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) Copyright © 2011 Mathalino.com All rights reserved.

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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 R

1

, R

2

, and M and only two equations (

Σ

M =

0 and

ΣF

v

= 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

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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.

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 R

1

and

R

2

. 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 _R 2 A B w (N/m) C x R1 V M A

together with a couple M to hold the left portion of

the bar in equilibrium under the action of R

1

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:

∑M

B

= 0

5R

D

+ 1(30) = 3(50)

R

D

= 24 kN

∑M

D

= 0

5R

B

= 2(50) + 6(30)

R

B

= 56 kN

Segment AB:

V

AB

= –30 kN

M

AB

= –30x kN

⋅

m

30 kN 50 kN 1 m 3 m 2 m A B C D Figure P-403 30 kN A x

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Segment BC:

V

BC

= –30 + 56

V

BC

= 26 kN

M

BC

= –30x + 56(x – 1)

M

BC

= 26x – 56 kN

⋅

m

Segment CD:

V

CD

= –30 + 56 – 50

V

CD

= –24 kN

M

CD

= –30x + 56(x – 1) – 50(x – 4)

M

CD

= –30x + 56x – 56 – 50x + 200

M

CD

= –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, M_AB = 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, M_BC = 48 kN⋅m. When M_BC = 0, x = 2.154 m, thus the moment is zero at 1.154 m from B. (3) M_CD = –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 Load Diagram RB = 56 kN RB = 24 kN 26 kN –30 kN –24 kN Shear Diagram Moment Diagram –30 kN⋅m 56 kN⋅m 1.154 m

Solution 404.

∑M

A

= 0

∑M

D

= 0

12R

D

+ 4800 = 3(2000)

12R

A

= 9(2000) + 4800

R

D

= 100 lb

R

A

= 1900 lb

Segment AB:

V

AB

= 1900 lb

M

AB

= 1900x lb

⋅

ft

Segment BC:

V

BC

= 1900 – 2000

V

BC

= –100 lb

M

BC

= 1900x – 2000(x – 3)

M

BC

= 1900x – 2000x + 6000

M

BC

= –100x + 6000

Segment CD:

V

CD

= 1900 – 2000

V

CD

= –100 lb

M

CD

= 1900x – 2000(x – 3) – 4800

M

CD

= 1900x – 2000x + 6000 – 4800

M

CD

= –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, M_AB = 0; at x = 3 ft, M_AB = 5700 lb⋅ft. (2) For segment BC, MBC = –100x + 6000 is linear; at x = 3 ft, MBC = 5700 lb⋅ft; at x = 9 ft, M_BC = 5100 lb⋅ft. (3) MCD = –100x + 1200 is again linear; at x = 9 ft, MCD = 300 lb⋅ft; at x = 12 ft, M_CD = 0.

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Problem 405.

Beam loaded as shown in Fig. P-405.

Solution 405.

∑M

A

= 0

∑M

C

= 0

10R

C

= 2(80) + 5[10(10)]

10R

A

= 8(80) + 5[10(10)]

R

C

= 66 kN

R

A

= 114 kN

Segment AB:

V

AB

= 114 – 10x kN

M

AB

= 114x – 10x(x/2)

M

AB

= 114x – 5x

2

kN

⋅

m

Segment BC:

V

BC

= 114 – 80 – 10x

V

BC

= 34 – 10x kN

M

BC

= 114x – 80(x – 2) – 10x(x/2)

M

BC

= 160 + 34x – 5x

2 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 – 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, M}

BC = 208 kN⋅m; at x = 10 m, M_BC = 0.

(3) Note that the maximum moment occurs at point of zero shear. Thus, at x = 3.4 m, M_BC = 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

Problem 406.

Beam loaded as shown in Fig. P-406.

Solution 406.

∑M

A

= 0

12R

C

= 4(900) + 18(400) + 9[(60)(18)]

R

C

= 1710 lb

∑M

C

= 0

12R

A

+ 6(400) = 8(900) + 3[60(18)]

R

A

= 670 lb

Segment AB:

V

AB

= 670 – 60x lb

M

AB

= 670x – 60x(x/2)

M

AB

= 670x – 30x

2

lb

⋅

ft

Segment BC:

V

BC

= 670 – 900 – 60x

V

BC

= –230 – 60x lb

M

BC

= 670x – 900(x – 4) – 60x(x/2)

M

BC

= 3600 – 230x – 30x

2

lb

⋅

ft

Segment CD:

V

CD

= 670 + 1710 – 900 – 60x

V

CD

= 1480 – 60x lb

M

CD

= 670x + 1710(x – 12)

– 900(x – 4) – 60x(x/2)

M

CD

= –16920 + 1480x – 30x

2

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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Problem 407.

