Chapter 6 Shear and Moments in Beams

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Simple beam Cantilever beam

Overhang beam

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Chapter

6

SHEAR AND MOMENT IN BEAMS6.1 Introduction

6.2 Clai!ication o! Bea"

6.# Calculation o! Bea" Reaction

6.$ Shear %orce and Bendin& Mo"ent

6.' (oad) Shear) and Mo"ent

Relationhip

6.6 Shear and Mo"ent Dia&ra"

6.* Dicontinuit+ %unction

6.1 Introduction,

The term

tran-ere

refers to

load and

sections that

are perpendicul

ar to their

longitudinal

axis of the

member.

• The behavior of slender members subjected to axial loads and

to torsional loadings was discussed in Chapters 4 and 5,

respectively. In this chapter, the consideration of beams that are

usually long, straight, prismatic members, which are support

transverse loads. They resist transverse applied loads by a

combination of internal shear force and bending moment.

• A beam may be defined as a member whose length is relatively

large in comparison with its thickness and depth, and which is

loaded with transverse loads that produce significant bending

effects as oppose to twisting or axial effects.

6.2 Clai!ication o! Bea",

embers that are slender and support loadings applied perpendicular to their

longitudinal axis are called beams. !eams can be classified by the manner in which

the supports are arranged into"a. #imply supported beam $simple beam%.

b. Cantilever beam $fixed end beam%.

c. !eam with an overhang.

6.# Calculation o! Bea" Reaction,

• &'uations of static e'uilibrium are used to determine the reaction forces of

a loaded beam.

• #elf weight of the beam is usually neglected unless otherwise specified.

T+pe o! load,

a. concentrated load $single force%

b. distributed load $measured by their intensity% " uniformly

distributed load $uniform load%, and linearly varying load.

c. couple



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• Reaction, to determine the reactions, consider the loaded beams in figure (.).

*. #imply supported beam"

e'uation of e'uilibrium in

hori+ontal direction"

). Cantilever beam"

. -verhanging beam"



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6.$ Internal Shear %orce and Bendin& Mo"ent,

Consider a cantilever beam shown in igure (.4 with a concentrated load

P applied at the end /, at the cross section mn,

the shear force and bending moment are found

as"

∑ F

y=0⇒V = P

∑ M =0⇒ M = Px

Sign conventions (deformation

sign conventions): the sign

conventions are shown in the

igure (.5.

i.

A /0

internal hear !orce V0,

♦ acts downward on the right0hand face of a beam.

♦ acts upward on the lift0hand face of a beam.

/ $1% internal shear force $V %" tends to rotate the material clockwise is defined

as positive.

ii. A /0 internal endin& "o"ent M0,

♦ acts countercloc2wise $CC3% on the right0hand face of a beam. ♦ acts cloc2wise$C3% on the lift0hand face of a beam.



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/ $1% internal bending

moment $ M % bends a beam

element concave upward

$tends to compress theupper part of the beam and

elongate the lower part%.

Ea"ple 6.1,

TheW 360× 79

rolled0steel beam /C is

simply supported and carries the uniformly

distributed load shown.

Deter"ine, write shear0force and bending0

moment e'uation for an arbitrary section in

the interval !C of the beam. Calculate the

magnitude of for a distance *.5 m

from the left support !.

Solution,



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Ea"ple 6.2,

The structure shown consists of a W 10

×112

rolled0steel beam /! and of two short members

welded together and to the beam.

Deter"ine, write shear0force and bending0

moment e'uation for an arbitrary section in the

interval C6 of the beam. Calculate the

magnitude of for at 6.

Solution,



Figure 6.7

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6.' (oad) Shear) and Mo"ent Relationhip,

Consider an element of a beam of length dx subjected to distributed loads w ,

equilibrium of forces in vertical direction"

∑ F y=0∴V +wdx−(V +dv )=0

⇒dv

dx=w

Integration above e'uation, give the following"

∫ A

B

dv=∫ A

B

wdx⇒V B−V A=∫ A

B

wdx=(area of the loading diagram between A∧B) ,

the area of loading diagram may be positive or negative.

