Neglect the mass of the beam in each problem.

Problem 403. Beam loaded as shown in Fig. P-403. Solution 403. From the load diagram:

∑MB = 0 5RD + 1(30) = 3(50) RD = 24 kN

∑MD = 0 5RB = 2(50) + 6(30) RB = 56 kN Segment AB: VAB = –30 kN

MAB = –30x kN⋅m

30 kN 50 kN

1 m 3 m 2 m

A B C D

Figure P-403

30 kN

A

x

169


Segment BC: VBC = –30 + 56 VBC = 26 kN

MBC = –30x + 56(x – 1)

MBC = 26x – 56 kN⋅m Segment CD: VCD = –30 + 56 – 50 VCD = –24 kN MCD = –30x + 56(x – 1) – 50(x – 4) MCD = –30x + 56x – 56 – 50x + 200 MCD = –24x + 144 Problem 404. Beam loaded as shown in Fig. P-404.

30 kN

1 m

A B

RB = 56 kN

x

30 kN 50 kN

1 m 3 m

A B C

RB = 56 kN

x

2000 lb

M = 4800 lb⋅ft

3 ft 6 ft 3 ft

A B C

D

Figure P-404

RA RD

To draw the Shear Diagram: (1) In segment AB, the shear is

uniformly distributed over the

segment at a magnitude of –30

kN.

(2) In segment BC, the shear is

uniformly distributed at a

magnitude of 26 kN.

(3) In segment CD, the shear is

uniformly distributed at a

magnitude of –24 kN.

To draw the Moment Diagram: (1) The equation MAB = –30x is

linear, at x = 0, MAB = 0 and at

x = 1 m, MAB = –30 kN⋅m.

(2) MBC = 26x – 56 is also linear.

At x = 1 m, MBC = –30 kN⋅m; at

x = 4 m, MBC = 48 kN⋅m. When

MBC = 0, x = 2.154 m, thus the

moment is zero at 1.154 m

from B.

(3) MCD = –24x + 144 is again

linear. At x = 4 m, MCD = 48

kN⋅m; at x = 6 m, MCD = 0.

30 kN 50 kN

1 m 3 m 2 m

A B C D

LoadDiagram

RB = 56 kN RB = 24 kN

26 kN

–30 kN –24 kN

ShearDiagram

MomentDiagram

–30 kN⋅m

56 kN⋅m

1.154 m


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Solution 404. ∑MA = 0 ∑MD = 0 12RD + 4800 = 3(2000) 12RA = 9(2000) + 4800 RD = 100 lb RA = 1900 lb Segment AB: VAB = 1900 lb

MAB = 1900x lb⋅ft Segment BC: VBC = 1900 – 2000 VBC = –100 lb MBC = 1900x – 2000(x – 3) MBC = 1900x – 2000x + 6000 MBC = –100x + 6000 Segment CD: VCD = 1900 – 2000 VCD = –100 lb MCD = 1900x – 2000(x – 3) – 4800 MCD = 1900x – 2000x + 6000 – 4800 MCD = –100x + 1200

2000 lb

B A

RA = 1900 lb

x

3 ft

6 ft C

2000 lb

B A

RA = 1900 lb x

3 ft

M = 4800 lb⋅ft

x

A

RA = 1900 lb

2000 lb M = 4800 lb⋅ft

3 ft 6 ft 3 ft

A B C

D

RA = 1900 lb RD = 100 lb

Load Diagram

1900 lb

–100 lb

Shear Diagram

Moment Diagram

5700 lb⋅ft

5100 lb⋅ft

300 lb⋅ft

To draw the Shear Diagram: (1) At segment AB, the shear is

uniformly distributed at 1900 lb.

(2) A shear of –100 lb is uniformly

distributed over segments BC

and CD.

To draw the Moment Diagram: (1) MAB = 1900x is linear; at x = 0,

MAB = 0; at x = 3 ft, MAB = 5700

lb⋅ft. (2) For segment BC, MBC = –100x +

6000 is linear; at x = 3 ft, MBC =

5700 lb⋅ft; at x = 9 ft, MBC =

5100 lb⋅ft. (3) MCD = –100x + 1200 is again

linear; at x = 9 ft, MCD = 300

lb⋅ft; at x = 12 ft, MCD = 0.

171


Problem 405. Beam loaded as shown in Fig. P-405.

Solution 405. ∑MA = 0 ∑MC = 0 10RC = 2(80) + 5[10(10)] 10RA = 8(80) + 5[10(10)] RC = 66 kN RA = 114 kN Segment AB: VAB = 114 – 10x kN MAB = 114x – 10x(x/2)

MAB = 114x – 5x2 kN⋅m Segment BC: VBC = 114 – 80 – 10x VBC = 34 – 10x kN MBC = 114x – 80(x – 2) – 10x(x/2) MBC = 160 + 34x – 5x2

80 kN

10 kN/m

2 m 8 m

A

B

C

Figure P-405

RA RC

10 kN/m

x

A

RA = 114 kN

80 kN

B 10 kN/m

A

RA = 114 kN

x

2 m

To draw the Shear Diagram: (1) For segment AB, VAB = 114 – 10x

is linear; at x = 0, VAB = 14 kN; at

x = 2 m, VAB = 94 kN.

(2) VBC = 34 – 10x for segment BC is

linear; at x = 2 m, VBC = 14 kN; at

x = 10 m, VBC = –66 kN. When VBC

= 0, x = 3.4 m thus VBC = 0 at 1.4

m from B.

To draw the Moment Diagram: (1) MAB = 114x – 5x

2 is a second

degree curve for segment AB; at x

= 0, MAB = 0; at x = 2 m, MAB =

208 kN⋅m.

(2) The moment diagram is also a

second degree curve for segment

BC given by MBC = 160 + 34x –

5x2; at x = 2 m, MBC = 208 kN⋅m;

at x = 10 m, MBC = 0.

(3) Note that the maximum moment

occurs at point of zero shear.

Thus, at x = 3.4 m, MBC = 217.8 kN⋅m.

80 kN

10 kN/m

2 m 8 m

A

B C

RA = 114 kN RC = 66 kN

Load Diagram

1.4 m

114 kN

94 kN

14 kN

–66 kN

Shear Diagram

208 kN⋅m

217.8 kN⋅m

Moment Diagram


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Solution 406. ∑MA = 0 12RC = 4(900) + 18(400) + 9[(60)(18)] RC = 1710 lb

∑MC = 0 12RA + 6(400) = 8(900) + 3[60(18)] RA = 670 lb Segment AB: VAB = 670 – 60x lb MAB = 670x – 60x(x/2)

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

MBC = 3600 – 230x – 30x2 lb⋅ft Segment CD: VCD = 670 + 1710 – 900 – 60x VCD = 1480 – 60x lb MCD = 670x + 1710(x – 12) – 900(x – 4) – 60x(x/2)

MCD = –16920 + 1480x – 30x2 lb⋅ft

Figure P-406

900 lb

60 lb/ft

4 ft 8 ft

A

B

D

RA 6 ft

C

RC

400 lb

x

A

RA = 670 lb

60 lb/ft

900 lb

x

A

RA = 670 lb

4 ft 60 lb/ft

B C

RC = 1710 lb

900 lb

A

RA = 670 lb

4 ft

x

8 ft

60 lb/ft

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173



Solution 407. ∑MA = 0 ∑MD = 0 6RD = 4[2(30)] 6RA = 2[2(30)] RD = 40 kN RA = 20 kN Segment AB: VAB = 20 kN

MAB = 20x kN⋅m

Figure P-407 30 kN/m

3 m

A

B

D

RA

2 m C

1 m

RD

x

A

RA = 20 kN

To draw the Shear Diagram: (1) VAB = 670 – 60x for segment AB is

linear; at x = 0, VAB= 670 lb; at x

= 4 ft, VAB = 430 lb.

(2) For segment BC, VBC = –230 – 60x

is also linear; at x= 4 ft, VBC = –

470 lb, at x = 12 ft, VBC = –950 lb.

(3) VCD = 1480 – 60x for segment CD

is again linear; at x = 12, VCD =

760 lb; at x = 18 ft, VCD = 400 lb.


2 for segment AB

is a second degree curve; at x =

0, MAB = 0; at x = 4 ft, MAB = 2200

lb⋅ft. (2) For BC, MBC = 3600 – 230x – 30x

2,

is a second degree curve; at x = 4

ft, MBC = 2200 lb⋅ft, at x = 12 ft, MBC = –3480 lb⋅ft; When MBC = 0,

3600 – 230x – 30x2 = 0, x = –

15.439 ft and 7.772 ft. Take x =

7.772 ft, thus, the moment is zero

at 3.772 ft from B.

(3) For segment CD, MCD = –16920 +

1480x – 30x2 is a second degree

curve; at x = 12 ft, MCD = –3480

lb⋅ft; at x = 18 ft, MCD = 0.

2200 lb⋅ft

3.772 ft

–3480 lb⋅ft

Moment Diagram

900 lb

60 lb/ft

4’ 8’

A

B D

RA = 670 lb

6’ C

RC = 1710 lb

400 lb

Load Diagram

670 lb 430 lb

–470 lb

–950 lb

760 lb

Shear Diagram

400 lb


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Segment BC: VBC = 20 – 30(x – 3) VBC = 110 – 30x kN MBC = 20x – 30(x – 3)(x – 3)/2 MBC = 20x – 15(x – 3)2 Segment CD: VCD = 20 – 30(2) VCD = –40 kN MCD = 20x – 30(2)(x – 4) MCD = 20x – 60(x – 4) Problem 408. Beam loaded as shown in Fig. P-408.

Solution 408. ∑MA = 0 ∑MD = 0 6RD = 1[2(50)] + 5[2(20)] 6RA = 5[2(50)] + 1[2(20)] RD = 50 kN RA = 90 kN

x

30 kN/m

A B

RA = 20 kN

2 m C

3 m

A B

RA = 20 kN

x

3 m

30 kN/m

Figure P-408 20 kN/m

2 m

A

B

D

RA RD

2 m 2 m C

50 kN/m

30 kN/m

3 m

A

B

D

RA = 20 kN

2 m C

1 m

RD = 40 kN

Load Diagram

20 kN

0.67 m –40 kN

Shear Diagram

60 kN⋅m 66.67 kN⋅m

40 kN⋅m Moment Diagram

To draw the Shear Diagram: (1) For segment AB, the shear is

uniformly distributed at 20 kN.

(2) VBC = 110 – 30x for segment BC;

at x = 3 m, VBC = 20 kN; at x = 5

m, VBC = –40 kN. For VBC = 0, x =

3.67 m or 0.67 m from B.

(3) The shear for segment CD is

uniformly distributed at –40 kN.

To draw the Moment Diagram: (1) For AB, MAB = 20x; at x = 0, MAB =

0; at x = 3 m, MAB = 60 kN⋅m.

(2) MBC = 20x – 15(x – 3)2 for

segment BC is second degree

curve; at x = 3 m, MBC = 60 kN⋅m;

at x = 5 m, MBC = 40 kN⋅m. Note

that maximum moment occurred

at zero shear; at x = 3.67 m, MBC

= 66.67 kN⋅m.