Beam loaded as shown in Fig. P-407.

Solution 407.

∑M

A

= 0

∑M

D

= 0

6R

D

= 4[2(30)]

6R

A

= 2[2(30)]

R

D

= 40 kN

R

A

= 20 kN

Segment AB:

V

AB

= 20 kN

M

AB

= 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, V_BC = –230 – 60x is also linear; at x= 4 ft, VBC = – 470 lb, at x = 12 ft, VBC = –950 lb. (3) V_CD = 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, M_CD = –16920 + 1480x – 30x2 _{is a second degree} curve; at x = 12 ft, MCD = –3480 lb⋅ft; at x = 18 ft, M_CD = 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

Segment BC:

V

BC

= 20 – 30(x – 3)

V

BC

= 110 – 30x kN

M

BC

= 20x – 30(x – 3)(x – 3)/2

M

BC

= 20x – 15(x – 3)

2

Segment CD:

V

CD

= 20 – 30(2)

V

CD

= –40 kN

M

CD

= 20x – 30(2)(x – 4)

M

CD

= 20x – 60(x – 4)

Problem 408.

Beam loaded as shown in Fig. P-408.

Solution 408.

∑M

A

= 0

∑M

D

= 0

6R

D

= 1[2(50)] + 5[2(20)]

6R

A

= 5[2(50)] + 1[2(20)]

R

D

= 50 kN

R

A

= 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, M_BC = 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.

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Segment AB:

V

AB

= 90 – 50x kN

M

AB

= 90x – 50x(x/2)

M

AB

= 90x – 25x

2

Segment BC:

V

BC

= 90 – 50(2)

V

BC

= –10 kN

M

BC

= 90x – 2(50)(x – 1)

M

BC

= –10x + 100 kN

⋅

m

Segment CD:

V

CD

= 90 – 2(50) – 20(x – 4)

V

CD

= –20x + 70 kN

M

CD

= 90x – 2(50)(x – 1)

– 20(x – 4)(x – 4)/2

M

CD

= 90x – 100(x – 1) – 10(x – 4)

2

M

CD

= –10x

2

+ 70x – 60 kN

⋅

m

Problem 409.

Cantilever beam loaded as shown in Fig. P-409.

x A R_A = 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) V_AB = 90 – 50x is linear; at x = 0, V_BC = 90 kN; at x = 2 m, VBC = –10 kN. When VAB = 0, x = 1.8 m. (2) VBC = –10 kN along segment BC. (3) VCD = –20x + 70 is linear; at x = 4 m, VCD = –10 kN; at x = 6 m, VCD = – 50 kN.

To draw the Moment Diagram: (1) M_AB = 90x – 25x2 _{is second degree;} at x = 0, MAB = 0; at x = 1.8 m, MAB = 81 kN⋅m; at x = 2 m, MAB = 80 kN⋅m. (2) M_BC = –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

Solution 409.

Segment AB:

V

AB

= –w

o

x

M

AB

= –w

o

x(x/2)

M

AB

= –

₂1

w

o

x

2

Segment BC:

V

BC

= –w

o

(L/2)

V

BC

= –

₂1

w

o

L

M

BC

= –w

o

(L/2)(x – L/4)

M

BC

= –

₂1

w

o

Lx +

₈1

w

o

L

2

Problem 410.

Cantilever beam carrying the uniformly varying load

shown in Fig. P-410.

Solution 410.

x

y

=

L

w

_o

y =

x

L

w

o

F

x

=

₂1

xy

F

x

=













x

L

w

x

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 1_w oL Shear Diagram Moment Diagram –8 1 woL2 –8 3 woL2

To draw the Shear Diagram:

(1) V_AB = –w_ox for segment AB is linear; at x = 0, V_AB = 0; at x = L/2, V_AB = –

2 1_w

oL.

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

1_w oL.

