cive1143 – structural analysis shear force and bending ......sep 17, 2017 · shear force and...

CIVE1143 – Structural Analysis

Shear Force and Bending Moment

Equations and Diagrams

Lecturers: Srikanth & Indu

Revision: Member Internal Forces

To determinate the internal forces at a

point such as B, we need to imaginarily cut

the beam through the point into two

segments. Thus the internal loading will be

exposed on the free-body diagram of the

segment.

A B

1P2P

xA

yAAM

xA

yAA B

AM

N

V

M

N

V

MB

1P2P

OR

Revision: Sign Convention

N N

N N

V V

V V

M M

M M

Positive axial force

(tension)Positive shear force Positive bending moment

N

V

M

N

V

M

xA

P

A BC

yAyB

B

yB

CNCV

CM

P

AxA

yA

CN

CV

CM

Revision: Calculation of the Internal Forces

xA

P

A BC

yAyB

P

AxA

yA

CN

CV

CM

• Identify a section in the structure where

you wish to determine the values of the

internal forces.

• Consider the free body diagram from

this section to the end of the beam

obtained by “cutting” the body through

this section. You may consider either the

right or the left part of the beam obtained,

whichever gives the easier calculation.

•Use the sign convention defined earlier

to provide the positive direction of these

forces. Treat the 3 internal forces at the

section as unknowns.

• 3 unknown internal forces can be

determined by 3 equilibrium equations.

∑ = 0xF+ ∑ = 0yF+ ∑ = 0BM+

B

yB

CNCV

CM

OR

V and M Functions

� Design of beam requires detailed knowledge of the

variations of V and M.

� Internal axial force N is generally not considered as

��The loads applied to a beam normally act perpendicular to the beThe loads applied to a beam normally act perpendicular to the beamam’’s s

axisaxis

��For design purpose, a beamFor design purpose, a beam’’s resistance to shear & bending is more s resistance to shear & bending is more

important than its ability to resist normal forceimportant than its ability to resist normal force

��An exception is when it is subjected to compressive axial force An exception is when it is subjected to compressive axial force where where

buckling may occurbuckling may occur

V and M Functions

� To obtain the variation of V and M, it is necessary to section the beam at

an arbitrary distance x from one end of beam. For example

5m

x

10kN/m

A B

The variation range of x:

m50 ≤≤ x

V and M Functions

5m

x

10kN/m

A B


m50 ≤≤ x

� The functions of V and M along the beam’s axis can be obtained by the

method of sections.

x

10kN/m

B

N

V

M

N

V

M

Sign Convention:

V

M

FBD:

(the right part)

The resultant force of distributed load is

equivalent to the area under the loading

diagram, and act through the geometric

center of this area.

xxarea 1010 =×=

10x

V and M Functions

5m

x

10kN/m

A B


m50 ≤≤ x

� The functions of V and M along the beam’s axis can be obtained by the

method of sections.

x

10kN/m

B

V

M

FBD:

(the right part)

10x

:0∑ =M+

xV 10=

25xM −=

02

10 =×−−x

xM

Equilibrium Equations:

:0∑ =yF+ 010 =− xV

V and M Diagrams

5m

x

10kN/m

A B


m50 ≤≤ x

� If the resulting functions of x are plotted, the graphs are termed the shear

force diagram and bending moment diagram.

xV 10=25xM −=

)pointat(0when Bx =

0010 =×=V

m5.2when =x

kN255.210 =×=V

)pointat(m5when Ax =

kN50510 =×=V

V

x 02.5m

25kN

5m

50kN

SFD:

Straight line

V and M Diagrams

5m

x

10kN/m

A B


m50 ≤≤ x

� If the resulting functions of x are plotted, the graphs are termed the shear

force diagram and bending moment diagram.

xV 10=25xM −=

)pointat(0when Bx =

005 2=×−=M

m5.2when =x

kNm25.315.25 2−=×−=M

)pointat(m5when Ax =

kNm12555 2−=×−=M

V

x 05m

50kN

SFD:

BMD: x

M

02.5m

-125kNm

5m

Parabola(open downward)

V and M Diagrams

5m

x

10kN/m

A B

� Bending moment diagram is

plotted on the compression side of

the beam.

