Shear Force And Bending Moment In Beams

1

Mechanics of Materials

Chaper4

Shear Force and Bending Moment in Beams

HENAN UNIVERSITY OF SCIENCE & TECHNOLOGY

Edited by YANG MIN-xian

2

§4–1 Concepts of planar bending and calculation sketch of the beam

§4–2 The shearing force and bending moment of the beam

§4–3 The shearing-force and bending-moment equations · the shearing-

force and bending-moment diagrams

§4–4 Relations among the shearing force 、 the bending moment and the

density of the distributed load and their applications

§4–5 Plot the bending-moment diagram by the theorem of

superpositiom

§4–6 The internal-force diagrams of the planar rigid theorem frames

CHAPTER 4 Shear Force and Bending Moment in BeamsCHAPTER 4 Shear Force and Bending Moment in Beams

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§4–1 CONCEPTS OF PLANAR BENDING AND

CALCULATION SKETCH OF THE BEAM

§4–1 CONCEPTS OF PLANAR BENDING AND

CALCULATION SKETCH OF THE BEAM

1 、 CONCEPTS OF BENDING

1). BENDING: The action of the external force or external the couple vector

perpendicular to the axis of the rod makes the axis of the rod change into

curve from original straight lines, this deformation is called bending.

2).BEAM ： The member of

which the deformation is

mainly bending is generally

called beam.

4

3).Practical examples in engineering about bending

5

6

4).Planar bending ： After deformation the curved axis of the beam is still in the same plane with the external forces.

Symmetric bending （ as shown in the following figure ）— a special example of the planar bending.

The plane of symmetry

M

P1 P2q

7

Unsymmetrical bending— if a beam does not possess any plane of

symmetry, or the external forces do not act in a plane of symmetry of the

beam with symmetric planes, this kind of bending is called unsymmetrical

bending. In later chapters we will mainly discuss the bending stresses and

deformations of the beam under symmetric bending.

8

2 、 Calculation sketch of the beam

In general supports and external forces of the beam are very complex.

We should do some necessary simplification for them for our convenient

calculation and obtain the calculation sketch.

1). Simplification of the beams

In general case we take the place of the beam by its axis.

2). Simplification of the loads

The loads (including the reaction) acting on the beam may be reduced into

three types ： concentrated force 、 concentrated force couple and distributed

force.3). Simplification of the supports

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①Fixed hinged support

2 constraints ， 1 degree of

freedom. Such as the fixed

hinged support under bridges ，

thrust ball bearing etc.

②Movable hinged support

1 constraint ， 2 degree of

freedom. Such as the movable

hinged support under the

bridge ， ball bearing etc.

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③Rigidly fixed end

3 constraints ， 0 degree of

freedom. Such as the support of

diving board at the swimming pool ，

support of the lower end of a

wooden pole.

XA

YA

MA

4) Three basis types of beams

①Simple beam(or simply

supported beam)

M —Concentrated force couple

q(x)—Distributed force

②Cantilever beam

A

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③Overhanging beam—Concentrated

forcePq— Uniformly

distributed force

5). Statically determinate and statically indeterminate beams

Statically determinate beams ： Reactions of the beam can be determined only

by static equilibrium equations ， such as the above three kinds of basic beams.

Statically indeterminate beams ： Reactions of the beam cannot be

determined or only part of reactions can be determined by static equilibrium

equations.

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Example 1 A stock tank is shown in the figure. Its length is L=5m ， its

inside diameter is D=1m ， thickness of its wall is t =10mm. Density of steel is

7.8g/cm³. Density of the liquid is 1g/cm³. Height of the liquid is 0.8m. Length of

overhanging end is 1m. Try to determine the calculation sketch of the stock tank.

Solution ：

q— UniformlyDistributed force

13

LgLAgLA

LgV

Lmgq 2211

rad85513106 0 ..

gRRgDt 222

1 )]sin(2

1[

gAgA 2211 (kN/m) 9

9.81000)]sin106.3(1.8550.521

0.5[3.148978000101143

2

2

...

q— UniformlyDistributed force

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§4–2 THE SHEARING FORCE AND BENDING MOMENT OF THE BEAM

§4–2 THE SHEARING FORCE AND BENDING MOMENT OF THE BEAM

1 、 Internal force in bending ： Example Knowing conditions are P，

a ， l , as shown in the figure.