Moment equilibrium of the element:

∴∑ M =0∴− M +wdx

(dx2 )−(V +dV ) (dx )+( M +dM )=0

-r,

dM

dx =V

Integration above e'uation, give the following"

∮ A

B

dM =∫ A

B

Vdx



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M B− M A=∫ A

B

Vdx=(area of the shear−force diagrambetween A∧B)

This e'uation is valid even when concentrated loads act on the beam between / and

!, but it is not valid if a couple acts between / and !.

6.6 Shear and Mo"ent Dia&ra",

• The bending couple M creates normal

stresses in the cross section, while the

shear force V creates shearing

stresses in that section.

• In most cases the dominant

criterion in the design of a beam for strength

is the maximum value of the normal

stress in the beam.

• The determination of the normal

stresses and shearing stresses in a beam

will be discussed in Chap. 7.

#ince the distribution of these

stresses in a given section depends only upon the values of the bending

moment and the shear force in that section.

• /s indicated above, the determination of the maximum absolute values

of the shear and of the bending moment in a beam are greatly

facilitated if and are plotted against the distance x measured from

one end of the beam.

• !esides, as you will see in Chap. 8, the 2nowledge of as a function

of x is essential to the determination of the deflection of a beam.

9ere we ta2e up the e'uilibrium and graphical methods of

concentrating shear and moment diagrams"

*. &'uilibrium ethod" /lso called as the direct approach, as discussed

earlier it consists of"

Calculating the reactions,



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3riting algebraic expressions for and , and

Constructing the curves from these e'uations.

). :raphical ethod" /lso called as the summation approach involves"

Treating the entire beam as a rigid body, determine the reaction

forces. #hear force and bending couple at a point are determined by passing a

section through the beam and applying an e'uilibrium analysis on the

beam portions on either side of the section.

6etermine the value of at the change 0of ;load points, successively

summing from the left end of the beam the vertical external forces.

6raw the shear diagram and locate the +ero points.

6etermine the value of at the change 0of ;load points and at a

points of +ero shear, either continuously summing from the left end of

the beam the external moments. 6raw the moment diagram.

#ee table (.* in your textboo2 $p)47%

• In the examples and sample problems of this section, the shear and

bending0moment diagrams will be obtained by determining the values

of V and M at selected points of the beam. These values will be found

in the usual way, i.e., by passing a section through the point where

they are to be determined and considering the e'uilibrium of the

portion of beam located on either side of the section.

• These conventions can be more easily remembered as discussed

earlier.

SHEAR AND MOMENT DIA3RAMS,

4rocedure !or anal+i,

a) Support reactions

• 6etermine all reactive forces and couple moments acting on beam

• <esolve all forces into components acting perpendicular and parallel

to beam=s axis

b) Shear and moment functions• #pecify separate coordinates x having an origin at beam=s left end,

and extending to regions of beam between concentrated forces and>or

couple moments, or where there is no discontinuity of distributed

loading.

• #ection beam perpendicular to its axis at each distance x

• 6raw free0body diagram of one segment

• a2e sure V and M are shown acting in positive sense, according to

sign convention

• #um forces perpendicular to beam=s axis to get shear

• #um moments about the sectioned end of segment to get moment



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c) Shear and moment diagrams

• ?lot shear diagram $ vs. x% and moment diagram $ vs. x%

• If numerical values are positive, values are plotted above axis,

otherwise, negative values are plotted below axis

•

It is convenient to show the shear and moment diagrams directly below the free0body diagram

Ea"ple 6.#,

6raw the shear and moment diagrams for beam shown below.

Support reactions: Shown in free-body

diagram.