To draw the Moment Diagram:

(1) MAB = –₂1 wox2 is a second degree curve; at x = 0, M_AB = 0; at x = L/2, M_AB = –₈1 w_oL2_. (2) MBC = –₂1woLx + 8 1 woL2 is a second degree; at x = L/2, MBC = –8 1 woL2; at x = L, MBC = – 8 3 woL2.

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F

x

=

2

2

L

x

w

_o

Shear equation:

V = –

2

2

L

x

w

_o

Moment equation:

M = –

₃1

_xF

x

= –













₂

2

3

1

x

L

w

x

o

M

= –

3

6

L

x

w

_o

Problem 411.

Cantilever beam carrying a distributed load with

intensity varying from w

o

at the free end to zero at the

wall, as shown in Fig. P-411.

Solution 411.

L

w

x

L

y

₌

o

−

y =

(

L

x

)

L

w

o

₋

Figure P-411 L wo L wo Load Diagram Shear Diagram – 2 1_w oL –6 1 woL2 Moment Diagram 2nd _degree 3rd _degree

To draw the Shear Diagram: V = – ox2

2L w

is a second degree curve; at x = 0, V = 0; at x = L, V = –

2 1 _w

oL.

To draw the Moment Diagram: M = – o x3

6L w

is a third degree curve; at x = 0, M = 0; at x = L, M =–₆1

F

1

=

₂1

x

(

w

o

−

y

)

F

1

=













−

−

(

)

2

1

x

L

L

w

w

x

o o

F

1

=













+

−

x

L

w

w

w

x

o o o

2

1

F

1

=

2

2

L

x

w

o

F

2

= xy =













−

)

(

L

x

L

w

x

o

F

2

=

(

Lx

x

2

)

L

w

o

₋

Shear equation:

V = –F

1

– F

2

= –

2

2

L

x

w

o

_–

₍

2

₎

x

Lx

L

w

o

₋

V

= –

2

2

L

x

w

_o

–

w

o

x

+

2

x

L

w

_o

V

=

2

2

L

x

w

o

_–

_w

_x

o

Moment equation:

M = –

₃2

_xF

1

–

21

xF

2

M

= –













₂

2

3

1

x

L

w

x

o

_–













−

)

(

2

1

_Lx

_x

2

L

w

x

o

M

= –

3

3

L

x

w

o

_–

2

2

x

w

o

₊

3

2

L

x

w

o

M

= –

2

2

x

w

o

₊

3

6

L

x

w

o

Problem 412.

Beam loaded as shown in Fig. P-412.

x wo y L – x 2 1 _x 3 2 x F1 _F 2 Figure P-406 800 lb/ft 2 ft 4 ft A B D RA 2 ft C RC To draw the Shear Diagram:

V = ox2 2L w

– wox is a concave upward second degree curve; at x = 0, V = 0; at x = L, V = –₂1

woL. To draw the Moment diagram:

M = – ox2 2 w + ox3 6L w is in third degree; at x = 0, M = 0; at x = L, M = –₃1 w_oL2_. L wo Load Diagram –1₂ woL Shear Diagram 2nd _degree –3 1 woL2 Moment Diagram 3rd _degree

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Solution 412.

∑M

A

= 0

∑M

C

= 0

6R

C

= 5[6(800)]

6R

A

= 1[6(800)]

R

C

= 4000 lb

R

A

= 800 lb

Segment AB:

V

AB

= 800 lb

M

AB

= 800x

Segment BC:

V

BC

= 800 – 800(x – 2)

V

BC

= 2400 – 800x

M

BC

= 800x – 800(x – 2)(x – 2)/2

M

BC

= 800x – 400(x – 2)

2

Segment CD:

V

CD

= 800 + 4000 – 800(x – 2)

V

CD

= 4800 – 800x + 1600

V

CD

= 6400 – 800x

M

CD

= 800x + 4000(x – 6) – 800(x – 2)(x – 2)/2

M

CD

= 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 R_C = 4000 lb 800 lb/ft B A R_A = 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, M_BC = 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

Problem 413.

Beam loaded as shown in Fig. P-413.

Solution 413.

