1

CHAPTER 7: Forces in Beams

Beams– Various types of loading and support– Shear and bending moment in a beam– Shear and bending moment diagrams– Relations among Load, Shear, and

Bending Moment

2

F

F

A

B

Compression

F

F

A

B

Tension

7.1 Internal Forces in Members

C

C

BF

F

A

C F

F

C

C

BF

F

A

C F

F

3

A

BC

D

EF

W

G

A

B

C

D

Ax

Ay

T Cx

CyFEB

J

D

T V

N

J

M

A

B

C

Ax

Ay

Cx

CyFEB

JVM

N

4

A

B

C

PP

V

NM

A

DP

M

N

V

PD

B

C

D

5

Example 7.1

In the frame shown, determine the internal forces (a) in member ACF atpoint J and in member BCD at point K. (b) determine the equation ofinternal forces of member ACD and draw its diagram. This frame hasbeen previously considered in sample

Guide : Assuming that Framecan be treated as a rigid body.The reactions and the forcesacting on each member of theframe are determined.Member ACF is cut at point Jand member BCD is cut atpoint K, the internal forces arerepresented by an equivalentforce-couple system and canbe determined by consideringthe equilibrium of either part.

2.7 m

4.8 m

3.6 m1.2 m

1.5 m2.7 m

E F

B CD

J

K

A

2400 N

1000 N

6

2.7 m

4.8 m

3.6 m1.2 m

1.5 m2.7 m

E F

B CD

J

K

A

2400 N

• Define the state of the problem

Assuming that Frame can be treated as a rigid body. Member ACF iscut at point J and member BCD is cut at point K, the internal forces arerepresented by an equivalent force-couple system.

(a)

1200 N

7

2.7 m

2.7 m

E

B

A

750 N

2400 N

1200 N

2400 N

450 N

2.7 m

4.8 m

3.6 m1.2 m

1.5 m2.7 m

E F

B CD

A

2400 N

Ex = 1200 N

Ey = 750 N F = 3150 N

2.7 m

4.8 m

2.7 mF

B C

A

2400 N

450 N

3600 N

3150 N

2400 N

• Construct the physical model, construct mathematical model and solve equations

1200 N

1200 N

Ex = 1200 N

1.2 m

CD

2400 N2.4 m

B

2400 N

2400 N3600 N1200 N

8

x

V (N)

-1200

2400

x

M (N•m)

A = -2880

A = +2880

-2880-1800

x

N (N)

1.2 m

CD

2400 N2.4 m

B

3600 N1200 N

x2

x1

Nx1 = +2400 NNx2 = 0

Vx1= -1200 NVx2= 2400 N

Mx1 = -1200x1 N•mMx2 = -2400x2 N•m

• The diagram of internal forces of member BCD

2400 N

2400 N

+2400

9

7.2 Various Types of Loading and SupportBEAMS

P1 P2

A B CD

Concentrated loads

A C

Distributed load

w

B

10

Simply supported beam

L

Overhanging beam

L

Cantilever beam

L

Continuous beam

L2L1

Beam fixed at one end and simplysupported at the other end

L

Fixed beam

L

Hinge

11

7.3 Shear and Bending Moment in a BeamInternal Loadings at a Specified Point

• Sign Convention

N N

V

V

M M

NM

V

V

N

M

Positive Sign Convention

12

Shear and Moment Functions

+ ΣMx1 = 0:Mx1 = (P/2)x1

ΣFy = 0: Vx1 = P/2+

P

A B

L/2 L/2

x1 x2

A

P/2Vx1

Mx1x1

BP/2

x2

Vx2

Mx2

+ ΣMx2 = 0:Mx2 = (P/2)x2

ΣFy = 0: Vx2 = -P/2+

P/2 P/2

V

x

P/2

-P/2M

x

PL/4

13

x1

x2

x3

x1 x2 x3

w

P

AB C

D

w

P

AB C

D

OR

14

. .B C

A D

F1 F3F2

w = w(x)