V

x 05m

50kN

SFD:

BMD: x

M

0

-125kNm

5m

Parabola(open downward)

T

C

� UDL causes a linear change (a

straight line) to the SFD.

� UDL results in the BMD

assuming a parabolic curved shape.

Example

6kN/m

A B

9m

Draw the shear force and bending moment

diagrams for the beam.

Example

6kN/m



Reaction Forces:

6kN/m

A B

9m

FBD:





kN272

m9kN/m6=

×=area

A B

Ay By

27kN

3m

Example

6kN/m



Reaction Forces:

6kN/m

A B

9m

FBD: A B

Ay By

27kN

3m

:0∑ =BM+

kN18=yB

kN9=yA

03279 =×+×− yA

:0∑ =yF+ 027 =+− yy BA

Example

18kN



V and M Functions:

6kN/m

A B

9m

FBD: Ax

9kN

9kN

x

N

V

M

N

V

M

Sign Convention:

V

M

Establish an arbitrary distance x from point A

where m90 ≤≤ x

?

6kN/m

?

x9m

96

? x=

3

2?

x=

2x3

Example

18kN



V and M Functions:

6kN/m

A B

9m

FBD: Ax

9kN

9kN

x

V

M


where m90 ≤≤ x

2x3





33

2

2

1 2xx

xarea =××=

x3

x2

3

Example

18kN



V and M Functions:

6kN/m

A B

9m

FBD: Ax

9kN

9kN

x

V

M


where m90 ≤≤ x

2x3

x3

x2

3

:0∑ =M+

39

2x

V −=

xx

M 99

3

+−=

033

92

=+×+×− Mxx

x

:0∑ =yF+ 03

92

=−− Vx

Example

18kN



V and M Diagrams:

6kN/m

A B

9m

SFD:

9kN

xm)90( ≤≤ x

39

2x

V −=

)pointat(0when Ax =

kN9=V

)pointat(m9when Bx =

kN18−=V

x

V

9kN

-18kN

,0when =V

03

92

=−x

m2.5=x

5.2m

0

Example

18kN



V and M Diagrams:

6kN/m

A B

9m

SFD:

9kN

xm)90( ≤≤ xx

xM 9

9

3

+−=

)pointat(0when Ax =

0=M

)pointat(m9when Bx =

0=M

x

V

9kN

-18kN

,m2.5when =x

kNm2.31max =M

5.2m

0

31.2kNm

Discontinuous V and M Functions

�� The type or magnitude of the distributed load changesThe type or magnitude of the distributed load changes

�� Concentrated forces or couple moments are appliedConcentrated forces or couple moments are applied

In general, the internal shear force and bending moment functions will be

discontinuous or their slope will discontinuous at points where:

Because of this discontinuous property, V and M functions must be determined

for each segment of the beam located between two discontinuities of loading.

P

w

A B C D

3m 3m 4m

x1

x2

x3

3m0 1 ≤≤ x

Range of variations:

m63m 2 ≤≤ x

m106m 3 ≤≤ x

(from A to B)

(from B to C)

(from C to D)

Discontinuous V and M Functions

�� The type or magnitude of the distributed load changesThe type or magnitude of the distributed load changes

�� Concentrated forces or couple moments are appliedConcentrated forces or couple moments are applied

In general, the internal shear force and bending moment functions will be

discontinuous or their slope will discontinuous at points where:

Because of this discontinuous property, V and M functions must be determined

for each segment of the beam located between two discontinuities of loading.

P

w

A B C D

3m 3m 4m

3m0 1 ≤≤ x


m30 2 ≤≤ x

m40 3 ≤≤ x

(from A to B)

(from B to C)

(from C to D)

x1 x2 x3

Example

Draw the shear force and bending moment diagrams for the beam.