∑M

B

= 0

6R

E

= 1200 + 1[6(100)]

R

E

= 300 lb

∑M

E

= 0

6R

B

+ 1200 = 5[6(100)]

R

B

= 300 lb

Segment AB:

V

AB

= –100x lb

M

AB

= –100x(x/2)

M

AB

= –50x

2

lb

⋅

ft

Segment BC:

V

BC

= –100x + 300 lb

M

BC

= –100x(x/2) + 300(x – 2)

M

BC

= –50x

2

+ 300x – 600 lb

⋅

ft

Segment CD:

V

CD

= –100(6) + 300

V

CD

= –300 lb

M

CD

= –100(6)(x – 3) + 300(x – 2)

M

CD

= –600x + 1800 + 300x – 600

M

CD

= –300x + 1200 lb

⋅

ft

Segment DE:

V

DE

= –100(6) + 300

V

DE

= –300 lb

M

DE

= –100(6)(x – 3) + 1200 + 300(x – 2)

M

DE

= –600x + 1800 + 1200 + 300x – 600

M

DE

= –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 R_B = 300 lb 4 ft C 1’ D x

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Problem 414.

Cantilever beam carrying the load shown in Fig.

P-414.

Solution 414.

Segment AB:

V

AB

= –2x kN

M

AB

= –2x(x/2)

M

AB

= –x

2

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) M_AB = –50x2 _{is a second degree} curve; at x= 0, MAB = 0; at x = ft, MAB = –200 lb⋅ft. (2) MBC = –50x2 + 300x – 600 is also second degree; at x = 2 ft; MBC = – 200 lb⋅ft; at x = 6 ft, MBC = –600 lb⋅ft; at x = 3 ft, MBC = –150 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

F

1

= 2x

F

2

=

₂1

(x – 2)y

F

2

=

₂1

(x – 2)[

₃2

(x – 2)]

F

2

=

₃1

(x – 2)

2

V

BC

= –F

1

– F

2

V

BC

= – 2x –

₃1

(x – 2)

2

M

BC

= –(x/2)F

1

–

₃1

(x – 2)F

2

M

BC

= –(x/2)(2x) –

₃1

(x – 2)[

₃1

(x – 2)

2

]

M

BC

= –x

2

–

₉1

(x – 2)

3

Problem 415.

Cantilever beam loaded as shown in Fig. P-415.

Solution 415.

Segment AB:

V

AB

= –20x kN

M

AB

= –20x(x/2)

M

AB

= –10x

2

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 = –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. 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

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Segment BC:

V

BC

= –20(3)

V

AB

= –60 kN

M

BC

= –20(3)(x – 1.5)

M

AB

= –60(x – 1.5) kN

⋅

m

Segment CD:

V

CD

= –20(3) + 40

V

CD

= –20 kN

M

CD

= –20(3)(x – 1.5) + 40(x – 5)

M

CD

= –60(x – 1.5) + 40(x – 5)

Problem 416.

Beam

carrying

uniformly

varying

load shown in Fig.

P-416.

Solution 416.

∑M

R2

= 0

LR

1

=

₃1

LF

R

1

=

₃1

(

₂1

Lw

o

)

R

1

=

₆1

Lw

o

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) M_AB = –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, M_BC = –90 kN⋅m; at x = 5 m, M_BC = –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.

∑M

R1

= 0

LR

2

=

₃2

LF

R

2

=

₃2

(

₂1

Lw

o

)

R

2

=

₃1

Lw

o

L

w

x

y

_o

=

y =

x

L

w

_o

F

x

=

₂1

xy =













x

L

w

x

o

2

1

F

x

=

2

2

L

x

w

o

V = R

1

– F

x

V

=

₆1

_Lw

o

–

2

2

L

x

w

o

M = R

1

x – F

x

(

₃1

x

)

M

=

₆1

_Lw

o

x –

2

2

L

x

w

_o

(

₃1

x

₎

M

=

₆1

_Lw

o

x –

3

6

L

x

w

o

x R₁ wo Fx = ½ xy y 2/3 x 1/3 x

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_/R 1 = L2/(R1 + R2) a2_{/(1/6 Lw} o) = L2/(1/6 Lwo + 1/3 Lwo) a2 _{= (1/6 L}3_w o)/(1/2 Lwo) = 1/3 L2 a = 0.5774L a =

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 L R1 w_o R2 –R2 Mmax Load Diagram Shear Diagram Moment Diagram a R1

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Problem 417.

Beam carrying the triangular loading shown in Fig.

P-417.

Solution 417.