Shear and moment functions: Since there is a

discontinuity of distributed load and a concentrated loadat beam’s center, two regions of must be considered.



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Ea"ple 6.$,

6raw the shear and bending0moment diagrams

for the beam and loading shown, and determine

the maximum absolute value $a% of the shear, $b%

of the bending moment.

SO(5TION,



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Ea"ple 6.$,

6etermine $a% the e'uations of the shear and

bending0moment curves for the beam and

loading shown, $b% the maximum absolute value

of the bending moment in the beam.

SO(5TION,

Ea"ple 6',

an overhanging beam is subjected to a

uniform load of ' @ * 2b>ft on /! and

a couple A @ *) 2ib0ft on midpoint of !C, construct the 0 and diagrams

for the beam.



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Ea"ple 66,



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6raw the shear and bending moment

diagrams for the beam and loading shown.

SO(5TION,

Ta2ing the entire beam as a free body, determine the reactions at A

and .

/pply the relationship between

shear and load to develop the shear

diagram.

/pply the relationship between

bending moment and shear to

develop the bending moment

diagram.

0 +ero slope between concentrated

loads

0 linear variation over uniform loadsegment



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Ea"ple 6*,



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6raw the shear and bending moment diagrams

for the beam and loading shown.

SO(5TION,

6.2 3RA4HICA( METHOD ,

/ simpler method to construct shear and moment diagram, one that is based

on two differential e'uations that exist among distributed load, shear and

moment.

Regions of distributed load:

!hange in shear force, "#$% & $rea under shear diagram

∆ M AB=∫Vdx

!hange in shear force, "#$% & $rea under shear diagram



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6.* Dicontinuit+ %unction,

'e can simplify the calculations by epressing the bending

moment in terms of discontinuity functions, also (nown as

)acaulay brac(et functions. *iscontinuity functions enable

us to write a single epression for the bending moment that

is +alid for the entire length of the beam, e+en if the loading

is

discontinuous.$s an eample,

consider the

beam loaded

as shown in Fig.

6.a. he

free-body

diagrams of the three segments of the beam are shown in

Figs. 6.b/d.

Note that : the reactions at $ and * ha+e already beencompleted using e0uilibrium analysis. sing the e0uilibrium



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e0uation

for each

segment

the

moment ista(en about the right end of the segment, we obtain the

net bending moments:

2ote that in each successi+e segment an etra term is added

to M, while the rest of the epression remains unchanged.

his pattern suggests using the epression:

for the entire beam, with the

understanding that the term -

3 disappears when x ≤2 and

( x−3)2 , disappears when

x ≤3 .

his idea is formali4ed by using

the )acaulay brac(et functionsdescribed below.

$ )acaulay brac(et function, often referred to as a 55brac(et

function,’’ is dened as:

⟨ x−a ⟩ n={ 0 if x<a

( x−a )nif x≥ a

where n is a nonnegative integer.

⟨ x−a ⟩0

={0when x<a1when x≥ a

⟨ x−a ⟩1

={ 0when x<a

x−awhenx≥a

The brac2ets ⟨.. ⟩ , identify the expression as a brac2et function.

Note that " a brac2et function is +ero by definition if the expression in the

brac2etsBnamely, ( x−a )

B is negative otherwise, it is evaluated aswritten. / brac2et function can be integrated by the same rule as an ordinary



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functionBnamely,

∫ ⟨ x−a ⟩n dx=⟨ x−a ⟩n

n+1forn≥0

Sin&ularit+ %unction,

Dsed for applied load"

w ( x )= P ⟨ x−a ⟩−1

={0 if x ! a

P if x=a

Dsed for applied moment"

w ( x )= M o ⟨ x−a ⟩−2

={ 0 if x! a

M o if x=a

The rule of integration for #"

∫ ⟨ x−a ⟩n dx=⟨ x−a ⟩ n+1forn<0(n=−1,−2,..)

Chapter 6 Shear and Moments in Beams

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