(3) MCD = 20x – 60(x – 4) for segment

BC is linear; at x = 5 m, MCD = 40

kN⋅m; at x = 6 m, MCD = 0.

175


Segment AB: VAB = 90 – 50x kN MAB = 90x – 50x(x/2) MAB = 90x – 25x2 Segment BC: VBC = 90 – 50(2) VBC = –10 kN MBC = 90x – 2(50)(x – 1)

MBC = –10x + 100 kN⋅m Segment CD: VCD = 90 – 2(50) – 20(x – 4) VCD = –20x + 70 kN MCD = 90x – 2(50)(x – 1) – 20(x – 4)(x – 4)/2 MCD = 90x – 100(x – 1) – 10(x – 4)2

MCD = –10x2 + 70x – 60 kN⋅m Problem 409. Cantilever beam loaded as shown in Fig. P-409.

x

A

RA = 90 kN

50 kN/m

B

A

50 kN/m

x

2 m

RA = 90 kN

20 kN/m

2 m C B

A

50 kN/m

x

2 m

RA = 90 kN

Figure P-409

L/2

A

B L/2

C

wo

To draw the Shear Diagram: (1) VAB = 90 – 50x is linear; at x = 0, VBC

= 90 kN; at x = 2 m, VBC = –10 kN.

When VAB = 0, x = 1.8 m.

(2) VBC = –10 kN along segment BC.

(3) VCD = –20x + 70 is linear; at x = 4

m, VCD = –10 kN; at x = 6 m, VCD = –

50 kN.


2 is second degree;

at x = 0, MAB = 0; at x = 1.8 m, MAB

= 81 kN⋅m; at x = 2 m, MAB = 80

kN⋅m.

(2) MBC = –10x + 100 is linear; at x = 2

m, MBC = 80 kN⋅m; at x = 4 m, MBC =

60 kN⋅m.

(3) MCD = –10x2 + 70x – 60; at x = 4 m,

MCD = 60 kN⋅m; at x = 6 m, MCD = 0.

20 kN/m

D

RD = 50 kN

2 m C B

A

50 kN/m

2 m

RA = 90 kN

2 m

90 kN

–10 kN

–50 kN

Load Diagram

Shear Diagram

Moment Diagram

81 kN⋅m

1.8 m

80 kN⋅m

60 kN⋅m


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Solution 409. Segment AB: VAB = –wox MAB = –wox(x/2)

MAB = – 21 wox2

Segment BC: VBC = –wo(L/2)

VBC = – 21 woL

MBC = –wo(L/2)(x – L/4)

MBC = – 21 woLx + 8

1 woL2

Problem 410. Cantilever beam carrying the uniformly varying load

shown in Fig. P-410.

Solution 410. x

y =

L

wo

y = xL

wo

Fx = 21 xy

Fx =

x

L

wx o

2

1

x

A

wo

B A

wo

x

L/2

Figure P-410

L

wo

x

y

3

1 x Fx

wo

L – x

L/2

A

B L/2

C

wo

Load Diagram

–2

1woL

Shear Diagram

Moment Diagram

–8

1 woL2

–8

3 woL2

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

= 0; at x = L/2, VAB = –2

1woL.

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

2

1woL.

To draw the Moment Diagram:

(1) MAB = –2

1wox

2 is a second degree curve; at x =

0, MAB = 0; at x = L/2, MAB = – 8

1 woL2.

(2) MBC = –2

1woLx + 8

1 woL2 is a second degree; at

x = L/2, MBC = – 8

1 woL2; at x = L, MBC = –

8

3 woL2.

177


Fx = 2

2x

L

wo

Shear equation:

V = – 2

2x

L

wo

Moment equation:

M = – 31 xFx = –

2

23

1x

L

wx o

M = – 3

6x

L

wo

Problem 411. Cantilever beam carrying a distributed load with

intensity varying from wo at the free end to zero at the wall, as shown in Fig. P-411.

Solution 411. L

w

xL

y o=−

y = )( xLL

wo −

Figure P-411

L

wo

L

wo

Load Diagram

Shear Diagram

–2

1woL

–6

1 woL2

Moment Diagram

2nd degree

3rd degree

To draw the Shear Diagram:

V = –2o

x

2L

w is a second degree curve;

at x = 0, V = 0; at x = L, V = –2

1woL.

To draw the Moment Diagram:

M = –3o

x

6L

w is a third degree curve; at

x = 0, M = 0; at x = L, M =–6

1 woL2.


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F1 = )(21 ywx o −

F1 =

−− )(2

1xL

L

wwx o

o

F1 =

+− xL

wwwx o

oo2

1

F1 = 2

2x

L

wo

F2 = xy =

− )( xLL

wx o

F2 = )( 2xLxL

wo −

Shear equation:

V = –F1 – F2 = – 2

2x

L

wo – )( 2xLxL

wo −

V = – 2

2x

L

wo – xwo + 2xL

wo

V = 2

2x

L

wo – xwo

Moment equation:

M = – 32 xF1 – 2

1 xF2

M = –

2

23

1x

L

wx o –

− )(2

1 2xLxL

wx o

M = – 3

3x

L

wo – 2

2x

wo + 3

2x

L

wo

M = – 2

2x

wo + 3

6x

L

wo


x

wo

y

L – x

2

1x

3

2 x

F1 F2

Figure P-406 800 lb/ft

2 ft 4 ft

A

B

D

RA 2 ft

C

RC

To draw the Shear Diagram:

V = 2o

x

2L

w – xw

o is a concave

upward second degree curve; at x

= 0, V = 0; at x = L, V = –2

1 woL.

To draw the Moment diagram:

M = –2o

x

2

w +

3ox

6L

w is in third

degree; at x = 0, M = 0; at x = L,

M = –3

1 woL2.

L

wo

Load Diagram

–2

1 woL Shear Diagram

2nd degree

–3

1 woL2

Moment Diagram

3rd degree

179


Solution 412. ∑MA = 0 ∑MC = 0 6RC = 5[6(800)] 6RA = 1[6(800)] RC = 4000 lb RA = 800 lb Segment AB: VAB = 800 lb MAB = 800x Segment BC: VBC = 800 – 800(x – 2) VBC = 2400 – 800x MBC = 800x – 800(x – 2)(x – 2)/2 MBC = 800x – 400(x – 2)2 Segment CD: VCD = 800 + 4000 – 800(x – 2) VCD = 4800 – 800x + 1600 VCD = 6400 – 800x MCD = 800x + 4000(x – 6) – 800(x – 2)(x – 2)/2 MCD = 800x + 4000(x – 6) – 400(x – 2)2

x

A

RA = 800 lb

x

800 lb/ft

B A

RA = 800 lb

2 ft

C

RC = 4000 lb

800 lb/ft

B A

RA = 800 lb

2 ft

x

4 ft

To draw the Shear Diagram: (1) 800 lb of shear force is uniformly

distributed along segment AB.

(2) VBC = 2400 – 800x is linear; at x = 2 ft,

VBC = 800 lb; at x = 6 ft, VBC = –2400

lb. When VBC = 0, 2400 – 800x = 0,

thus x = 3 ft or VBC = 0 at 1 ft from B.

(3) VCD = 6400 – 800x is also linear; at x =

6 ft, VCD = 1600 lb; at x = 8 ft, VBC = 0.

To draw the Moment Diagram: (1) MAB = 800x is linear; at x = 0, MAB = 0;

at x = 2 ft, MAB = 1600 lb⋅ft. (2) MBC = 800x – 400(x – 2)

2 is second

degree curve; at x = 2 ft, MBC = 1600

lb⋅ft; at x = 6 ft, MBC = –1600 lb⋅ft; at x = 3 ft, MBC = 2000 lb⋅ft.

(3) MCD = 800x + 4000(x – 6) – 400(x – 2)2

is also a second degree curve; at x = 6

ft, MCD = –1600 lb⋅ft; at x = 8 ft, MCD =

0.

800 lb/ft

2 ft 4 ft

A

B D

RA = 800 lb

2 ft C

RC = 4000 lb

800 lb

–2400 lb

1600 lb

1 ft

2000 lb⋅ft

1600 lb⋅ft

–1600 lb⋅ft

Moment Diagram

Shear Diagram

Load Diagram


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Solution 413. ∑MB = 0 6RE = 1200 + 1[6(100)] RE = 300 lb

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

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

MBC = –50x2 + 300x – 600 lb⋅ft Segment CD: VCD = –100(6) + 300 VCD = –300 lb MCD = –100(6)(x – 3) + 300(x – 2) MCD = –600x + 1800 + 300x – 600

MCD = –300x + 1200 lb⋅ft Segment DE: VDE = –100(6) + 300 VDE = –300 lb MDE = –100(6)(x – 3) + 1200 + 300(x – 2) MDE = –600x + 1800 + 1200 + 300x – 600 MDE = –300x + 2400

Figure P-413 100 lb/ft

2 ft

A

B

E

RB 4 ft

C

1 ft RE

M = 1200 lb⋅ft

1 ft

D

x

A

100 lb/ft

100 lb/ft

2 ft

A

B

RB = 300 lb

x

100 lb/ft

2 ft

A

B

RB = 300 lb

4 ft C

x

1200 lb⋅ft 100 lb/ft

2 ft

A

B

RB = 300 lb

4 ft C

1’ D

x

181


Problem 414. Cantilever beam carrying the load shown in Fig. P-

414. Solution 414. Segment AB: VAB = –2x kN MAB = –2x(x/2)

MAB = –x2 kN⋅m Segment BC:

3

2

2=

−x

y

y = 32 (x – 2)

2 kN/m

x

A

2 kN/m

2 m

A B

2 kN/m

y

3 m

x

F1

F2

x – 2

2 kN/m

2 m 3 m

A B C

4 kN/m Figure P-414

To draw the Shear Diagram: (1) VAB = –100x is linear; at x = 0, VAB =

0; at x = 2 ft, VAB = –200 lb.

(2) VBC = 300 – 100x is also linear; at x

= 2 ft, VBC = 100 lb; at x = 4 ft, VBC

= –300 lb. When VBC = 0, x = 3 ft,

or VBC =0 at 1 ft from B.

(3) The shear is uniformly distributed at

–300 lb along segments CD and DE.

To draw the Moment Diagram: (1) MAB = –50x

2 is a second degree

curve; at x= 0, MAB = 0; at x = ft,

MAB = –200 lb⋅ft. (2) MBC = –50x

2 + 300x – 600 is also

second degree; at x = 2 ft; MBC = –

200 lb⋅ft; at x = 6 ft, MBC = –600

lb⋅ft; at x = 3 ft, MBC = –150 l⋅ft. (3) MCD = –300x + 1200 is linear; at x =

6 ft, MCD = –600 lb⋅ft; at x = 7 ft, MCD = –900 lb⋅ft.

(4) MDE = –300x + 2400 is again linear;

at x = 7 ft, MDE = 300 lb⋅ft; at x = 8 ft, MDE = 0.