By symmetry:

R

1

= R

2

=

1₂

(

₂1

Lw =

o

)

41

Lw

o

2

/

L

w

x

y

₌

o

_{; y =}

_x

L

w

o

2

F = xy

₂1

₌

_











x

L

w

x

2

o

2

1

F =

x

2

L

w

_o

V = R

1

– F

V =

₄1

Lw

_o

_–

_x

2

L

w

o

M = R

1

x – F

(

₃1

x

)

M =

₄1

Lw

_o

x

_–

₍

₎

3 1 2

_x

x

L

w

_o













M =

41

Lw

o

x

–

3

3

L

x

w

o 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 o 4 1 Lw − o 4 1 Lw o 2 12 1_Lw Moment Diagram

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 = L2

wo/12. The other half of the diagram can be drawn by the concept of symmetry.

Problem 418.

Cantilever beam loaded as shown in Fig. P-418.

Solution 418.

Segment AB:

V

AB

= –20 kN

M

AB

= –20x kN

⋅

m

Segment BC:

V

AB

= –20 kN

M

AB

= –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

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Solution 419.

[ ∑M

C

= 0 ]

9R

1

= 5(810)

R

1

= 450 lb

[ ∑M

A

= 0 ]

9R

2

= 4(810)

R

2

= 360 lb

Segment AB:

6

270

=

x

y

y = 45x

F = xy

₂1

₌

₍

₄₅

₎

2 1

_x

_x

F = 22.5x

2

V

AB

= R

1

– F

V

AB

= 450 – 22.5x

2

lb

M

AB

= R

1

x – F( x

₃1

)

M

AB

= 450x – 22.5x

2

( x

31

)

M

AB

= 450x – 7.5x

3

lb

⋅

ft

Segment BC:

V

BC

= 450 – 810

V

BC

= –360 lb

M

BC

= 450x – 810(x – 4)

M

BC

= 450x – 810x + 3240

M

BC

= 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

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

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w = 30(1000)/12

w = 2500 lb/ft

∑F

V

= 0

R = W

20r = 30(1000)

r = 1500 lb/ft

First segment (from 0 to 4 ft from left):

V

1

= 1500x

M

1

= 1500x(x/2)

M

1

= 750x

2

Second segment (from 4 ft to mid-span):

V

2

= 1500x – 2500(x – 4)

V

2

= 10000 – 1000x

M

2

= 1500x(x/2) – 2500(x – 4)(x – 4)/2

M

2

= 750x

2

– 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.

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:

Q

x

= Q sin

θ

Q

y

= Q cos

θ

P

x

= P sin (90

°

–

θ

)

P

x

= P (sin 90

°

cos

θ

– cos 90

°

sin

θ

)

P

x

= P cos

θ

P

y

= P cos (90

°

–

θ

)

P

y

= P (cos 90

°

cos

θ

+ sin 90

°

sin

θ

)

P

y

= P sin

θ

Shear:

V = ∑F

y

V = Q

y

– P

y

V = Q cos θθθθ

–

P sin θθθθ

Moment arms:

d

Q

= R sin

θ

d

P

= R – R cos

θ

d

P

= R (1 – cos

θ

)

Moment:

M = ∑M

counterclockwise

– ∑M

clockwise

M = Q(d

Q

) – P(d

P

)

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

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For

θ

that is greater than 90

°

Components of Q and P:

Q

x

= Q sin (180

°

–

θ

)

Q

x

= Q (sin 180

°

cos

θ

– cos 180

°

sin

θ

)

Q

x

= Q cos

θ

Q

y

= Q cos (180

°

–

θ

)

Q

y

= Q (cos 180

°

cos

θ

+ sin 180

°

sin

θ

)

Q

y

= –Q sin

θ

P

x

= P sin (

θ

– 90

°

)

P

x

= P (sin

θ

cos 90

°

– cos

θ

sin 90

°

)

P

x

= –P cos

θ

P

y

= P cos (

θ

– 90

°

)

P

y

= P (cos

θ

cos 90

°

+ sin

θ

sin 90

°

)

P

y

= P sin

θ

Shear:

V = ∑F

y

V = –Q

y

– P

y

V = –(–Q sin θ

) – P sin

θ

V = Q sin θθθθ

–

P sin θθθθ

Moment arms:

d

Q

= R sin (180

°

–

θ

)

d

Q

= R (sin 180

°

cos

θ

– cos 180

°

sin

θ

)

d

Q

= R sin

θ

d

P

= R + R cos (180

°

–

θ

)

d

P

= R + R (cos 180

°

cos

θ

+ sin 180

°

sin

θ

)

d

P

= R – R cos

θ

d

P

= R(1 – cos

θ

)

Moment:

M = ∑M

counterclockwise

– ∑M

clockwise

M = Q(d

Q

) – P(d

P

)

M = QR sin θθθθ

–

PR(1 – cos θθθθ

)

θ P Q V dQ R dP 180° – θ 180° – θ θ – 90°

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.