Determine the internal forces on the

section at the distance x to the end A.

Pa

P

l

YA

XA

RB

A

A B

B

Solution ：① Determine external forces

l

alPYY

l

PaRm

XX

ABA

A

)( , 0 , 0

0 , 0

15

A BP

YA

XA

RBm

m

x

②Determine internal forces— method of section

xYMm

l

alPYQY

AC

A

, 0

)( , 0

A

YA

Q

M

RB

P

M

Q

Internal forces of the beam in bending

Shearing force

Bending moment

1). Bending moment ： M

Moment of the internal force couple with

the acting plane in the cross-section

perpendicular to the section when the beam is

bending.

C

C

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2). Shearing force ： Q

Internal force which the acting line in the cross-section parallel to the section, when the beam is

bending.

3).Sign conventions for the internal forces:

① Shearing force Q: It is positive when it results in a clockwise rotation with respect to the object under consideration, otherwise it is negative.

②Bending moment M ： It is positive when it tends to bend the portion concave upwards, otherwise it is negative.

Q(+) Q(–)

Q(–)Q(+)

M(+) M(+)

M(–) M(–)

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Example 2 ： Determine the internal forces acting on sections 1—1 and 2—2

section as shown in fig.(a).

qLQ

QqLY

1

1

0

Solution ： Determine internal

forces by the method of section.

Free body diagram of the left portion of

section 1—1 is shown in fig. （ b ） .

Fig. （ a ）

11

11

0)(

qLxM

MqLxFm iA

2 、 Examples

qqL

a b1

12

2

qL

Q1

A

M1

Fig. （ b ）x1

18

L)axq Q 22 (

axqMqLx

Fm iB

0)(21

, 0)(

2222

Free body diagram of the left portion of section 2—2 is shown in fig. （ b ） .

)ax(qQqLY 022

22

22 )(21 qLxaxqM

xy

图（ a ）

qqL

a b1

12

2

qL

Q2

B

M2

x2

图（ c ）

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1. Internal-force equations: Expressions that show the internal forces as functions of the position x of the section..

2. The shearing-force and bending-moment diagrams:

)(xQQ Shearing force equation

)(xMM Bending moment equation

)(xQQShearing-force diagram sketch of the shearing-force equation

)(xMM

Bending Moment diagram sketch of the bending-moment equation

§4–3 THE SHEARING-FORCE AND BENDING-MOMENT EQUATIONS THE SHEARING-FORCE AND BENDING-MOMENT DIAGRAMS§4–3 THE SHEARING-FORCE AND BENDING-MOMENT EQUATIONS THE SHEARING-FORCE AND BENDING-MOMENT DIAGRAMS

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Example 3 Determine the internal-force equations and plot the diagrams of the

beam shown in the following figure.

PY)x(Q O

Solution ：① Determine the reactions of the supports

)Lx(P

MxY)x(M OO

②Write out the internal- force equations

PL MPY OO ;

P

③Plot the internal- force diagrams

Q(x)

M(x)

x

x

P

–PL

YO

L

M(x)

xQ(x)

MO

⊕

○

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Solution ：① Write out the internal-force equations

②Plot the internal-force diagram

qx)x(Q

2

21 qx)x(M

L

q

M(x)

xQ(x)

Q(x)

x

– qL

2

2qL

M(x) x

⊕

○

22

)3(6

220 xLL

q)x(Q

Solution ：① Determine the reactions of the supports

② Write out the internal- force equations

3 ;

600 Lq

RLq

R BA

q0

RA

③ Plot the internal- force diagrams

RB

L

)xL(Lxq

xM 220

6)(

x

L33

Q(x)

x6

20Lq

30Lq

273 2

0Lq

M(x)

⊕

⊕

○

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1 、 Relations among the shearing force 、the bending moment and the the distributed load