M1

.M2

.

w

x ∆x

w(x)∆x

ε∆x

∆x

w(x)

O . M + ∆∆∆∆M

V + ∆∆∆∆V

ΣFy = 0:+

0)()( =∆+−∆− VVxxwV

+ ΣMO = 0:

0)()()( =∆++∆∆+−∆− MMxxxwMxV ε2)()( xxwxVM ∆−∆=∆ ε

xxwV ∆−=∆ )(

M

V

7.4 Relations Among Load, Shear, and BendingMoment

15----------(7-4)

----------(7-3)

----------(7-2)

----------(7-1)

2)()( xxwxVM ∆−∆=∆ ε

xxwV ∆−=∆ )(

Dividing by ∆x and taking the limit as ∆x 0, these equation become

Vdx

dM =

Slope of Moment Diagram = Shear

)(xwdxdV −=

Slope of Shear Diagram = -Intensity of Distributed Load

Equations (7-1) and (7-2) can be “integrated” from one point to another between concentrated forces or couples, in which case

dxxwV ∫−=∆ )(Change in Shear = -Area under Distributed Loading Diagram

and

∫=∆ dxxVM )(

Change in Moment = Area under Shear Diagram

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∆x

M

V

M + ∆∆∆∆M

V + ∆∆∆∆V

.M´

O

F

∆x

M

V

M + ∆∆∆∆M

V + ∆∆∆∆V

+ ΣMO = 0: 'MM =∆

ΣFy = 0:+ FV −=∆

Thus, when F acts downward on the beam, ∆V is negative so that the shear diagram shows a “jump” downward. Likewise, if F acts upward, the jump (∆V) is upward.

In this case, if an external couple moment M´ is applied clockwise, ∆M is positive, so that themoment diagram jumps upward, and when M acts counterclockwise, the jump (∆M) must bedownward.

17

ML MRM´0

P

VLVR

ML MR

VLVR

ML MRw1

w2

VLVR

ML MRw1 w2

VLVR

ML MRw0

VL VR

Slope = VL

Slope = VR0

0

ML

MR

ML MR

0

0

VL

VR

MR-wo Slope = VL

Slope = VR

ML

VR

VL

ML MR

Slope = -w1

Slope = -w2

Slope = VR

Slope = VL

ML

VL

VR

Slope = w1

Slope = -w2 MR

Slope = VR

Slope = VL

18

Example 7.2

From the structure show. Draw the shear and bending moment diagramfor the beam and loading.(a) use the equations each section.(b) use the relations among load, shear, and bending moment.

2.5 m 3 m 2 m

20 kN 40 kN

A B

CD

19

2.5 m 3 m 2 m

20 kN 40 kN

A B

CD

kNBy 465

)5.7(20)2(40 =+=

Dy = 20 + 40 - 46 = 14 kN

(a) use the equations each section.

x1 x3

x1

20 kN

A

Vx1

Mx1

+ ΣMx1 = 0:Mx1 = -20x1

ΣFy = 0: Vx1 = -20+

x2

+ ΣMx2 = 0: Mx2 = 46x2 - 20(2.5 + x2) Mx2 = 26x2 - 50

ΣFy = 0: Vx2 = 46-20 = 26+x2

2.5 m

20 kN

A B

Vx2

Mx2

46 kN

Dx3

14 kN

Mx3

Vx3+ ΣMx3 = 0:Mx3 = 14x3

ΣFy = 0: Vx3 = -14+

20

(a) use the equations each section.

x1 x3x2

x

V (kN)

-14

26

-20

x

V (kN•m)

-50

28

2.5 m 3 m 2 m

20 kN 40 kN

A B

CD

46 kN14 kN

Vx1 = -20

Mx1 = -20x1

Vx2 = 26

Mx2 = 26x2 - 50

Vx3 = -14

Mx3 = 14x3

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x

V (kN)

-20

26

-14 -14

26

-20A = -50

A = +78

A = -28

x

M (kN•m)

-50

28

(b) use the relations among load, shear, and bending moment.