∑M

C

= 0

2R(R

A

) = RP

R

A

=

₂1

P

For

θ

that is less than 90

°

Shear:

V

AB

= R

A

cos (90

°

–

θ

)

V

AB

=

₂1

P (cos 90°

cos

θ

+ sin 90

°

sin

θ

)

V

AB

=

21

P sin θθθθ

Moment arm:

d = R – R cos θ

d = R(1 – cos θ

)

Moment:

M

AB

= R

a

(d)

M

AB

=

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

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For

θ

that is greater than 90

°

Components of P and R

A

:

P

x

= P sin (

θ

– 90

°

)

P

x

= P (sin

θ

cos 90

°

– cos

θ

sin 90

°

)

P

x

= –P cos

θ

P

y

= P cos (

θ

– 90

°

)

P

y

= P (cos

θ

cos 90

°

+ sin

θ

sin 90

°

)

P

y

= P sin

θ

R

Ax

= R

A

sin (

θ

– 90

°

)

R

Ax

=

₂1

P (sin θ

cos 90

°

– cos

θ

sin 90

°

)

R

Ax

= –

21

P cos θ

R

Ay

= R

A

cos (

θ

– 90

°

)

R

Ay

=

21

P (cos θ

cos 90

°

+ sin

θ

sin 90

°

)

R

Ay

=

₂1

P sin θ

Shear:

V

BC

= ∑F

y

V

BC

= R

Ay

– P

y

V

BC

=

₂1

P sin θ

– P sin

θ

V

BC

= –

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:

M

BC

= ∑M

counterclockwise

– ∑M

clockwise

M

BC

= R

A

(R + d) – Pd

M

BC

=

₂1

P(R – R cos θ

) – P(–R cos

θ

)

M

BC

=

₂1

PR –

₂1

PR cos θ

+ PR cos

θ

M

BC

=

₂1

PR +

₂1

PR cos θ

M

BC

=

₂1

PR(1 + cos θθθθ

)

θ P O A B RA R 180° – θ V d θ – 90° θ – 90° R

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

V

C

= (

ΣF

v

)

L

= R

1

– wx

where R

1

= R

2

= wL/2

V

C

=

2

wL

– wx

The moment at C is

M

C

= (

ΣM

C

) =

wL

x

2

–













2

x

wx

M

C

=

2

wLx

–

2

2

wx

If we differentiate M with respect to x:

dx

dM

=

2

wL

dx

dx

–

dx

dx

x

w

2

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

.

Shear and Moment in Beams

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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

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.)

Problem 425.

Beam loaded as shown in Fig. P-425.

Positive Bending Negative Bending

Positive Shear Negative Shear

R1 R2

60 kN 30 kN

2 m 4 m 1 m

Shear and Moment in Beams

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Solution 425.

∑M

A

= 0

6R

2

= 2(60) + 7(30)

R

2

= 55 kN

∑M

C

= 0

6R

1

+ 1(30) = 4(60)

R

1

= 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

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

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free contents. Thank you for your

support.

Problem 426.

Cantilever beam acted upon by a uniformly

distributed load and a couple as shown in Fig. P-426.

Solution 426.

Problem 427.

Beam loaded as shown in Fig. P-427.

1 m Figure P-426 2 m 2 m 5 kN/m M = 60 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 –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

Shear and Moment in Beams

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Solution 427.

∑M

C

= 0

12R

1

= 100(12)(6) + 800(3)

R

1

= 800 lb

∑M

A

= 0

12R

2

= 100(12)(6) + 800(9)

R

2

= 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

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

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. 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’

Solution 428.

∑M

D

= 0

5R

1

= 50(0.5) + 25

R

1

= 10 kN

∑M

A

= 0

5R

2

+ 25 = 50(4.5)

R

2

= 40 kN

Problem 429.

Beam loaded as shown in Fig. P-429.

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 = 1 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 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 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.