By analysis of the equilibrium of the

infinitesimal length dx ， we can get

0dd

0

)x(Q)x(Qx)x(q)x(Q

Y

)x(Qx)x(q dd

§4–4 RELATIONS AMANG THE SHEARING FORCE, THE BENDING MOMENT

AND THE INDENSITY OF THE DISTRIBUTED LOAD AND THEIR APPLICATIONS §4–4 RELATIONS AMANG THE SHEARING FORCE, THE BENDING MOMENT

AND THE INDENSITY OF THE DISTRIBUTED LOAD AND THEIR APPLICATIONS

dxx

q(x)

q(x)

M(x)+d M(x)

Q(x)+d Q(x)

Q(x)

M(x)

dx

A

y xqx

xQ

d

d

Slope of the tangential line at a point in the

shearing-force diagram is equal to the intensity

of the distributed load at the same point.

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q(x)

M(x)+d M(x)

Q(x)+d Q(x)

Q(x)

M(x)

dx

A

y

0)](d)([)())(d(2

1)d( 0,)( 2 xMxMxMxxqxxQFm iA

)(d

)(dxQ

x

xM

Slope of the tangential line at a point in the bending-moment diagram is

equal to the magnitude of the shearing force at the same point.

)(d

)(d2

2

xqx

xM

Relation between the bending moment and the indensity of the distributed load ：

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2 、 Relations between the shearing force 、 the bending moment and the external load

Extern

al force

No external-force segment

Uniform-load segment

Concentrated force Concentrated couple

q=0 q>0 q<0

Ch

aracteristics of Q-

diagram

Ch

aracteristics of M-

diagram

C

P

C

m

Horizontal straight line

x

Q

Q>0

Q

Q<0

x

Inclined straight line

Increasing function

x

Q

x

Q

Decreasing function

x

Q

CQ1

Q2

Q1–Q2=P

Sudden change from the left to right

x

Q

C

No change

Inclined straight line

x

MIncreasing function

x

MDecreasing function

curves

x

MTomb-like

x

MBasin-like

Flex from the left to the right

Sudden change from the left to the right

Op

posite tom

x

MFlex opposite to P

M

xM1M

2

mMM 21

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Simple method to plot the diagram: The method to plot the diagrams by

using the relation between the internal forces and the external forces and values

of the internal forces at some special points.

Example 4 Plot the internal force diagrams of the beams shown in the

following figures by the simple method to plot the diagram.

Solution:

Special points:a a

qa q

A

Plot the diagram by using the relation between the internal forces and the external forces and the internal force values at

some special points of the beam.

End point 、 partition point （ the point at which external forces changed ） and

stationary point etc.

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2

2

30 qaM;Q

0 ; MqaQ

2 ; qaMqaQ

2

2

3; 0 qaMQ

a a

qa q

ALeft end ：Shape of the curve is determined according to

)(d

)(d xQx

xM )(d

)(d2

2

xqx

xM

； xqxxQ

dd

；

And the law of the point acted by concentrated force.

Partition

point A ：Stationary

point of M ：

Right end ：

Q x

2

23 qa

qa2

–

qa

–

xM

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Example 5 Plot the internal-force diagrams of the beams shown in the

following figures by the simple method to plot the diagram.

Solution ：Determine reactions

2

; 2

qaR

qaR DA

0;2

Mqa

QLeft end A ：

2

2

1;

2qaM

qaQ

2

2

1;

2qaM

qaQ Right of

point B ：2

2

1;

2qaM

qaQ Left of

point C ：Stationary

point of M ：2

8

3; 0 qaMQ

2

2

1;

2qaM

qaQ Right of

point C ：0 ; 2

1 MqaQRight end D ：

q qa2

qaRA RD

Q x

qa/2 qa/2

qa/2

– –

+

A B

C D

qa2/2

xM qa2/2

qa2/23qa2/8–

+

Left of point B ：

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§4–5 PLOT THE DIAGRAM OF BENDING MOMENT BY THE THEOREM OF SUPERPOSITIOM

§4–5 PLOT THE DIAGRAM OF BENDING MOMENT BY THE THEOREM OF SUPERPOSITIOM

1 、 Theorem of superposition ： Internal forces in the structure due to simultaneous action of many forces are

equal to algebraic sum of the internal forces due to separate action of each force.