100 lb/ft

2 ft

A

B

E

RB = 300 lb

4 ft C

1’

RE = 300 lb

M = 1200 lb⋅ft

1’ D

–300 lb

300 lb⋅ft

100 lb

–200 lb

1 ft

–200 lb⋅ft –150 lb⋅ft

–600 lb⋅ft

–900 lb⋅ft

Load Diagram

Shear Diagram

Moment Diagram


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F1 = 2x

F2 = 21 (x – 2)y

F2 = 21 (x – 2)[ 3

2 (x – 2)]

F2 = 31 (x – 2)2

VBC = –F1 – F2

VBC = – 2x – 31 (x – 2)2

MBC = –(x/2)F1 – 31 (x – 2)F2

MBC = –(x/2)(2x) – 31 (x – 2)[ 3

1 (x – 2)2]

MBC = –x2 – 91 (x – 2)3

Problem 415. Cantilever beam loaded as shown in Fig. P-415. Solution 415. Segment AB: VAB = –20x kN MAB = –20x(x/2)

MAB = –10x2 kN⋅m

3 m 2 m

A B C

Figure P-415

2 m D

20 kN/m

40 kN

A

20 kN/m

x

To draw the Shear Diagram: (1) VAB = –2x is linear; at x = 0, VAB = 0; at x = 2 m, VAB =

–4 kN.

(2) VBC = – 2x – 31 (x – 2)2 is a second degree curve; at x

= 2 m, VBC = –4 kN; at x = 5 m; VBC = –13 kN.

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

2 is a second degree curve; at x = 0, MAB = 0;

at x = 2 m, MAB = –4 kN⋅m.

(2) MBC = –x2 – 9

1 (x – 2)3 is a third degree curve; at x = 2

m, MBC = –4 kN⋅m; at x = 5 m, MBC = –28 kN⋅m.

2 kN/m

2 m 3 m

A B C

4 kN/m

1st degree

2nd degree

2nd degree

3rd degree

Moment Diagram

–4 kN

Shear Diagram

Load Diagram

–13 kN

–4 kN⋅m

–28 kN⋅m

183


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

MAB = –60(x – 1.5) kN⋅m Segment CD: VCD = –20(3) + 40 VCD = –20 kN MCD = –20(3)(x – 1.5) + 40(x – 5) MCD = –60(x – 1.5) + 40(x – 5) Problem 416. Beam carrying

uniformly varying load shown in Fig. P-416.

Solution 416. ∑MR2 = 0

LR1 = 31 LF

R1 = 21

31 ( Lwo)

R1 = 61 Lwo

Figure P-416

L

R1

wo

R2

L

R1

wo

R2

F = ½ Lwo

2/3 L 1/3 L

3 m

A B

20 kN/m

x

3 m 2 m

A B C

20 kN/m

40 kN x

3 m 2 m

A B C

2 m D

20 kN/m

40 kN

–20 kN

–60 kN

–90 kN⋅m

–210 kN⋅m –250 kN⋅m

Load Diagram

Shear Diagram

Moment Diagram

To draw the Shear Diagram

(1) VAB = –20x for segment AB is linear; at

x = 0, V = 0; at x = 3 m, V = –60 kN.

(2) VBC = –60 kN is uniformly distributed

along segment BC.

(3) Shear is uniform along segment CD at

–20 kN.

To draw the Moment Diagram (1) MAB = –10x

2 for segment AB is second

degree curve; at x = 0, MAB = 0; at x =

3 m, MAB = –90 kN⋅m.

(2) MBC = –60(x – 1.5) for segment BC is

linear; at x = 3 m, MBC = –90 kN⋅m; at

x = 5 m, MBC = –210 kN⋅m.

(3) MCD = –60(x – 1.5) + 40(x – 5) for

segment CD is also linear; at x = 5 m,

MCD = –210 kN⋅m, at x = 7 m, MCD = –

250 kN⋅m.


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∑MR1 = 0

LR2 = 32 LF

R2 = 21

32 ( Lwo)

R2 = 31 Lwo

L

w

x

y o=

y = xL

wo

Fx = 21 xy =

x

L

wx o

2

1

Fx = 2

2x

L

wo

V = R1 – Fx

V = 61 Lwo – 2

2x

L

wo

M = R1x – Fx( x31 )

M = 61 Lwox – 2

2x

L

wo ( x31 )

M = 61 Lwox – 3

6x

L

wo

x

R1

wo

Fx = ½ xy

y

2/3 x 1/3 x

To draw the Shear Diagram: V = 1/6 Lwo – wox

2/2L is a second degree curve; at x =

0, V = 1/6 Lwo = R1; at x = L, V = –1/3 Lwo = –R2; If

a is the location of zero shear from left end, 0 = 1/6 Lwo

– wox2/2L, x = 0.5774L = a; to check, use the squared

property of parabola:

a2/R1 = L2/(R1 + R2)

a2/(1/6 Lwo) = L2/(1/6 Lwo + 1/3 Lwo)

a2 = (1/6 L3wo)/(1/2 Lwo) = 1/3 L2

a = 0.5774L a =

To draw the Moment Diagram M = 1/6 Lwox – wox

3/6L is a third degree curve; at x =

0, M = 0; at x = L, M = 0; at x = a = 0.5774L, M =

Mmax

Mmax = 1/6 Lwo(0.5774L) – wo(0.5774L)3/6L

Mmax = 0.0962L2wo – 0.0321L

2wo

Mmax = 0.0641L2wo

L

R1

wo

R2

–R2

Mmax

Load Diagram

Shear Diagram

Moment Diagram

a

R1

185


Problem 417. Beam carrying the triangular loading shown in Fig. P-417.

Solution 417. By symmetry:

R1 = R2 = )( 21

21

oLw = oLw41

2/L

w

x

y o= ; y = xL

wo2

F = xy21 =

x

L

wx o2

2

1

F = 2xL

wo

V = R1 – F

V = oLw41 – 2x

L

wo

M = R1x – F )( 31 x

M = xLwo41 – )( 3

12 xxL

wo

M = xLwo41 – 3

3x

L

wo

L/2 L/2

R1 R2

wo

Figure P-417

R1

L/2

wo

x

y

F

1/3 x

L/2 L/2

R1 R2

wo

Load Diagram

Shear Diagram

o4

1Lw−

o4

1Lw

o

2

12

1wL

Moment Diagram

To draw the Shear Diagram: V = Lwo/4 – wox

2/L is a second degree curve;

at x = 0, V = Lwo/4; at x = L/2, V = 0. The

other half of the diagram can be drawn by the

concept of symmetry.

To draw the Moment Diagram M = Lwox/4 – wox

3/3L is a third degree curve;

at x = 0, M = 0; at x = L/2, M = L2wo/12. The

other half of the diagram can be drawn by the concept of symmetry.


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Problem 418. Cantilever beam loaded as shown in Fig. P-418. Solution 418. Segment AB: VAB = –20 kN

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

MAB = –20x + 80 kN⋅m Problem 419. Beam loaded as shown in Fig. P-419.

20 kN

4 m

A

B

Figure P-418

2 m C

M = 80 kN⋅m

20 kN

x

A 20 kN

4 m

A

B

x

M = 80 kN⋅m

–80 kN⋅m

–40 kN⋅m

20 kN

4 m

A

B

2 m C

M = 80 kN⋅m

–20 kN

Load Diagram

To draw the Shear Diagram: VAB and VBC are equal and constant

at –20 kN.

To draw the Moment Diagram: (1) MAB = –20x is linear; when x = 0,

MAB = 0; when x = 4 m, MAB = –

80 kN⋅m.

(2) MBC = –20x + 80 is also linear;

when x = 4 m, MBC = 0; when x = 6 m, MBC = –60 kN⋅m

Shear Diagram

Moment Diagram

6 ft 3 ft

R1 R2

Figure P-419

A B

270 lb/ft

C

187


Solution 419.

[ ∑MC = 0 ] 9R1 = 5(810) R1 = 450 lb

[ ∑MA = 0 ] 9R2 = 4(810) R2 = 360 lb Segment AB:

6

270=x

y

y = 45x

F = xy21 = )45(2

1 xx

F = 22.5x2 VAB = R1 – F VAB = 450 – 22.5x2 lb

MAB = R1x – F( x31 )

MAB = 450x – 22.5x2( x31 )

MAB = 450x – 7.5x3 lb⋅ft Segment BC: VBC = 450 – 810 VBC = –360 lb MBC = 450x – 810(x – 4) MBC = 450x – 810x + 3240

MBC = 3240 – 360x lb⋅ft

5 ft

6 ft 3 ft

R1 R2

A B

270 lb/ft

4 ft

810 lb

C

6 ft

R1

A

270 lb/ft y

x

1/3 x F

6 ft

x

R1 = 450 lb

A B

270 lb/ft

4 ft 810 lb


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Problem 420. A total distributed load of 30 kips supported by a

uniformly distributed reaction as shown in Fig. P-420. Solution 420.

1080 lb⋅ft

450 lb

6 ft 3 ft

R1 = 450 lb R2 = 360 lb

A B

270 lb/ft

C

Load Diagram

a =

–360 lb

Shear Diagram

Moment Diagram

a = √20

1341.64 lb⋅ft

3rd degree

linear

To draw the Shear Diagram: (1) VAB = 450 – 22.5x2 is a second degree

curve; at x = 0, VAB = 450 lb; at x = 6

ft, VAB = –360 lb.

(2) At x = a, VAB = 0,

450 – 22.5x2 = 0

22.5x2 = 450

x2 = 20

x = √20

To check, use the squared property of

parabola.

a2/450 = 62/(450 + 360)

a2 = 20

a = √20

(3) VBC = –360 lb is constant.

To draw the Moment Diagram: (1) MAB = 450x – 7.5x3 for segment AB is

third degree curve; at x = 0, MAB = 0;

at x = √20, MAB = 1341.64 lb⋅ft; at x =

6 ft, MAB = 1080 lb⋅ft. (2) MBC = 3240 – 360x for segment BC is

linear; at x = 6 ft, MBC = 1080 lb⋅ft; at

x = 9 ft, MBC = 0.

4 ft 12 ft 4 ft

W = 30 kips

Figure P-420

4 ft 12 ft 4 ft

W = 30 kips

r lb/ft

w lb/ft

189


w = 30(1000)/12 w = 2500 lb/ft

∑FV = 0 R = W 20r = 30(1000) r = 1500 lb/ft First segment (from 0 to 4 ft from left): V1 = 1500x M1 = 1500x(x/2) M1 = 750x2 Second segment (from 4 ft to mid-span): V2 = 1500x – 2500(x – 4) V2 = 10000 – 1000x M2 = 1500x(x/2) – 2500(x – 4)(x – 4)/2 M2 = 750x2 – 1250(x – 4)2

r = 1500 lb/ft

x

4 ft

2500 lb/ft

r = 1500 lb/ft

x

12,000 lb⋅ft

30,000 lb⋅ft

4 ft 12 ft 4 ft

25 lb/ft

1500 lb/ft

Load Diagram

–6000 lb

6000 lb

Shear Diagram

6 ft

Moment Diagram

12,000

To draw the Shear Diagram: (1) For the first segment, V1 = 1500x is

linear; at x = 0, V1 = 0; at x = 4 ft, V1 =

6000 lb.