6kN

A B C

2m 4m

Example


6kN

A B C

2m 4m Reaction Force:

Ay Cy

:0∑ =CM+

kN2=yC

kN4=yA

0466 =×+×− yA

:0∑ =yF+ 06 =+− yy CA

Example


6kN

A B C

2m 4m

V and M Functions:

4kN 2kN

For Segment AB, establish an arbitrary distance

x1 from point A where m20 1 ≤≤ x

x1

A

x14kN

V

M

N

V

M

N

V

M

Sign Convention:

FBD::0∑ =M+

14xM =

kN4=V

04 1 =+×− Mx

:0∑ =yF+ 04 =−V

Example


6kN

A B C

2m 4m

V and M Functions:

4kN 2kN



x1

A

x24kN

V

M

N

V

M

N

V

M

Sign Convention:

FBD::0∑ =M+

14xM =

kN4=V

04 1 =+×− Mx

:0∑ =yF+ 04 =−V

For Segment BC, establish an arbitrary distance

x2 from point A where m6m2 2 ≤≤ x

x2

2m (x2 – 2)

:0∑ =M+

122 2 +−= xM

kN2−=V

0)2(64 22 =+−×+×− Mxx

:0∑ =yF+ 064 =−− V

6kN

Example


6kN

A B C

2m 4m

V and M Functions:

4kN 2kN

For Segment AB m)20( 1 ≤≤ x

x1

4kN

14xM =kN4=V

For Segment BC m)6m2( 2 ≤≤ x

x2

122 2 +−= xMkN2−=V

0

V and M Diagrams:

SFD:

-2kN

Example


6kN

A B C

2m 4m

V and M Functions:

4kN 2kN


x1

4kN

14xM =kN4=V

For Segment BC m)6m2( 2 ≤≤ x

x2

122 2 +−= xMkN2−=V

0

V and M Diagrams:

SFD:

-2kN

BMD: 0

)pointat(m2when 1 Bx = kNm8=M

)pointat(m6when 1 Cx =

kNm8=M)pointat(m2when 2 Bx =

0=M

)pointat(0when 1 Ax = 0=M

8kNm

Example – Another option


6kN

A B C

2m 4m

V and M Functions:

4kN 2kN



x1

FBD::0∑ =M+

14xM =

kN4=V

04 1 =+×− Mx

:0∑ =yF+ 04 =−V

For Segment BC, establish an arbitrary distance

x2 from point C where m4m0 2 ≤≤ x

x2

:0∑ =M+

22xM =

kN2−=V

02 2 =×+− xM

:0∑ =yF+ 02 =+V2kN

C

V

M

x2

N

V

M

N

V

M

Sign Convention:



6kN

A B C

2m 4m

V and M Functions:

4kN 2kN


x1

4kN

14xM =kN4=V

For Segment BC m)40( 2 ≤≤ x

22xM =kN2−=V

0

V and M Diagrams:

SFD:

-2kN

x2



6kN

A B C

2m 4m

V and M Functions:

4kN 2kN


x1

4kN

14xM =kN4=V

For Segment BC m)40( 2 ≤≤ x

22xM =kN2−=V

0

V and M Diagrams:

SFD:

-2kN

BMD: 0

)pointat(0when 1 Ax = 0=M

)pointat(m2when 1 Bx = kNm8=M

)pointat(m4when 1 Bx =

0=M)pointat(0when 2 Cx =

kNm8=M8kNm

x2

Example


6kN

A B C

2m 4m

Features of SFD:

4kN 2kN

4kN

0SFD:

-2kN

BMD: 0

8kNm

SFD follows the force arrows – if a

load or reaction points upward, move

upwards on SFD.

SFD should be a horizontal line where

no load is applied such as segment AB

and segment BC separately.

SFD jumps at where a point load is

applied, and the step equal to the value

of the load.

Example


6kN

A B C

2m 4m

Features of BMD:

4kN 2kN

4kN

0SFD:

-2kN

BMD: 0

8kNm

BMD follows the deformation shape

of the beam – plot BMD on the

compression side.

BMD should be a straight line where

no load is applied such as segment AB

and segment BC.

BMD has a sudden change in the slope

at where a point force is applied.

C

T

BMD always has a zero value at a end

pin or roller support, unless a moment

is applied at that point.

Complicate Examples

F1

A B Cx1

x2

F2

D

x3

F1

A B Cx1

x2

F2

D

x3

F1

A B Cx1

x2

F2

D

x3

OR

Summary

� Design of beam requires detailed knowledge of the variations of V

and M. Internal axial force N is generally not considered.