2.5 m 3 m 2 m

20 kN 40 kN

A B

CD

46 kN14 kN

20 kN 40 kN

46 kN14 kN

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Example 7.3

Draw the shear and bending moment diagram for the the mechanical linkAB. The distributed load of 40 N/mm extends over 12 mm of the beam,from A to C, and 400 N load is applied at E.

4 mm12 mm 6 mm 10 mm

32 mm

40 N/mm

A BC D

400 N

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400 N

4 mm12 mm 6 mm 10 mm

32 mm

480 N

A BC D

400 N kN

By

36532

)22(400)6(480

=

+=Ay = 480 + 400 - 365 = 515 kN

480 N

365 N515 N

1600 N•mm

V (N)

x

515

35

-365 -365V (N•mm)

x

A = +3300

A = 5110A = 210

33005110

3510

1600 N•mm

365 N515 N 400 N

480 N

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Example 7.4

Draw the shear and moment diagrams for the beam shown in the figure.

9 m

20 kN/m

25

9 m

20 kN/m

+

(2/3)9 = 6 m(1/2)(9)(20) = 90 kN

90-60 = 30 kN 90(6)/9 = 60 kN

x

V (kN)

30

+

60

-

x

M (kN•m)

V = 0

M

)9

20)()(21( xx

3x

ΣFy = 0:+

x = 5.20 m

0)9

20)()(21(30 =− xxx

+ ΣMx = 0:

0)2.5(30)32.5)](

92.520)(2.5)(

21[( =−+M

M = 104 kN•m

104

V = 0

= 5.20 m

x

)9

20( x

30 kN

26

Example 7.5

Draw the shear and moment diagrams for the beam shown in the figure.

3 kN5 kN•m

A BC D

3 m 1.5 m 1.5 m

27

3 kN5 kN•m

A BC D

3 m 1.5 m 1.5 m0.67 kN 2.33 kN

V (N)x (m)

0.67+

-2.33

-

M (kN•m)x (m)

2.01+

-1.49

3.52

-+

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Example 7.6

Draw the shear and moment diagrams for the compound beam shown inthe figure. Assume the supports at A is fix C is roller and B is pinconnections.

12 m12 m 15 m

8 kN 30 kN•m

A B C

hinge

29

8 kN

30 kN•m

Ay

Ax

MA

By

Bx

Bx

By

Cy

= 48 kN•m

= 0

= 6 kN

= 2 kN

= 0

= 2 kN

0 =

= 2 kN

12 m12 m 15 m

8 kN 30 kN•m

A B C

hinge

30

12 m12 m 15 m

8 kN

A B C

V (kN)x (m)

6 6

-2 -2

x (m)M (kN•m)

-48

24

-30

8 m

30 kN•m30 kN•m

6 kN

48 kN•m

2 kN

31

Example 7.7

Draw the shear and moment diagrams for the compound beam shown inthe figure. Assume the supports at A and C are rollers and B and D arepin connections.

5 kN 3 kN/m2 kN/m60 kN • mHinge

10 m 6 m 4 m 6 m 6 m

AB C D

32

By

Bx

Bx

ByCy Dy

Dx

Ay

= 45 kN

= 0 kN

= -6 kN

= 0 kN= 16 kN

= 16 kN

0 =

= 4 kN5 kN 9 kN 9 kN

4 m 4 m

60 kN • m20 kN

5 kN 3 kN/m2 kN/m60 kN • mHinge

10 m 6 m 4 m 6 m 6 m

AB C D

33

V (kN) x (m)

-16-21 -21

24

6

M (kN • m) x (m)

2 m

6064

-96

-180

4

5 kN2 kN/m60 kN • m

Hinge

10 m 6 m 4 m 6 m 6 mB C D

3 kN/m

A4 kN

45 kN

6 kN