(2) For the second segment, V2 = 10000 –

1000x is also linear; at x = 4 ft, V1 =

6000 lb; at mid-span, x = 10 ft, V1 = 0.

(3) For the next half of the beam, the shear

diagram can be accomplished by the

concept of symmetry.

To draw the Moment Diagram: (1) For the first segment, M1 = 750x2 is a

second degree curve, an open upward

parabola; at x = 0, M1 = 0; at x = 4 ft,

M1 = 12000 lb⋅ft. (2) For the second segment, M2 = 750x2 –

1250(x – 4)2 is a second degree curve,

an downward parabola; at x = 4 ft, M2 =

12000 lb⋅ft; at mid-span, x = 10 ft, M2 =

30000 lb⋅ft. (2) The next half of the diagram, from x =

10 ft to x = 20 ft, can be drawn by using

the concept of symmetry.


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Problem 421. Write the shear and moment equations as functions of

the angle θ for the built-in arch shown in Fig. P-421.

Solution 421. For θ that is less than 90° Components of Q and P:

Qx = Q sin θ

Qy = Q cos θ

Px = P sin (90° – θ)

Px = P (sin 90° cos θ – cos 90° sin θ)

Px = P cos θ

Py = P cos (90° – θ)

Py = P (cos 90° cos θ + sin 90° sin θ)

Py = P sin θ Shear:

V = ∑Fy V = Qy – Py

V = Q cos θθθθ – P sin θθθθ Moment arms:

dQ = R sin θ

dP = R – R cos θ

dP = R (1 – cos θ) Moment:

M = ∑Mcounterclockwise – ∑Mclockwise M = Q(dQ) – P(dP)

M = QR sin θθθθ – PR(1 – cos θθθθ)

θ

R

Figure P-421 P

Q

B A

θ

P

Q

V

θ

90° – θ

dQ R

dP

191


For θ that is greater than 90° Components of Q and P:

Qx = Q sin (180° – θ)

Qx = Q (sin 180° cos θ – cos 180° sin θ)

Qx = Q cos θ

Qy = Q cos (180° – θ)

Qy = Q (cos 180° cos θ + sin 180° sin θ)

Qy = –Q sin θ

Px = P sin (θ – 90°)

Px = P (sin θ cos 90° – cos θ sin 90°)

Px = –P cos θ

Py = P cos (θ – 90°)

Py = P (cos θ cos 90° + sin θ sin 90°)

Py = P sin θ Shear:

V = ∑Fy V = –Qy – Py

V = –(–Q sin θ) – P sin θ

V = Q sin θθθθ – P sin θθθθ Moment arms:

dQ = R sin (180° – θ)

dQ = R (sin 180° cos θ – cos 180° sin θ)

dQ = R sin θ

dP = R + R cos (180° – θ)

dP = R + R (cos 180° cos θ + sin 180° sin θ)

dP = R – R cos θ

dP = R(1 – cos θ) Moment:

M = ∑Mcounterclockwise – ∑Mclockwise M = Q(dQ) – P(dP)

M = QR sin θθθθ – PR(1 – cos θθθθ)

θ

P

Q

V

dQ R

dP

180° – θ

180° – θ

θ – 90°


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Problem 422. Write the shear and moment equations for the semicircular arch as shown in Fig. P-422 if (a) the load P is vertical as shown, and (b) the load is applied horizontally to the left at the top of the arch.

Solution 422. ∑MC = 0 2R(RA) = RP

RA = 21 P

For θ that is less than 90° Shear:

VAB = RA cos (90° – θ)

VAB = 21 P (cos 90° cos θ + sin 90° sin θ)

VAB = 21 P sin θθθθ

Moment arm:

d = R – R cos θ

d = R(1 – cos θ) Moment: MAB = Ra (d)

MAB = 21 PR(1 – cos θθθθ)

θ

R

Figure P-422

P

C

O

A

B

θ

P

C

O

A

B

RA

R

θ

O

A

RA

R

V

d

90° – θ

193


For θ that is greater than 90° Components of P and RA:

Px = P sin (θ – 90°)

Px = P (sin θ cos 90° – cos θ sin 90°)

Px = –P cos θ

Py = P cos (θ – 90°)

Py = P (cos θ cos 90° + sin θ sin 90°)

Py = P sin θ

RAx = RA sin (θ – 90°)

RAx = 21 P (sin θ cos 90° – cos θ sin 90°)

RAx = – 21 P cos θ

RAy = RA cos (θ – 90°)

RAy = 21 P (cos θ cos 90° + sin θ sin 90°)

RAy = 21 P sin θ

Shear:

VBC = ∑Fy VBC = RAy – Py

VBC = 21 P sin θ – P sin θ

VBC = – 21 P sin θθθθ

Moment arm:

d = R cos (180° – θ)

d = R (cos 180° cos θ + sin 180° sin θ)

d = –R cos θ Moment:

MBC = ∑Mcounterclockwise – ∑Mclockwise

MBC = RA(R + d) – Pd

MBC = 21 P(R – R cos θ) – P(–R cos θ)

MBC = 21 PR – 2

1 PR cos θ + PR cos θ

MBC = 21 PR + 2

1 PR cos θ

MBC = 21 PR(1 + cos θθθθ)

θ

P

O

A

B

RA

R

180° – θ

V

d

θ – 90°

θ – 90°

R


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RELATIONSHIP BETWEEN LOAD, SHEAR, AND MOMENT The vertical shear at C in the figure shown in

previous section (Shear and Moment Diagram) is taken as

VC = (ΣFv)L = R1 – wx where R1 = R2 = wL/2

VC = 2

wL – wx

The moment at C is

MC = (ΣMC) = xwL

2 –

2

xwx

MC = 2

wLx –

2

2wx

If we differentiate M with respect to x:

dx

dM =

2

wL

dx

dx –

dx

dxx

w2

2

dx

dM =

2

wL – wx = shear

thus,

dx

dM = V

Thus, the rate of change of the bending moment with

respect to x is equal to the shearing force, or the slope of the moment diagram at the given point is the shear at that point.

195


Differentiate V with respect to x gives

dx

dV = 0 – w = load

dx

dV = Load

Thus, the rate of change of the shearing force with

respect to x is equal to the load or the slope of the shear diagram at a given point equals the load at that point.

PROPERTIES OF SHEAR AND MOMENT DIAGRAMS

The following are some important properties of shear and moment diagrams:

1. The area of the shear diagram to the left or to the

right of the section is equal to the moment at that section.

2. The slope of the moment diagram at a given point is the shear at that point.

3. The slope of the shear diagram at a given point equals the load at that point.

4. The maximum moment occurs at the point of zero shears. This is in reference to property number 2, that when the shear (also the slope of the moment diagram) is zero, the tangent drawn to the moment diagram is horizontal.

5. When the shear diagram is increasing, the moment diagram is concave upward.

6. When the shear diagram is decreasing, the moment diagram is concave downward.

SIGN CONVENTIONS

The customary sign conventions for shearing force and bending moment are represented by the figures below. A force that tends to bend the beam


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downward is said to produce a positive bending moment. A force that tends to shear the left portion of the beam upward with respect to the right portion is said to produce a positive shearing force.

An easier way of determining the sign of the bending

moment at any section is that upward forces always cause positive bending moments regardless of whether they act to the left or to the right of the exploratory section.

SOLVED PROBLEMS INSTRUCTION: Without writing shear and moment equations, draw

the shear and moment diagrams for the beams specified in the following problems. Give numerical values at all change of loading positions and at all points of zero shear. (Note to instructor: Problems 403 to 420 may also be assigned for solution by semi graphical method describes in this article.)


Positive Bending Negative Bending

Positive Shear Negative Shear

R1 R2

60 kN 30 kN

2 m 4 m 1 m

Figure P-425

197


Solution 425. ∑MA = 0 6R2 = 2(60) + 7(30) R2 = 55 kN

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

35 kN

–25 kN

30 kN

Shear Diagram

70 kN⋅m

–30 kN⋅m

Moment Diagram

R1 = 35 kN R2 = 55 kN

60 kN 30 kN

2 m 4 m 1 m

Load Diagram

A

B

C

D

To draw the Shear Diagram: (1) VA = R1 = 35 kN

(2) VB = VA + Area in load diagram – 60 kN

VB = 35 + 0 – 60 = –25 kN

(3) VC = VB + area in load diagram + R2

VC = –25 + 0 + 55 = 30 kN

(4) VD = VC + Area in load diagram – 30 kN

VD = 30 + 0 – 30 = 0

To draw the Moment Diagram: (1) MA = 0

(2) MB = MA + Area in shear diagram

MB = 0 + 35(2) = 70 kN⋅m

(3) MC = MB + Area in shear diagram

MC = 70 – 25(4) = –30 kN⋅m

(4) MD = MC + Area in shear diagram MD = –30 + 30(1) = 0

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Problem 426. Cantilever beam acted upon by a uniformly distributed load and a couple as shown in Fig. P-426.

Solution 426.


1 m

Figure P-426

2 m 2 m

5 kN/m

M = 60 kN⋅m


(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


MB = 0 – ½ (2)(10)

MB = –10 kN⋅m


MC = –10 – 10(2)

MC = –30 kN⋅m

MC2 = –30 + M = –30 + 60 = 30 kN⋅m

(4) MD = MC2 + Area in shear diagram

MD = 30 – 10(1)

MD = 20 kN⋅m –10 kN⋅m

–30 kN⋅m

30 kN⋅m

20 kN⋅m

Moment Diagram

2nd deg

1st deg

Shear Diagram

–10 kN

1 m 2 m 2 m

5 kN/m

M = 60 kN⋅m

Load Diagram

A

B C D

Figure P-427

9 ft

100 lb/ft

3 ft

800 lb

R1 R2

199


Solution 427. ∑MC = 0 12R1 = 100(12)(6) + 800(3) R1 = 800 lb

∑MA = 0 12R2 = 100(12)(6) + 800(9) R2 = 1200 lb Problem 428. Beam loaded as shown in Fig. P-428.

Figure P-428

10 kN/m

1 m 1 m 3 m 2 m

R1 R2

25 kN⋅m


(1) VA = R1 = 800 lb


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


(1) MA = 0

(2) Mx = MA + Area in shear diagram

Mx = 0 + ½ (8)(800) = 3200 lb⋅ft (3) MB = Mx + Area in shear diagram

MB = 3200 – ½ (1)(100) = 3150 lb⋅ft (4) MC = MB + Area in shear diagram

MC = 3150 – ½ (900 + 1200)(3) = 0

(5) The moment curve BC is downward

parabola with vertex at A’. A’ is the

location of zero shear for segment BC.

9 ft

100 lb/ft

3 ft

800 lb

R1 = 800 lb R2 = 1200 lb

A B C

Load Diagram

x = 8 ft

800 lb

–100 lb

–900 lb

–1200 lb Shear Diagram

3200 lb⋅ft 3150 lb⋅ft

Moment Diagram

A’


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Solution 428. ∑MD = 0 5R1 = 50(0.5) + 25 R1 = 10 kN

∑MA = 0 5R2 + 25 = 50(4.5) R2 = 40 kN Problem 429. Beam loaded as shown in Fig. P-429.