� To obtain the variation of V and M, it is necessary to section the

beam at an arbitrary distance x from one end of beam.

� If the resulting functions are plotted, the graphs are termed the

shear force diagram and bending moment diagram.

� Because of this discontinuous property, V and M functions must be

determined for each segment of the beam located between two

discontinuities of loading using x1, x2, ….(Note the variation range).

CIVE1143 – Structural Analysis

Shear Force and Bending Moment Equations and Diagrams

Lecturers: Srikanth & Indu

Revision: Discontinuous V and MFunctions

The type or magnitude of the distributed load changesThe type or magnitude of the distributed load changesConcentrated forces or couple moments are appliedConcentrated forces or couple moments are applied

In general, the internal shear force and bending moment functions will be discontinuous or their slope will discontinuous at points where:

Because of this discontinuous property, V and M functions must be determined for each segment of the beam located between two discontinuities of loading.

P

wA B C D

3m 3m 4m

x1

x2x3

3m0 1 ≤≤ x


m63m 2 ≤≤ x

m106m 3 ≤≤ x

(from A to B)

(from B to C)

(from C to D)

Revision: V and M Functions and DiagramsGeneral procedure for SFD and BMD

Determine all the Determine all the reactive forcesreactive forces and couple momentsand couple moments if necessary.if necessary.

Specify proper coordinators (Specify proper coordinators (xx11,, xx22,,……)) , draw the , draw the freefree--body diagramsbody diagramsfor corresponding segments.for corresponding segments.

Applying the equilibrium equations to obtain the Applying the equilibrium equations to obtain the shear force shear force functionsfunctions and and bending moment functionsbending moment functions for each segment (for each segment (VV((xx11), ), VV((xx22) ) …… and and MM((xx11), ), MM((xx22), ), ……). ).

Plot the Plot the shear force diagrams shear force diagrams andand bending moment diagramsbending moment diagramssegment by segment below the freesegment by segment below the free--body diagram of the structure. body diagram of the structure.

Features of SFD and BMD

StepHorizontalApplied moment

CubicParabolaLinearly varying distributed load

ParabolaStraight lineUDL

KinkStepPoint load

Straight lineHorizontalNo load

Shape of BMDShape of SFDType of load

Mathematical relationship between SFD and BMDConsider the small segment of beam subjected to only a uniform distributed load

w

V V+ΔV

M+ΔMM

Δx

wΔx Applying the force equation of equilibrium to segment, we have

:0∑ =yF+xwV Δ−=Δ

0)( =Δ+−Δ− VVxwV

wdxdV

−=

Dividing by Δx, and letting Δx → 0, we get

Slope of shear diagram = Negative of distributed load intensity


w

V V+ΔV

M+ΔMM

Δx

wΔx

Applying the moment equation of equilibrium to segment, we have

:0∑ =M+

2)( xwxVM Δ−Δ=Δ

0)(2

=Δ++Δ

×Δ+−Δ− MMxxwMxV

wdxdV

−=

Dividing by Δx, and letting Δx → 0 and treating 2nd order terms as negligible, we get


Vdx

dM=

Slope of moment diagram = Shear


w

V V+ΔV

M+ΔMM

Δx

wΔx wdxdV

−=


Vdx

dM=


Hence, the mathematical relationship among load, shear force and bending moment is

Load Shear Force

Bending Moment

Differentiate Differentiate

Integrate Integrate


w

V V+ΔV

M+ΔMM

Δx

wΔx wdxdV

−=


Vdx

dM=


Location of the point on the beam where the bending moment is a maxium or minimum

M is achieves a maximum or minimum where

0=dx

dM

Vdx

dM=Since

the maximum or minimum M will occur where V=0

Mathematical relationship between SFD and BMDConsider the small segment of beam subjected to only a single load

V V+ΔV

M+ΔMM

Δx

F Applying the force equation of equilibrium to segment, we have

:0∑ =yF+FV −=Δ

0)( =Δ+−− VVFV

The change in shear is negative, so that on the SFD the shear will jump downwards on the beam. Likewise, the jump in shear ΔV is upward when F acts upward.