)()()()( 221121 nnn PQPQPQPPPQ

)()()()( 221121 nnn PMPMPMPPPM

Applying condition ： Relation between the parameters

(internal forces 、 stresses 、 displacements ） and the

external forces must be linear, that is they satisfy Hooke’s

law.

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2 、 Structural members in mechanics of material is of small

deformation and linear elasticity, and must obey this principle

—— method of superposition

Steps ： ①Plot respectively the diagram of the bending moment of

the beam under the separate action of each external load ； ②Sum up the corresponding longitudinal coordinates

(Attention: do not simply piece together figures. ）

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Example 6 Plot the diagram of bending moment by the principle of superposition.

(AB=2a ， force P is acting at the middle point of the beam AB. ） P

q

q

P =+

A

A

A

B

B

B x

M2

x

M1

x

M

2Pa

+

+

+

2

2qa

22

2qaPa =+

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3 、 Applications of symmetry and antisymmetry ： For the symmetric structure under the action of symmetric loads

the diagram of its shearing stress Q is antisymmetric and the

diagram of the bending moment M is symmetric. For the symmetric

structure under the action of antisymmetric loads the diagram of its

shearing stress Q is symmetric and the diagram of the bending

moment M is antisymmetric.

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Example 7 Plot internal-force diagrams of the beams shown in the

following figure.PPL

P

PL

L L

L L

L L

0.5P

0.5P

0.5P

0.5P

P0

Qx

Q1 x

Q2 x

–

0.5P

0.5P

0.5P

–+

–P

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PPL

P

PL

L L

L L

L L

0.5P

0.5P

0.5P

0.5P

P0 M

x

M1

x

M2

x

0.5PL

PL

0.5PL–

+

+

0.5PL

+

35

Example 8 Correct the mistakes in the following internal-force diagrams.

a 2aa

qqa2

AB

Qx

x

M

– –

+

+qa/4 qa/4

3qa/4

7qa/4

qa2/4

49qa2/32

3qa2/2

5qa2/4

4

7

;4qa

R

qaR

B

A

RA RB

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Example 9 Knowing Q-diagram, determine external loads and M-diagram (Therefore no concentrated force couples acted on the beam).

M(kN·m)

Q(kN)x

1m 1m2m

2

3

1

5kN 1kN

q=2kN/m

+

–

+

x+

1

1

1.25

–

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§4–6 THE INTERNAL-FORCE DIAGRAMS OF THE PLANAR RIGID FRAMES§4–6 THE INTERNAL-FORCE DIAGRAMS OF THE PLANAR RIGID FRAMES

1 、 Planar rigid frame

1). Planar rigid frame ： Structure made from rods of different direction

that are mutually connected in rigidity at their ends in the same plane.

Characteristics ： There are internal forces Q, M and N in each rod.2). Conventions to plot diagram of internal forces ： Bending-moment diagram ： Plot it at the side where fibers are

elongated and not mark the sign of positive or negative.

Shearing-force and axial-force diagrams ： May be plotted

at any side of the frame （ In common the diagram with positive value is

plotted outside the frame ）， but must mark the signs of positive and

negative.

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Example 10 Try to plot the internal-force diagrams of the rigid frame

shown in the figure.P1P2

a

l

A

B C

– N-diagram

Q -diagram

P2 +

P1

+

P1

M -diagram

P1a

P1a

P1a

+ P

2 l

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1 、 Method to determine directly the internal forces ： When we determine the internal forces in an arbitrary section A, we can take

the left part of section A as our study object and use the following formulas to

calculate internal forces. where Pi and Pj are respectively upward and

downward external forces acted on the left part.