(1) VA = R1 = 10 kN


VB = 10 + 0 = 10 kN


VC = 10 + 0 = 10 kN


VD = 10 – 10(3) = –20 kN

VD2 = –20 + R2 = 20 kN

(5) VE = VD2 + Area in load diagram

VE = 20 – 10(2) = 0

(6) Solving for x:

x / 10 = (3 – x) / 20

20x = 30 – 10x

x = 1 m


(1) MA = 0


MB = 0 + 1(10) = 10 kN⋅m

MB2 = 10 – 25 = –15 kN⋅m

(3) MC = MB2 + Area in shear diagram

MC = –15 + 1(10) = –5 kN⋅m

(4) Mx = MC + Area in shear diagram

Mx = –5 + ½ (1)(10) = 0

(5) MD = Mx + Area in shear diagram

MD = 0 – ½ (2)(20) = –20 kN⋅m

(6) ME = MD + Area in shear diagram

ME = –20 + ½ (2)(20) = 0

1 m 1 m

B

10 kN/m

3 m 2 m

R1 = 10 kN R2 = 40 kN

A

C D E

50 kN

0.5 m

25 kN⋅m

Load Diagram

–20 kN⋅m

–5 kN⋅m

Moment Diagram

–15 kN⋅m

10 kN⋅m

Shear Diagram

10 kN

–20 kN

20 kN

x = 1 m

100 lb

2 ft 2 ft 2 ft

R1 R2

120 lb/ft 120 lb/ft

Figure P-429



201


Solution 429. ∑MC = 0 4R1 + 120(2)(1) = 100(2) + 120(2)(3) R1 = 170 lb

∑MA = 0 4R2 = 120(2)(1) + 100(2) + 120(2)(5) R2 = 410 lb Problem 430. Beam loaded as shown in P-430.

1000 lb

5 ft R1 R2

400 lb/ft 200 lb/ft

Figure P-430

2000 lb

10 ft 10 ft


(1) VA = R1 = 170 lb


VB = 170 – 120(2) = –70 lb

VB2 = –70 – 100 = –170 lb


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


(1) MA = 0

(2) Mx = MA + Area in shear diagram

Mx = 0 + ½ (17/12)(170)

Mx = 1445/12 = 120.42 lb⋅ft (3) MB = Mx + Area in shear diagram

MB = 1445/12 – ½ (2 – 17/12)(70)

MB = 100 lb⋅ft (4) MC = MB + Area in shear diagram

MC = 100 – 170(2) = –240 lb⋅ft (5) MD = MC + Area in shear diagram

MD = –240 + ½ (2)(240) = 0

–240 lb⋅ft

100 lb⋅ft

120.42 lb⋅ft

100 lb

2 ft 2 ft 2 ft

R1 = 170 lb R2 = 410 lb

120 lb/ft 120 lb/ft

A B C D

Load Diagram

–70 lb

170 lb

–170 lb

240 lb

Shear Diagram

x = 1.42 ft

Moment Diagram


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Solution 430. ∑MD = 0 20R1 = 1000(25) + 400(5)(22.5) + 2000(10) + 200(10)(5) R1 = 5000 lb

∑MB = 0 20R2 + 1000(5) + 400(5)(2.5) = 2000(10) + 200(10)(15) R2 = 2000 lb Problem 431. Beam loaded as shown in Fig. P-431.

1000 lb

5 ft R1 = 5000 lb R2 = 2000 lb

400 lb/ft 200 lb/ft

2000 lb

10 ft 10 ft

A

B C

D

Load Diagram

–10000 lb⋅ft

10000 lb⋅ft

5 ft

Moment Diagram

To draw the Shear Diagram (1) VA = –1000 lb


VB = –1000 – 400(5) = –3000 lb

VB2 = –3000 + R1 = 2000 lb


VC = 2000 + 0 = 2000 lb

VC2 = 2000 – 2000 = 0


VD = 0 + 200(10) = 2000 lb


(1) MA = 0


MB = 0 – ½ (1000 + 3000)(5)

MB = –10000 lb⋅ft (3) MC = MB + Area in shear diagram

MC = –10000 + 2000(10) = 10000 lb⋅ft (4) MD = MC + Area in shear diagram

MD = 10000 – ½ (10)(2000) = 0

(5) For segment BC, the location of zero

moment can be accomplished by

symmetry and that is 5 ft from B.

(6) The moment curve AB is a downward

parabola with vertex at A’. A’ is the

location of zero shear for segment AB at point outside the beam.

–3000 lb –1000 lb

2000 lb

–2000 lb

Shear Diagram

A’

A’

Figure P-431

7 m 3 m

10 kN/m

40 kN 50 kN 20 kN/m

1 m 2 m

R1 R2

203


Solution 431. ∑MD = 0 7R1 + 40(3) = 5(50) + 10(10)(2) + 20(4)(2) R1 = 70 kN

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


(1) VA = R1 = 70 kN


VB = 70 – 10(2) = 50 kN

VB2 = 50 – 50 = 0


VC = 0 – 10(1) = –10 kN


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


(1) MA = 0


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


ME = –165 + ½ (70 + 40)(3) = 0

(6) Moment curves AB, CD and DE are

downward parabolas with vertices at

A’, B’ and C’, respectively. A’, B’ and

C’ are corresponding zero shear

points of segments AB, CD and DE.

115 kN⋅m 120 kN⋅m

5 m 3 m

10 kN/m

40 kN 50 kN

20 kN/m 1 m 2 m

R1 = 70 kN R2 = 200 lb

2 m

Load Diagram

A

B C

D E

50 kN

–10 kN

70 kN

–130 kN

70 kN 40 kN

Shear Diagram

4 m

–165 kN⋅m

Moment Diagram

B’

A’

C’

x

y

a = 1/3 m

(7) Solving for point of zero moment:

a / 10 = (a + 4) / 130

130a = 10a + 40

a = 1/3 m

y / (x + a) = 130 / (4 + a)

y = 130(x + 1/3) / (4 + 1/3)

y = 30x + 10

MC = 115 kN⋅m

Mzero = MC + Area in shear

0 = 115 – ½ (10 + y)x

(10 + y)x = 230

(10 + 30x + 10)x = 230

30x2 + 20x – 230 = 0

3x2 + 2x – 23 = 0

x = 2.46 m

zero moment is at 2.46 m from C

Another way to solve the

location of zero moment

is by the squared

property of parabola (see

Problem 434). This point

is the appropriate location

for construction joint of

concrete structures.


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Solution 432. ∑ME = 0 5R1 + 120 = 6(60) + 40(3)(3.5) R1 = 132 kN

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

Figure P-432

R2

1 m 3 m 1 m 1 m

R1

60 kN 40 kN/m

M = 120 kN⋅m

R2 = 48 kN

1 m 3 m 1 m 1 m

R1 = 132 kN

60 kN 40 kN/m

M = 120 kN⋅m

Load Diagram

A B C

D

E

–60 kN

72 kN

–48 kN

Shear Diagram

x = 1.8 m


(1) VA = –60 kN


VB = –60 + 0 = –60 kN

VB2 = VB + R1 = –60 + 132 = 72 kN


VC = 72 – 3(40) = –48 kN


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


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


MD = –24 – ½ (24 + 72)(1) = –72 kN⋅m

MD2 = –72 + 120 = 48 kN⋅m

(6) ME = MD2 + Area in shear diagram

ME = 48 – 48(1) = 0

(7) The location of zero moment on

segment BC can be determined using

the squared property of parabola. See the solution of Problem 434.

–60 kN⋅m

4.8 kN⋅m

–24 kN⋅m

–72 kN⋅m

48 kN⋅m

Moment Diagram

205


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

Solution 433. ∑MC = 0 5R1 + 2(750) = 3000 R1 = 300 lb

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

Figure P-433

R2

2 ft

R1

750 lb

3000 lb⋅ft

3 ft 2 ft

R2 = 450 lb

2 ft

R1 = 300 lb

750 lb

3000 lb⋅ft

3 ft 2 ft

B C D A

Load Diagram

600 lb⋅ft

–2400 lb⋅ft

–1500 lb⋅ft

Moment Diagram

750 lb

300 lb

Shear Diagram


(1) VA = R1 = 300 lb


VB = 300 + 0 = 300 lb


VC = 300 + 0 = 300 lb

VC2 = VC + R2 = 300 + 450 = 750 lb


VD = 750 + 0 = 750

VD2 = VD – 750 = 750 – 750 = 0


(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


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Solution 434. ∑ME = 0 6R1 + 120 = 20(4)(6) + 60(4) R1 = 100 kN

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

Figure P-434

R2

2 m

R1

60 kN

20 kN/m

M = 120 kN⋅m

2 m 2 m 2 m


(1) VA = 0


VB = 0 – 20(2) = –40 kN

VB2 = VB + R1 = –40 + 100 = 60 kN


VC = 60 – 20(2) = 20 kN

VC2 = VC – 60 = 20 – 60 = –40 kN


VD = –40 + 0 = –40 kN

(5) VE = VD + Area in load diagram

VE = –40 + 0 = –40 kN

VE2 = VE + R2 = –40 + 40 = 0


(1) MA = 0


MB = 0 – ½ (40)(2) = –40 kN⋅m


MC = –40 + ½ (60 + 20)(2) = 40 kN⋅m


MD = 40 – 40(2) = –40 kN⋅m

MD2 = MD + M = –40 + 120 = 80 kN⋅m

(5) ME = MD2 + Area in shear diagram

ME = 80 – 40(2) = 0

(6) Moment curve BC is a downward parabola

with vertex at C’. C’ is the location of zero

shear for segment BC.

(7) Location of zero moment at segment BC:

By squared property of parabola:

(3 – x)2 / 50 = 32 / (50 + 40)

3 – x = 2.236

x = 0.764 m from B

–40 kN⋅m –40 kN⋅m

80 kN⋅m

Moment Diagram

C’

R2 = 40 kN

2 m

R1 = 100 kN

60 kN

20 kN/m

M = 120 kN⋅m

2 m 2 m 2 m

A B C D E

Load Diagram

–40 kN

60 kN

20 kN

–40 kN

Shear Diagram

1 m

50 kN⋅m 40 kN⋅m

3 – x

x

1 m

207


Problem 435. Beam loaded and supported as shown in Fig. P-435.