Load SFD

Mathematical relationship between SFD and BMDConsider the small segment of beam subjected to only a single moment

V V+ΔV

M+ΔMM

Δx

Applying the moment equation of equilibrium to segment (assuming Δx → 0), we have

0MM −=Δ

The change in moment is negative, ie. BMD will jump downward if M0 is anti-clockwise. Likewise, the jump in ΔM is upward when M0 is clockwise.

Load BMD

M0

:0∑ =M+ 0)(0 =Δ+++− MMMM

Example 1

AB

CD

E

Draw shear force and moment diagrams for the beam.

10kN 10kN 6kN2m 2m 2m 2m

Example 1Draw shear force and moment diagrams for the beam.

Reaction Forces:

AB

CD

E

10kN 10kN 6kN

By Dy

:0∑ =BM+

kN17=yB

kN9=yD

0664210210 =×−×+×−× yD

:0∑ =yF+061010 =−+−+− yy DB

AB

CD

E

10kN 10kN2m 2m 2m 2m 6kN


AB

CD

E


9kN17kN

Shear Force Diagram:

SFD:

SFD follows the force arrows – if a load or reaction points upward, move upwards on SFD.

-10kN

0


AB

CD

E


9kN17kN


SFD:


-10kN17kN0

7kN


AB

CD

E


9kN17kN


SFD:


-10kN

07kN

-3kN

6kN


AB

CD

E


9kN17kN

Bending Moment Diagram:

SFD:

1. The moment of Zero at Point A.

-10kN

07kN

-3kN

6kN

BMD: 0


AB

CD

E


9kN17kN


SFD:


-10kN

07kN

-3kN

6kN

BMD: 0

2. The moment of Point B

kNm202)10(0 −=×−+=

MMM AB Δ+=

ΔM - the area of SFD.

-20kNm


AB

CD

E


9kN17kN


SFD:


-10kN

07kN

-3kN

6kN

BMD: 0


kNm62720 −=×+−=

MMM BC Δ+=


-20kNm

3. The moment of Point C

-6kNm


AB

CD

E


9kN17kN


SFD:


-10kN

07kN

-3kN

6kN

BMD: 0


kNm122)3(6 −=×−+−=

MMM CD Δ+=


-20kNm


-6kNm

4. The moment of Point D.

-12kNm


AB

CD

E


9kN17kN


SFD:


-10kN

07kN

-3kN

6kN

BMD: 0


02612 =×+−=MMM DE Δ+=


-20kNm


-6kNm

4. The moment of Point D.

-12kNm

5. The moment of Point E.

Example 2

A BC

20kN/m

4m 4m


Example 2

A BC

20kN/m

4m 4m


Reaction Forces:

ByAy

80kN

:0∑ =BM+

kN60=yB

kN20=yA

02808 =×+×− yA

:0∑ =yF+080 =+− yy BA

Example 2

A BC

20kN/m

4m 4m



60kN20kN

SFD: 0

20kN


Example 2

A BC

20kN/m

4m 4m



60kN20kN

SFD: 0

20kN


The shear force at point B

kN604)20(20 −=×−+=

VVV CB Δ+=

ΔV - the area of UDL.-60kN

Example 2

A BC

20kN/m

4m 4m



60kN20kN

SFD: 0

20kN


-60kN

x x

4m

20

80

48020 x

= m1=x

Example 2

A BC

20kN/m

4m 4m



60kN20kN

SFD: 0

20kN

-60kN

1m

BMD: 0


Example 2

A BC

20kN/m

4m 4m



60kN20kN

SFD: 0

20kN

-60kN

1m

BMD: 0



kNm804200 =×+=

MMM AC Δ+=


80kNm

Example 2

A BC

20kN/m

4m 4m



60kN20kN

SFD: 0

20kN

-60kN

1m

BMD: 0



kNm902

12080 =×

+=

MMM C Δ+=max


3. The maximum moment

80kNm90kNm

Example 2

A BC

20kN/m

4m 4m



60kN20kN

SFD: 0

20kN

-60kN

1m

BMD: 0



02/3)60(90 =×−+=

MMM B Δ+= max


3. The maximum moment

80kNm90kNm


cive1143 – structural analysis shear force and bending ......sep 17, 2017 · shear force and...

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