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5-1Internal Forces in Members

Chapter 5Analysis and Design of Beams for Bending

INTRODUCTION

Internal Forces in Members

Shear and Bending-Moment Diagrams

Design of Prismatic Beams for Bending

A

B

A

B

C

P

A E

F

DB

C

5'

10'

4'

8'

80' 40'40'

BA

P P

C D

80' 40'40'

BA

P P

C D

Cross-section

A

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5-2Internal Forces in Members

A

B

7 14 7

7

10

A

B

D

C

E

F

A

B

C

P

A E

F

DB

C

5'

10'

4'

8'

C

E

INTERNAL FORCES IN MEMBERS

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5-3

A

B

ExampleDetermine the internal forcesat point J. P= 5000 lb.Units: Lb, ft.

P

A E

F

DB

C

5

10

4

8

J

1.5

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5-4 Shear and Bending Moment Diagrams

Shear and Bending Moment Diagrams

M

V

Sign Convention

M V

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5-5Shear and Bending Moment Diagrams

Why do we Need to Draw Shear and Bending Diagrams?

80' 40'40'

BA

P P

C D

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5-6

Support Reactions

Segment

A to C

C to D

D to B

Shear Moment

ExampleDraw the shear and bending-moment diagrams for the beam and loadingshown. Label all points of change, maximums and minimums, and theaxes. Units: Lb, ft.

8 44

BA

P P

C D

x

A

P

x

A

P

x

A

P

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5-7

Support Reactions

Segment Shear Moment

ExampleDraw the shear and bending-moment diagrams for the beam andloading shown. Label all points of change, maximums andminimums, and the axes.

L

BA

W

x

A

W

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5-8

Caution:

L

BA

W

L

BA

When drawing FBDs, always use the original loading and not theequivalent.

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5-9Relations Among Load, Shear, and Bending Moment

RELATIONS AMONG LOAD, SHEAR, AND BENDING-MOMENT

L

BA

w w

dVw

dx=

dMV

dx=

The change in shear isequal to the area under theload curve.

The change in moment isequal to the area under the

shear curve.

The slope of the sheardiagram is equal to thevalue of the w load.

The slope of the momentdiagram is equal to thevalue of the shear.

M V x =

V w x =

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5-10 Relationship Between Load, Shear, and Bending Moment

A B

Observations about the Shape of Shear/ Moment DiagramsShear Diagrams:

-Are a plot of forces (note the units).

-Discontinuities occur at concentrated forces.

Moment Diagrams:-Are a plot of moments (note the units).

-Discontinuities occur at concentrated moments.

Miscellaneous:-Check your work by noting that you always start and end at zero.

-Always use the original FBD and not the equivalent.

15 kN 25 kN 20 kN

30 kN

4 kips/ft

M

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5-11

Support Reactions

ExampleDraw the shear and bending-moment diagrams for the beam andloading shown. Label all points of change, maximums andminimums, and the axes. Units: kN, m

3

BA

8 8

C D

10

10

3 3 3

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5-12

Support Reactions

ExampleDraw the shear and bending-moment diagrams for the beam andloading shown. Label all points of change, maximums andminimums, and the axes. Units: kN, m

3.2

BA

2 kN/m

4

0.8

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5-13

ExampleDraw the shear and bending-moment diagrams for the beamand loading shown. Label all points of change, maximums andminimums, and the axes. Units: Lb, ft.

8 44

BA

30 lb/ft20 lb/ft

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5-14

ExampleSketch the shear and bending-moment diagrams for the beamand loading shown. Label all points of change, maximums andminimums, and the axes. Units: Lb, ft.

8 4

BA

30 lb/ft

4

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5-15

ExampleSketch the shear and bending-moment diagrams for the beamand loading shown. Label all points of change, maximums andminimums, and the axes. Units: Lb, ft.

5 434

BA

30 lb/ft

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5-16

ExampleSketch the shear and bending-moment diagrams for the beamand loading shown. Label all points of change, maximums andminimums, and the axes. Units: Lb, ft.

5 434

BA

30 lb/ft

20 lb/ft

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5-17

ExampleHere is an example of how the shape of the girder reflects theshear and bending diagrams.

A

B

L

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5-18

So why did they put that gap in the bridge?

BA

BA

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5-19

Pins

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5-20

ExampleDraw the shear and bending-moment diagrams for the beam andloading shown. Label all points of change, maximums andminimums, and the axes. The addition of the internal pin at thecenter of the beam allows additional head room because rather

than the moment being a maximum in the center it becomes zero.This design is used at Wings Air West in SLO. Total span= 75 ft.

500 lb/ft

A B

500 lb/ft

AB

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5-21

ExampleDraw the shear and bending-moment diagrams for the beam andloading shown. Label all points of change, maximums and minimums,and the axes. This example demonstrates that with the addition of aninternal pin we get an additional equation, otherwise we would have too

many unknowns.

Support ReactionsC

A B

3

3.5 1 2

C

B

3

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5-22 Design of Prismatic Beams for Bending

DESIGN OF PRISMATIC BEAMS FOR BENDING

ExampleDesign the cross sections minimum height of the beam, knowing thatthe grade of wood used has an allowable bending stress of 12 MPa.

Units: N, m.

6

BA

3000

1.5 1.5

3000

h

100

V(N)

M(Nm)

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5-23

Cross-section

ExampleKnowing that the grade of copper used has an allowable bendingstress of 15 ksi, determine the minimum wall thickness for the 6"diameter pipe. Units: lb, ft.

3

700500 lb/ft

V(lb)

M

(kin)

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5-24

ExampleKnowing that the allowable bending stress for steel is 160 MPa,determine the most economical W410-shape to support the load.Units: kN, m.

20 kN/m

6

BA

100

1.5 1.5

100

Cross-section

X X

V

(kN)

M(kNm)

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5-25

Cross-section

AB

6.5

50050 lb/ft

4

ExampleKnowing that the allowable bending stress for steel is 24 ksi, determinethe most economical C7-shape to support the load.Units: lb, ft.

Y Y

V(lb)

M(lbft)

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5-26

6

BA

30 kN/m

ExampleKnowing that the allowable bending stress for steel is 160 MPa,determine the most economical W310-shape to support the load.Units: kN, m.

80

3Cross-section

X X

V(kN)

M(kNm)

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5-27

ExampleTwo rolled-steel C150 channels are welded back to back. Knowingthat the allowable bending stress for steel is 160 MPa, determine themost economical channels to support the load.Units: kN, m.

6

BA

8

1.5

5 4

1.5 1.5

Cross-section

X X

V

(kN)

M

(kNm)

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SUMMARY

Internal Forces in Members

Shear and Bending-Moment Diagrams

Design of Prismatic Beams for Bending

A

B

A

B

C

P

A E

F

DB

C

5'

10'

4'

8'

80' 40'40'

BA

P P

C D

80' 40'40'

BA

P P

C D

Cross-section

A

Sign Convention

V

M

M V x = V w x =