Solution 435. ∑MB = 0 2wo (5) = 10(4)(0) + 20(2) + 40(3) wo = 16 kN/m

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

R1 = 68 kN

Figure P-435

2 m

R1

20 kN

10 kN/m 40 kN

2 m 1 m wo

1 m 2 m

64 kN⋅m

–32 kN

–20 kN

48 kN

28 kN

8 kN

2 m

R1 = 68 kN

20 kN

10 kN/m 40 kN

2 m 1 m wo = 16 kN/m

1 m 2 m

A B C D E

F

Load Diagram

Shear Diagram

–20 kN⋅m

56 kN⋅m

32 kN⋅m

C’

Moment Diagram


(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


(1) MA = 0


MB = 0 – ½ (20)(2) = –20 kN⋅m


MC = –20 + ½ (48 + 28)(2)

MC = 56 kN⋅m


MD = 56 + 8(1) = 64 kN⋅m


ME = 64 – 32(1) = 32 kN⋅m

(6) MF = ME + Area in shear diagram

MF = 32 – ½ (32)(2) = 0

(7) The location and magnitude of moment

at C’ are determined from shear

diagram. By squared property of

parabola, x = 0.44 m from B.

x

2.8 m

95.2 kN⋅m


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Problem 436. A distributed load is supported by two distributed reactions as shown in Fig. P-436.

Solution 436. ∑Mmidpoint of CD = 0 4w1 (11) = 440(8)(5) w1 = 400 lb/ft

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

R2 = w2 lb/ft

Figure P-436

R1 = w1 lb/ft

8 ft

4 ft 2 ft

440 lb/ft

R2 = 960 lb/ft R1 = 400 lb/ft

8 ft

4 ft 2 ft

440 lb/ft

A B C

D

Load Diagram

1600 lb

–1920 lb Shear Diagram


(1) VA = 0


VB = 0 + 400(4) = 1600 lb


VC = 1600 – 440(8) = –1920 lb


VD = –1920 + 960(2) = 0

(5) Location of zero shear:

x / 1600 = (8 – x) / 1920

x = 40/11 ft = 3.636 ft from B


(1) MA = 0


MB = 0 + ½ (1600)(4) = 3200 lb⋅ft (3) Mx = MB + Area in shear diagram

Mx = 3200 + ½ (1600)(40/11)

Mx = 6109.1 lb⋅ft (4) MC = Mx + Area in shear diagram

MC = 6109.1 – ½ (8 – 40/11)(1920)

MC = 1920 lb⋅ft (5) MD = MC + Area in shear diagram

MD = 1920 – ½ (1920)(2) = 0

x = 3.636 ft

Moment Diagram

3200 lb⋅ft

6109.1 lb⋅ft

1920 lb⋅ft

209


Problem 437. Cantilever beam loaded as shown in Fig. P-437 Solution 437.

4 ft 2 ft 2 ft

1000 lb

500 lb

400 lb/ft

Figure P-437

To draw the Shear Diagram (1) VA = –1000 lb

VB = VA + Area in load diagram

VB = –1000 + 0 = –1000 lb

VB2 = VB + 500 = –1000 + 500

VB2 = –500 lb


VC = –500 + 0 = –500 lb


VD = –500 – 400(4) = –2100 lb



MB = 0 – 1000(2) = –2000 lb⋅ft (3) MC = MB + Area in shear diagram

MC = –2000 – 500(2) = –3000 lb⋅ft (4) MD = MC + Area in shear diagram

MD = –3000 – ½ (500 + 2100)(4) MD = –8200 lb⋅ft

4 ft 2 ft 2 ft

1000 lb

500 lb

400 lb/ft

A B C D

Load Diagram

–3000 lb⋅ft

–2000 lb⋅ft

Moment Diagram

–8200 lb⋅ft

Shear Diagram

–1000 lb

–500 lb

–2100 lb


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Problem 438. The beam loaded as shown in Fig. P-438 consists of two segments joined by a frictionless hinge at which the bending moment is zero.

Solution 438.

∑MH = 0 4R1 = 200(6)(3) R1 = 900 lb

Figure P-438

4 ft 2 ft 2 ft

R1

200 lb/ft

Hinge

4 ft 2 ft

R1

200 lb/ft

A B H


(1) VA = 0


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


x / 500 = (4 – x) / 300

300x = 2000 – 500x

x = 2.5 ft


(1) MA = 0


MB = 0 – ½ (400)(2) = –400 lb⋅ft (3) Mx = MB + Area in load diagram

Mx = –400 + ½ (500)(2.5)

Mx = 225 lb⋅ft (4) MH = Mx + Area in load diagram

MH = 225 – ½ (300)(4 – 2.5) = 0 ok!

(5) MC = MH + Area in load diagram

MC = 0 – ½ (300 + 700)(2)

MC = –1000 lb⋅ft (6) The location of zero moment in segment

BH can easily be found by symmetry.

–400 lb

–300 lb

–700 lb

500 lb

4 ft 2 ft 2 ft

R1 = 900 lb

200 lb/ft

A B H C

Load Diagram

Moment Diagram

x = 2.5 ft

–400 lb⋅ft

225 lb⋅ft

–1000 lb⋅ft

Shear Diagram

1.0 ft

211


Problem 439. A beam supported on three reactions as shown in Fig. P-439 consists of two segments joined by frictionless hinge at which the bending moment is zero.

Solution 439.

∑MH = 0 ∑MA = 0 8R1 = 4000(4) 8VH = 4000(4) R1 = 2000 lb VH = 2000 lb

∑MD = 0 10R2 = 2000(14) + 400(10)(5) R2 = 4800 lb

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

400 lb/ft

Hinge

4000 lb

R1 R2 R3 4 ft 4 ft 4 ft 10 ft

Figure P-439

400 lb/ft

Hinge

R2 R3 4 ft 10 ft

C D H

VH = 2000 lb

Hinge

4000 lb

R1 4 ft 4 ft

VH

A B H

400 lb/ft

Hinge

4000 lb

R1 = 2000 lb R2 = 4800 lb R3 = 1200 lb

4 ft 4 ft 4 ft 10 ft

A B C D H

Load Diagram

8000 lb ft⋅

–8000 lb⋅ft

1800 lb⋅ft

3 ft

Moment Diagram

2000 lb

–2000 lb

2800 lb

–1200 lb

Shear Diagram

x = 7 ft


(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


x / 2800 = (10 – x) / 1200

1200x = 28000 – 2800x

x = 7 ft


(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


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Problem 440. A frame ABCD, with rigid corners at B and C, supports the concentrated load as shown in Fig. P-440. (Draw shear and moment diagrams for each of the three parts of the frame.)

Solution 440.

Member BC

B

L/2

P

PL/2

Lo

ad

Dia

gra

m

P

C PL/2

0

Sh

ea

r Dia

gra

m

Mo

me

nt D

iag

ram

– P

L/2

A

P

B

C D

L/2

L/2

L

Figure P-440

Member AB

A

P

B L/2

P

PL/2

P

–PL/2

Load Diagram

Shear Diagram

Moment Diagram

Member CD

PL/2

P

D C

Load Diagram

L

PL/2

–PL/2

Moment Diagram

Shear Diagram

–P



213


Problem 441. A beam ABCD is supported by a roller at A and a hinge at D. It is subjected to the loads shown in Fig. P-441, which act at the ends of the vertical members BE and CF. These vertical members are rigidly attached to the beam at B and C. (Draw shear and moment diagrams for the beam ABCD only.)

Solution 441. FBH = 14 kN to the right MB = 14(2)

MB = 28 kN⋅m counterclockwise FCH = 3/5 (10) FCH = 6 kN to the right FCV = 4/5 (10) FCV = 8 kN upward

MC = FCH (2) = 6(2)

MC = 12 kN⋅m clockwise

∑MD = 0 6RA + 12 + 8(2) = 28 RA = 0

∑MA = 0 6RDV + 12 = 28 + 8(4) RDV = 8 kN

∑FH = 0 RDH = 14 + 6 RDH = 20 kN

2 m 2 m 2 m

2 m

2 m

E 14 kN

10 kN

A B C D

F Figure P-441 3

4

2 m 2 m 2 m

2 m

2 m

E 14 kN

10 kN

A B C D

F 3

4 5

2 m 2 m 2 m

A B C D

28 kN⋅m 12 kN⋅m

14 kN 6 kN

8 kN

RA = 0 RDV = 8 kN

RDH = 20 kN


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Problem 442. Beam carrying the uniformly varying load shown in

Fig. P-442.

Solution 442. ∑MR2 = 0

LR1 = 31 L ( 2

1 Lwo)

R1 = 61 Lwo

∑MR1 = 0

LR2 = 32 L ( 2

1 Lwo)

R2 = 31 Lwo

2 m 2 m 2 m

A B C D

28 kN⋅m 12 kN⋅m

14 kN 6 kN

8 kN

RA = 0 RDV = 8 kN

RDH = 20 kN

Load Diagram

8 kN

Shear Diagram

Moment Diagram

–28 kN⋅m

–16 kN⋅m


(1) Shear in segments AB and BC is

zero.

(2) VC = 8


VD = 8 + 0 = 8 kN

VD2 = VD – RDV

VD2 = 8 – 8 = 0


(1) Moment in segment AB is zero

(2) MB = –28 kN⋅m


MC = –28 + 0 = –28 kN⋅m

MC2 = MC + 12 = –28 + 12

MC2 = –16 kN⋅m

(4) MD = MC2 + Area in shear diagram

MD = –16 + 8(2)

MD = 0

R1 R2

L

wO wO

Figure P-442

R1 R2

L

wo

½ Lwo

2/3 L 1/3 L

215


Problem 443. Beam carrying the triangular loads shown in Fig. P-

443.


(1) VA = R1 = 1/6 Lwo


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.


(2) MC = MA + Area in shear diagram

MC = 0 + 2/3 (L/√3)(1/6 Lwo)

MC = 0.06415L2wo = Mmax

(3) MB = MC + Area in shear diagram

MB = MC – A1 � see figure for solving A1

For A1:

A1 = 1/3 L(1/6 Lwo + 1/3 Lwo)

– 1/3 (L/√3)(1/6 Lwo)

– 1/6 Lwo (L – L/√3)

A1 = 0.16667L2wo – 0.03208L2wo

– 0.07044L2wo

A1 = 0.06415L2wo

MB = 0.06415L2wo – 0.06415L2wo = 0

(4) The shear diagram is second degree curve,

thus the moment diagram is a third degree

curve. The maximum moment (highest point)

occurred at C, the location of zero shear. The

value of shears in AC is positive then the

moment in AC is increasing; at CB the shear is

negative, then the moment in CB is

decreasing.

R1 = 1/6 Lwo R2 = 1/3 Lwo

L

wo

1/6 Lwo

–1/3 Lwo

Load Diagram

Shear Diagram

Moment Diagram

Mmax = 0.06415L2wo

A B

C

x = L / √3

Figure P-443

R1 R2

wO

L/2 L/2

1/6 Lwo

–1/3 Lwo

L / √3 A1

Figure for solving A1


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Solution 443. By symmetry:

R1 = R2 = )( 21

21

oLw

R1 = R2 = oLw41

Problem 444. Beam loaded as shown in Fig. P-444. Solution 444. Total load

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

= 21 Lwo

By symmetry

R1 = R2 = 21 × total load

R1 = R2 = 41 Lwo

To draw the Shear Diagram (1) VA = R1 = ¼ Lwo


VB = ¼ Lwo – ½ (L/2)(wo) = 0


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.