DIAGRAMS OF SHEARING STRESSES AND BENDING MOMENTS

EXERCISE LESSONS ABOUT INTERNAL FORCES OF BENDING EXERCISE LESSONS ABOUT INTERNAL FORCES OF BENDING

jiA PPQ

)( )( jAiAA PmPmM

40

)(d

)(d2

2

xqx

xM

Relations among the shearing force 、 the bending moment and the external load ：

q(x)

xqxxQ

dd

)(d

)(d xQx

xM

2 、 Simple method to plot the diagram:

The method to plot the diagrams by using the relation between the internal forces

and the external forces and using values of the internal forces at some special

points.

41

33 、、 Principle of superpositionPrinciple of superposition ： Internal forces in the structure due to simultaneous action of many forces are

equal to the algebra sum of the internal forces due to separate action of each

force.

)()()()( 221121 nnn PQPQPQPPPQ

)()()()( 221121 nnn PMPMPMPPPM 4 、 Applications of symmetry and antisymmetry ： For the symmetric structure under the action of symmetric loads the diagram of its

shearing stress is antisymmetric and the diagram of bending moment is symmetric.

For the symmetric structure under the action of antisymmetric loads the diagram of

its shearing stress is symmetric and the diagram of bending moment is

antisymmetric

42

5 、 Relations between the shearing force, the bending moment and the external load

Extern

al force

No external-force segment

Uniform-load segment

Concentrated force Concentrated couple

q=0 q>0 q<0

Ch

aracteristics of Q-

diagram

C

P

C

m

Horizontal straight line

x

Q

Q>0

Q

Q<0

x

Inclined straight line

Increasing function

x

Q

x

Q

Decreasing function

x

Q

CQ1

Q2

Q1–Q2=P

Sudden change from the left to right

x

Q

C

No change

Inclined straight line

x

MIncreasing function

x

MDecreasing function

curves

x

MTomb-like

x

MBasin-like

Flex from the left to the right

Sudden change from the left to the right

Op

posite tom

x

MFlex opposite to P

M

xM1M

2

mMM 21

43

Example 1 Plot the bending-moment diagrams of the beam shown in the

following figure.2P

a aP

=

2P

P

+

xM

xM1

x

M2

=+

–

+

+

2Pa

2Pa

Pa

(1)

44

(2)a

a

q

q

q

q

=+

xM1

=

xM

+

–

+

–

xM2

3qa2/2

qa2/2

qa2

45

(3) P

L/2 L/2

PL/2

=+

P

xM2

xM

=+

PL/2

PL/4

PL/2

xM1

–

+

–PL/2

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(4) 50kN

2m 2m

20kNm

=+

xM2

xM

=+

20kNm

50kNm

xM1

20kNm

50kN

20kNm20kNm

+

+

–

20kNm

30kNm

20kNm

47y

z

h

b )4

(2

22

yhI

QbIQS

zz

z

Solution: （ 1 ） Shearing stress on the cross section is

Example 2 The structure is shown in the figure. Try to prove ： (1 ） resultant of

the shearing stresses in an arbitrary cross section is equal to the shearing force in

the same section ；（ 2 ） Resultant moment of the normal stresses in an arbitrary

cross section is equal to the bending moment in the same section ；（ 3 ） which

force can balance the resultant of the shearing stress in the longitudinal section at

middle height balanced ?.

q

Normal stress on the cross section is

zI

My

48

h

h zA

ybyh

I

QA

5.0

5.0

22

d )4

(2

d

MIIMA

IMyM z

z

h.

h. zz

50

50

2

d

(2) Resultant shearing force in the cross section is ：

Qhh

I

Qb

z

])2

(3

2

4[

23

3

(3) Resultant force couple

49

)(bhqx.

AxQ. 51)(51max

h

qLxqx

hAQ

LL

AB 4

3d)(

2

3d

2

00

zA W

AMAN

22

1 1max1max 1

1AAB NQ

(4)Shearing stress in the

middle longitudinal section is ：

Resultant of the shearing stress in the longitudinal section is balanced by resultant of the normal stress in the right-side section.

(5)Resultant of the shearing stress in the longitudinal section is ：

max

h

qLbh

bh

qL

4

3

2

6

22

1 2

2

2

x L

50