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Mechanics of Materials

Chapter 4

Shear and Moment In Beams

Introduction

The term beam refers to a slender bar that carries

transverse

loading

; that is, the applied force are perpendicular to the bar.

In a beam, the internal force system consist of a shear force and a bending moment acting on the cross section of the bar. Theshear force and the bending moment usually vary continuouslyalong the length of the beam.

The internal forces give rise to

two

kinds of stresses on a

transverse section of a beam: (1) normal stress that is caused bybending moment and (2) shear stress due to the shear force.

Knowing the distribution of the shear force and the bending moment in a beam is essential for the computation of stressesand deformations. Which will be investigated in subsequentchapters.

Figure 4.

Statically determinate beams

A

cantilever beam

is built into a rigid support at one end, with the

other end being free, as shown in Fig.4.1(b). The built-in supportprevents displacements as well as rotations of the end of the beam.

An

overhanging beam

, illustrated in Fig.4.1(c), is supported by a

pin and a roller support, with one or both ends of the beamextending beyond the supports.

The three types of beams are

statically determinate

because the

support reactions can be found from the equilibrium equations.

( )

g

g

( )

g

g

(c) 2003 Brooks/Cole Publishing / Thomson Learning™(c) 2003 Brooks/Cole Publishing / Thomson Learning™

A

concentrated load

, such as

P

in Fig. 4.1(a). In contrast a

distributed load

is applied over a finite area. If the distributed

load acts on a very narrow area, the load may be approximated bya

line load

The

intensity

w

of this loading is expressed as force per unit

length (lb/ft, N/m, etc.) The load distribution may be uniform, asshown in Fig.4.1(b), or it may vary with distance along the beam,as in Fig.4.1(c).

The weight of the beam is an example of distributed loading, butits magnitude is usually small compared to the loads applied tothe beam.

( )

g

g

( )

g

g

(c) 2003 Brooks/Cole Publishing / Thomson Learning™(c) 2003 Brooks/Cole Publishing / Thomson Learning™

4.3 Shear- Moment Equations and Shear-Moment Diagrams

‰

The determination of the internal force system acting at a

given

section of a beam : draw a free-body diagram that expose theseforces and then compute the forces using equilibrium equations.

The goal of the beam analysis

determine the shear force

V

and

the bending moment

M

at

every

cross section of the beam.

To derive the expressions for

V

and

M

in terms of the distance x

measured along the beam. By plotting these expressions to scale,obtain the

shear force and bending moment diagrams

for the

beam.

The shear force and bending moment diagrams are convenientvisual references to the internal forces in a beam; in particular,they identify the maximum values of

V

and

M

a.

Sign conventions

(

)

k /

l

bli hi

/

h

i

(

)

k /

l

bli hi

/

h

i

Figure 4.3 Sign conventions for external loads; shear force,

and bending moment.

Determine the expressions for

V

and

M

from the equilibrium

equations obtainable from the FBD. These expressions, whichare usually functions of x, are the shear force and bendingmoment equations for the segment.

Plot the expressions for V and M for the segment. It is visually desirable to draw the

V

-diagram below the FBD of the entire

beam, and then draw the

M

• diagram below the

V

-diagram.

The bending moment and shear force diagrams of the beam arecomposites of the

V

and

M

diagrams of the segments. These

diagrams are usually discontinuous, or have discontinuousslopes. At the end-points of the segments due to discontinuitiesin loading.

Sample Problem

The simply supported beam in Fig. (a) carries two concentratedloads. (1) Derive the expressions for the shear force and the bendingmoment for each segment of the beam. (2) Sketch the shear forceand bending moment diagrams. Neglect the weight of the beam.Note that the support reactions at

A

and

D

have been computed and

are shown in Fig. (a).

Solution Part 1 The determination ofthe expressions for

V

and

M

for each of the

three beam segments(

AB

BC

, and

CD

) is

explained below.

Segment AB

＜x＜

5 m)

F

y

V

V

= +18-14 = +4 kN

Answer

M

E

• 18x + 14(x-2) +

M

M

= +18x-14(x-2) = 4x+28 kN· m

Answer

Segment CD

(5 m

＜x＜

7 m)

F

y

V

V

= +18-14-28 = -24 kN

Answer

M

G

• 18x+ 14(x-2)+28(x-5)+

M

M

= +18x-14(x-2) – (x-5) = -24x+168 kN· m

Answer

Sample problem

The simply supported beam in Fig. (a) is loaded by the clockwisecouple

C

0

at

B

. (1) Derive the shear and bending moment

equations. And (2) draw the shear force and bending momentdiagrams. Neglect the weight of the beam. The support reactions Aand C have been computed, and their values are shown in Fig. (a).^ SolutionPart 1^ Due to the presenceof the couple

C

0,

We

must analyzesegments

AB

and

BC

separately.

Segment AB (

0＜x＜

L/4)

Answer

Answer

Segment BC (

L

4＜x＜L)

Answer

Answer 0

0

0

= − − ↑ + = ∑

V

L C

F

y

L C

V

0

−

=

∑

=

0

D

M

x

L C

M

0

−

=

0

0

=

M

x

L C

0

0

0

= − − ↑ + = ∑

V

L C

F

y

L C

V

0

−

=

∑

=

0

E

M

0

0

0

=

−

M

C

x

L C

0

0

C

x

L C

M

−

=

Sample Problem

The cantilever beam inFig.(a) carries a triangularload. The intensity ofwhich varies from zero atthe left end to 360 lb/ft atthe right end. In addition, a1000-lb upward verticalload acts at the free end ofthe beam. (1) Derive theshear force and bendingmoment equations. And(2) draw the shear forceand bending momentdiagrams. Neglect theweight of the beam.

Solution Note that the triangular load has beenreplaced by is resultant, which is theforce 0.5 (12) (360) = 2160 lb (areaunder the loading diagram) acting atthe centroid of the loading diagram.

Part 1 Σ

F

y

x

2

V

V

x

2

lb

Answer

M

C

x

x

2

x

M

M = 1000

x

x

3

lb· ft

Answer

Because the loadingis continuous, thebeam does not haveto be divided intosegment.

w

x

= 360/12, or

w

x

lb/ft.