MB = 0 + 2/3 (L/2)(1/4 Lwo) = 1/12 Lwo


MC = 1/12 Lwo – 2/3 (L/2)(1/4 Lwo) = 0

(4) MAC is third degree because the shear diagram in AC

is second degree.

(5) The shear from A to C is decreasing, thus the slope

of moment diagram from A to C is decreasing. Moment Diagram

o

2

12

1wL

Shear Diagram o4

1Lw−

o4

1Lw

Load Diagram

R2 = ¼ Lwo R1 = ¼ Lwo

wO

L/2 L/2 A B C

Figure P-444

R1 R2

wo

L/2 L/2

wo

217


Problem 445. Beam carrying the loads shown in Fig. P-445.

Solution 445. ∑MR2 = 0 5R1 = 80(3) + 90(2) R1 = 84 kN

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


(1) VA = R1 = ¼ Lwo


VB = ¼ Lwo – ½ (L/2)(wo) = 0


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)



MB = 0 + 1/3 (L/2)(¼ Lwo) = 1/24 L2wo


MC = 1/24 L2wo – 1/3 (L/2)(¼ Lwo) = 0

(4) The shear diagram from A to C is decreasing,

thus, the moment diagram is a concave

downward third degree curve.

o4

1Lw

o4

1Lw−Shear Diagram

o

2

24

1wL

Moment Diagram

R1 = ¼ Lwo

wo

L/2 L/2

wo

R2 = ¼ Lwo

Load Diagram

A B

C

Figure P-445

R1 R2

1 m 3 m 1 m

80 kN/m

20 kN/m

R1 R2

60 kN/m

1 m 3 m

20 kN/m

1 m

F1 = 80 kN

F2 = 90 kN 1 m 2 m


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(1) VA = R1 = 84 kN


VB = 84 – 20(1) = 64 kN


VC = 64 – ½ (20 + 80)(3) = –86 kN


VD = –86 + 0 = –86 kN

VD2 = VD + R2 = –86 + 86 = 0


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) MA = 0


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


MD = 86 – 86(1) = 0

R1 = 84 kN

80 kN/m

1 m

20 kN/m

1 m

R2 = 86 kN

Load Diagram

3 m

y

A B C D

84 kN

Shear Diagram

64 kN

–86 kN

E z

x =1.72 m

A1

A2

73.97 64

1 m 1.72 m

1.28 m

9.97

Figure for solving A1 and A2

86

4 m

86 kN⋅m

137.5 kN⋅m

74 kN⋅m

Moment Diagram

2nd deg

3rd deg

1st deg

219


Problem 446. Beam loaded and supported as shown in Fig. P-446.

Solution 446. ∑FV = 0

4wo + 2[ 21 wo(1)] = 20(4) + 2(50)

5wo = 180 wo = 36 kN/m

20 kN/m

4 m 50 kN 50 kN

1 m 1 m

Figure P-446


(1) VA = 0


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


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.


(1) MA = 0


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


MD = 6 – 1/3 (1)(18) = 0

(6) The moment diagram at AB and CD are 3rd

degree curve while at BC is 2nd degree curve.

6 Kn⋅m

–26 kN⋅m

6 Kn⋅m

Moment Diagram

18 kN

–32 kN

–18 kN

32 kN

Shear Diagram

20 kN/m

4 m 50 kN 50 kN

1 m 1 m

wo = 36 kN/m

Load Diagram

A D

B C


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INSTRUCTION: In the following problems, draw moment and load diagrams corresponding to the given shear diagrams. Specify values at all change of load positions and at all points of zero shear.

Problem 447. Shear diagram as shown in Fig. P-447. Solution 447.

Figure P-447 2400

400

–4000

1000

V (lb)

x (ft)

2 3 2 2

To draw the Load Diagram

(1) A 2400 lb upward force is acting at

point A. No load in segment AB.

(2) A point force of 2400 – 400 = 2000

lb is acting downward at point B.

No load in segment BC.

(3) Another downward force of

magnitude 400 + 4000 = 4400 lb

at point C. No load in segment CD.

(4) Upward point force of 4000 + 1000

= 5000 lb is acting at D. No load

in segment DE.

(5) A downward force of 1000 lb is

concentrated at point E.


(1) MA = 0


MB = 0 + 2400(2) = 4800 lb⋅ft MAB is linear and upward


MC = 4800 + 400(3) = 6000 lb⋅ft MBC is linear and upward


MD = 6000 – 4000(2) = –2000 lb⋅ft MCD is linear and downward


ME = –2000 + 1000(2) = 0 MDE is linear and upward

4800 lb⋅ft 6000 lb⋅ft

–2000 lb⋅ft Moment Diagram

R1 = 2400 lb

2000 lb 4400 lb

R2 = 5000 lb

1000 lb

Load Diagram

2 ft 3 ft 2 ft 2 ft

A B C E

2400 lb

400 lb

–4000 lb

1000 lb

Given Shear Diagram

D

221



2 1 1.6 2.4 2 V

(kN)

x (m)

36

16

–40

–24

20

Figure P-448


(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.


(1) MA = 0


MB = 0 – ½ (40)(2) = –40 kN⋅m

MAB is downward parabola with vertex at A.


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.


ME = –20 + ½ (20)(2) = 0 MDE is downward parabola with vertex at E.

–4 kN⋅m

–40 kN⋅m

8.8 kN⋅m

–20 kN⋅m

Moment Diagram

R1 = 76 kN R2 = 44 kN

20 kN/m

10 kN/m

20 kN

2 m 1 m 4 m 2 m

Load Diagram

F

36 kN

16 kN

–40 kN

–24 kN

20 kN

Given Shear Diagram

1.6 m 2.4 m

A B

C D E


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–4000

3700

1700

–3100

V

(lb)

x (ft)

3 2 3 4 8

Figure P-449


(1) Downward 4000 lb force is concentrated at

A and no load in segment AB.

(2) The shear in BC is uniformly increasing,

thus a uniform upward force is acting at a

magnitude of (3700 + 4000)/2 = 3850

lb/ft. No load in segment CD.

(3) Another point force acting downward with

3700 – 1700 = 1200 lb at D and no load in

segment DE.

(4) The shear in EF is uniformly decreasing,

thus a uniform downward force is acting

with magnitude of (1700 + 3100)/8 = 600

lb/ft.

(5) Upward force of 3100 lb is concentrated at

end of span F.


(1) The locations of zero shear (points G and

H) can be easily determined by ratio and

proportion of triangle.

(2) MA = 0


MB = 0 – 4000(3) = –12,000 lb⋅ft (4) MG = MB + Area in shear diagram

MG = –12,000 – ½ (80/77)(4000)

MG = –14,077.92 lb⋅ft (5) MC = MG + Area in shear diagram

MC = –14,077.92 + ½ (74/77)(3700)

MC = –12,300 lb⋅ft (6) MD = MC + Area in shear diagram

MD = –12,300 + 3700(3) = –1200 lb⋅ft (7) ME = MD + Area in shear diagram

ME = –1200 + 1700(4) = 5600 lb⋅ft (8) MH = ME + Area in shear diagram

MH = 5600 + ½ (17/6)(1700)

MH = 8,008.33 lb⋅ft (9) MF = MH + Area in shear diagram

MF = 8,008.33 – ½ (31/6)(3100) = 0

8,008.33 lb⋅ft

–14,077.92 lb⋅ft –12,300 lb⋅ft

–1,200 lb⋅ft

Moment Diagram

–12,000 lb⋅ft

5,600 lb⋅ft

3 ft 2 ft 3 ft 4 ft 8 ft

4000 lb 2000 lb

3100 lb 3850 lb/ft

600 lb/ft

Load Diagram

A B C D E F

80/77 ft

–4000 lb

3700 lb

1700 lb

–3100 lb 74/77 ft

17/6 ft

31/6 ft

Given Shear Diagram

G H

223


Problem 450. Shear diagram as shown in Fig. P-450. Solution 450. Solution

V

(lb)

x (ft)

4 2 4 4 4

900

–900

–1380

480

Figure P-450


(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.


(1) MA = 0


MB = 0 + ½ (4)(900) = 1800 lb⋅ft (3) MC = MB + Area in shear diagram

MC = 1800 + 900(2) = 3600 lb⋅ft (4) MD = MC + Area in shear diagram

MD = 3600 + 0 = 3600 lb⋅ft (5) ME = MD + Area in shear diagram

ME = 3600 – ½ (900 + 1380)(4)

ME = –960 lb⋅ft (6) MF = ME + Area in shear diagram

MF = –960 + ½ (480)(4) = 0

(7) The shape of moment diagram in AB is upward

parabola with vertex at A, while linear in BC and

horizontal in CD. For segment DE, the diagram

is downward parabola with vertex at G. G is the

point where the extended shear in DE intersects

the line of zero shear.

(8) The moment diagram in EF is a downward

parabola with vertex at F.

1800 lb⋅ft

4 ft 2 ft 4 ft 4 ft 4 ft

225 lb/ft

120 lb/ft 900 lb

900 lb

1860 lb

Load Diagram

900 lb

–1380 lb

–900 lb

480 lb

Given Shear Diagram

3600 lb⋅ft

–960 lb⋅ft

Moment Diagram

A B C D E F

G


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Figure P-451


(1) Upward concentrated load at A is 10 kN.

(2) The shear in AB is a 2nd-degree curve, thus

the load in AB is uniformly varying. In this

case, it is zero at A to 2(10 + 2)/3 = 8 kN at

B. No load in segment BC.

(3) A downward point force is acting at C in a

magnitude of 8 – 2 = 6 kN.

(4) The shear in DE is uniformly increasing, thus

the load in DE is uniformly distributed and

upward. This load is spread over DE at a

magnitude of 8/2 = 4 kN/m.


(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


MC = 18 – 2(1) = 16 kN⋅m


MD = 16 – 8(1) = 8 kN⋅m


ME = 8 – ½ (2)(8) = 0

(8) The moment diagram in AB is a second

degree curve, at BC and CD are linear and

downward. For segment DE, the moment

diagram is parabola open upward with vertex at E.

3 m 1 m 1 m 2 m

10 kN 4 kN/m

8 kN/m

6 kN

Load Diagram

A B C D

E

V (kN)

x (m)

10

–2

–8

3 1 1 2

2nd-degree curve

10 kN

–2 kN

–8 kN

Given Shear Diagram

F

x = 2.74 m

18.26 kN⋅m 18 kN⋅m 16 kN⋅m

8 kN⋅m

Moment Diagram



225


MOVING LOADS From the previous section, we see that the maximum moment occurs at a point of zero shears. For beams loaded with concentrated loads, the point of zero shears usually occurs under a concentrated load and so the maximum moment.

Beams and girders such as in a bridge or an overhead

crane are subject to moving concentrated loads, which are at fixed distance with each other. The problem here is to determine the moment under each load when each load is in a position to cause a maximum moment. The largest value of these moments governs the design of the beam.

SINGLE MOVING LOAD

For a single moving load, the maximum moment occurs when the load is at the midspan and the maximum shear occurs when the load is very near the support (usually assumed to lie over the support).

Mmax = 4

PL and Vmax = P

TWO MOVING LOADS

For two moving loads, the maximum shear occurs at the reaction when the larger load is over that support. The maximum moment is given by

P P Position for

maximum shear

Position for

maximum moment

L

L/2 L/2

Pb Ps d

L

Pb > Ps




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Mmax = PL

dPPL s

4

)( 2−

where Ps is the smaller load, Pb is the bigger load, and

P is the total load (P = Ps + Pb). THREE OR MORE MOVING LOADS In general, the bending moment under a particular

load is a maximum when the center of the beam is midway between that load and the resultant of all the loads then on the span. With this rule, we compute the maximum moment under each load, and use the biggest of the moments for the design. Usually, the biggest of these moments occurs under the biggest load.

The maximum shear occurs at the reaction where the

resultant load is nearest. Usually, it happens if the biggest load is over that support and as many a possible of the remaining loads are still on the span.

In determining the largest moment and shear, it is

sometimes necessary to check the condition when the bigger loads are on the span and the rest of the smaller loads are outside.

SOLVED PROBLEMS Problem 453. A truck with axle loads of 40 kN and 60 kN on a

wheel base of 5 m rolls across a 10-m span. Compute the maximum bending moment and the maximum shearing force.

Solution 453. R = 40 + 60 = 100 kN xR = 40(5) x = 200/R x = 200/100 x = 2 m

5 m

40 kN 60 kN

R

x 5 – x

227


For maximum moment under 40 kN wheel:

∑MR2 = 0 10R1 = 3.5(100) R1 = 35 kN MTo the left of 40 kN = 3.5R1 MTo the left of 40 kN = 3.5(35)

MTo the left of 40 kN = 122.5 kN⋅m For maximum moment under 60 kN wheel:

∑MR1 = 0 10R2 = 4(100) R2 = 40 kN MTo the right of 60 kN = 4R2 MTo the right of 60 kN = 4(40)

MTo the right of 60 kN = 160 kN⋅m

Thus, Mmax = 160 kN⋅⋅⋅⋅m

The maximum shear will occur when the 60 kN is over a support.

∑MR1 = 0 10R2 = 100(8) R2 = 80 kN Thus, Vmax = 80 kN Problem 454. Repeat Prob. 453 using axle loads of 30 kN and 50 kN

on a wheel base of 4 m crossing an 8-m span. Solution 454. R = 30 + 50 = 80 kN xR = 4(30) x = 120/R x = 120/80 x = 1.5 m

40 kN 60 kN

R = 100 kN

2 m 3 m

C L

10 m

3.5 m 1.5 m 3.5 m R1 R2

40 kN 60 kN

R = 100 kN

2 m 3 m

10 m

4 m 1 m 4 m R1 R2

C L

8 m R1 R2

40 kN 60 kN

R = 100 kN

2 m 3 m

4 m

30 kN 50 kN

R

x 4 – x


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Maximum moment under 30 kN wheel:

∑MR2 = 0 8R1 = 2.75(80) R1 = 27.5 kN MTo the left of 30 kN = 2.75R1 MTo the left of 30 kN = 2.75(27.5)

MTo the left of 30 kN = 75.625 kN⋅m Maximum moment under 50 kN wheel:

∑MR1 = 0 8R2 = 3.25(80) R2 = 32.5 kN MTo the right of 50 kN = 3.25R2 MTo the right of 50 kN = 3.25(32.5)

MTo the right of 50 kN = 105.625 kN⋅m

Thus, Mmax = 105.625 kN⋅⋅⋅⋅m

The maximum shear will occur when the 50 kN is over a support.

∑MR1 = 0 8R2 = 6.5(80) R2 = 65 kN Thus, Vmax = 65 kN Problem 455. A tractor weighing 3000 lb, with a wheel base of 9 ft,

carries 1800 lb of its load on the rear wheels. Compute the maximum moment and maximum shear when crossing a 14 ft-span.

Solution 455. R = Wr + Wf 3000 = 1800 + Wf Wf = 1200 lb Rx = 9Wf 3000x = 9(1200) x = 3.6 ft

30 kN 50 kN R = 80 kN

1.5 m 2.5 m

C L

2.75 m

8 m

1.25 m 2.75 m R1 R2

30 kN 50 kN R = 80 kN

1.5 m 2.5 m

C L

3.25 m

8 m

0.75 m 3.25 m R1 R2

30 kN 50 kN

R = 80 kN 1.5 m

2.5 m

6.5 m

8 m R1 R2

Wr = 1800 lb

Wf

R = 3000 lb

9 – x x

9 ft

229


9 – x = 5.4 ft

When the midspan is midway between Wr and R, the front wheel Wf will be outside the span (see figure). In this case, only the rear wheel Wr = 1800 lb is the load. The maximum moment for this condition is when the load is at the midspan.

R1 = R2 = ½ (1800) R1 = 900 lb Maximum moment under Wr MTo the left of rear wheel = 7R1 MTo the left of rear wheel = 7(900)

MTo the left of rear wheel = 6300 lb⋅ft Maximum moment under Wf

∑MR1 = 0 14R2 = 4.3R

14R2 = 4.3(3000) R2 = 921.43 lb MTo the right of front wheel = 4.3R2 MTo the right of front wheel = 4.3(921.43)

MTo the right of front wheel = 3962.1 lb⋅ft Thus, Mmax = MTo the left of rear wheel

Thus, Mmax = 6300 lb⋅⋅⋅⋅ft

The maximum shear will occur when the

rear wheel (wheel of greater load) is directly over the support.

∑MR2 = 0 14R1 = 10.4R 14R1 = 10.4(3000) R1 = 2228.57 lb Thus, Vmax = 2228.57 lb

14 ft

R1 R2

Wr = 1800 lb Wf = 1200 lb R = 3000 lb

5.4 ft 3.6 ft

5.2 ft

C L

5.2 ft 1.8 ft

14 ft R1 R2

Wr = 1800 lb

7 ft

14 ft

R1 R2

Wr = 1800 lb Wf = 1200 lb R = 3000 lb

5.4’ 3.6’

4.3 ft

C L

4.3 ft 2.7 ft

R2

14 ft

R1

Wr = 1800 lb

Wf = 1200 lb

R = 3000 lb

5.4’ 3.6’

10.4 ft


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Problem 456. Three wheel loads roll as a unit across a 44-ft span. The loads are P1 = 4000 lb and P2 = 8000 lb separated by 9 ft, and P3 = 6000 lb at 18 ft from P2. Determine the maximum moment and maximum shear in the simply supported span.

Solution 456. R = P1 + P2+ P3 R = 4k + 8k + 6k R = 18 kips R = 18,000 lbs xR = 9P2 + (9 + 18)P3 x(18) = 9(8) + (9 + 18)(6) x = 13 ft � the resultant R is 13 ft from P1 Maximum moment under P1

∑MR2 = 0 44R1 = 15.5R 44R1 = 15.5(18) R1 = 6.34091 kips R1 = 6,340.91 lbs MTo the left of P1 = 15.5R1 MTo the left of P1 = 15.5(6340.91)

MTo the left of P1 = 98,284.1 lb⋅ft Maximum moment under P2

∑MR2 = 0 44R1 = 20R 44R1 = 20(18) R1 = 8.18182 kips R1 = 8,181.82 lbs MTo the left of P2 = 20R1 – 9P1 MTo the left of P2 = 20(8,181.82) – 9(4000)

MTo the left of P2 = 127,636.4 lb⋅ft

P1 = 4k P2 = 8k P3 = 6k

R = 18k

x = 13’

18’ 9’

R1 R2

C L

20’

44’

2’ 20’

P1 = 4k P2 = 8k P3 = 6k

9’ 18’

R

x

P1 = 4k P2 = 8k P3 = 6k

R = 18k

x = 13’

R1 R2

C L

15.5’

44’

6.5’ 15.5’

18’ 9’

231


Maximum moment under P3

∑R1 = 0 44R2 = 15R 44R2 = 15(18) R2 = 6.13636 kips R2 = 6,136.36 lbs MTo the right of P3 = 15R2

MTo the right of P3 = 15(6,136.36)

MTo the right of P3 = 92,045.4 lb⋅ft Thus, Mmax = MTo the left of P2

Thus, Mmax = 127,636.4 lb⋅⋅⋅⋅ft

The maximum shear will occur when P1 is over the support.

∑MR2 = 0 44R1 = 35R 44R1 = 31(18) R1 = 12.6818 kips R1 = 12,681.8 lbs Thus, Vmax = 12,681.8 lbs Problem 457. A truck and trailer combination crossing a 12-m span

has axle loads of 10, 20, and 30 kN separated respectively by distances of 3 and 5 m. Compute the maximum moment and maximum shear developed in the span.

Solution 457. R = 10 + 20 + 30 R = 60 kN xR = 3(20) + 8(30) x(60) = 3(20) + 8(30) x = 5 m

P1 = 4k P2 = 8k P3 = 6k

R = 18k

x = 13’

18’ 9’

R1 R2

C L

15’

44’

7’ 15’

P1 = 4k P2 = 8k P3 = 6k

R = 18k

x = 13’

18’ 9’

R1 R2

44’

31’

5 m

10 kN 20 kN 30 kN

R

x

3 m


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Maximum moment under 10 kN

∑MR2 = 0 12R1 = 3.5R 12R1 = 3.5(60) 12R1 = 210 R1 = 12.7 kN MTo the left of 10 kN = 3.5R1 MTo the left of 10 kN = 3.5(12.7)

MTo the left of 10 kN = 61.25 kN⋅m Maximum moment under 20 kN

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

5 m

10 kN 20 kN 30 kN

R = 60 kN

x = 5 m

3 m

C L

R1 R2

3.5 m

12 m

3.5 m 2.5 m

5 m

10 kN 20 kN 30 kN

R = 60 kN

x = 5 m

3 m

C L

R1 R2

5 m

12 m

5 m

1 m

233


MTo the left of 20 kN = 5R1 – 3(10) MTo the left of 20 kN = 5(25) – 30

MTo the left of 20 kN = 95 kN⋅m When the centerline of the beam is midway between

reaction R = 60 kN and 30 kN, the 10 kN comes off the span.

R = 20 + 30 R = 50 kN xR = 5(30) x(50) = 150 x = 3 m from 20 kN

∑MR1 = 0 12R2 = 5R 12R2 = 5(50) R2 = 20.83 kN MTo the right of 30 kN = 5R2 MTo the right of 30 kN = 5(20.83)

MTo the right of 30 kN = 104.17 kN⋅m Thus, the maximum moment will occur when only

the 20 and 30 kN loads are on the span. Mmax = MTo the right of 30 kN

Mmax = 104.17 kN⋅⋅⋅⋅m

5 m

10 kN 20 kN 30 kN

R = 50 kN

x = 3 m

3 m

C L

R1 R2

5 m

12 m

5 m

1 m


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The maximum shear will occur when the three loads are on the span and the 30 kN load is directly over the support.

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

5 m

10 kN 20 kN 30 kN

R = 60 kN

x = 5 m

3 m

R1 R2

12 m 9 m

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Documents

Shear and Moment