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MECHANICS OF MATERIALS

Bending Moment and Shear Force Diagrams

MECHANICS OF MATERIALSIntroduction

Beams - structural members supporting loads at various points along the member

Transverse loadings of beams are classified as concentrated loads or distributed loads

Applied loads result in internal forces consisting of a shear force (from the shear stress distribution) and a bending couple (from the normal stress distribution)

MECHANICS OF MATERIALSTypes of Beam supports

MECHANICS OF MATERIALSShear and Bending Moment Diagrams

• We need maximum internal shear force and bending couple in order to find maximum normal and shear stresses in beams.

• Shear force and bending couple at a point are determined by passing a section through the beam and applying an equilibrium analysis on the beam portions on either side of the section.

• Sign conventions for shear forces V and V’and bending couples M and M’

MECHANICS OF MATERIALSExample 1

For the timber beam and loading shown, draw the shear and bend-moment diagrams

SOLUTION:

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

• Identify the maximum shear and bending-moment from plots of their distributions.

• Section the beam at points near supports and load application points. Apply equilibrium analyses on resulting free-bodies to determine internal shear forces and bending couples


SOLUTION:

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

kN14kN40:0 from DBBy RRMF

• Section the beam and apply equilibrium analyses on resulting free-bodies

00m0kN200

kN200kN200

111

11

MMM

VVFy

mkN500m5.2kN200

kN200kN200

222

22

MMM

VVFy

0kN14

mkN28kN14

mkN28kN26

mkN50kN26

66

55

44

33

MV

MV

MV

MV


• Identify the maximum shear and bending-moment from plots of their distributions.

mkN50kN26 Bmm MMV


5- 8

The structure shown is constructed of a W 250x167 rolled-steel beam. Draw the shear and bending-moment diagrams for the beam and the given loading.

SOLUTION:

Replace the 45 kN load with an equivalent force-couple system at D. Find the reactions at B by considering the beam as a rigid body.

Section the beam at points near the support and load application points. Apply equilibrium analyses on resulting free-bodies to determine internal shear forces and bending couples.


5- 9

SOLUTION:

Replace the 45 kN load with equivalent force-couple system at D. Find reactions at B.

Section the beam and apply equilibrium analyses on resulting free-bodies.

kNm5.220450kN450450

:

221

1 xMMxxMxVVxF

CtoAFrom

y

kNm1086.12902.11080kN10801080

:

2 xMMxMVVF

DtoCFrom

y

kNm1531.305kN153:

xMVBtoDFrom

MECHANICS OF MATERIALSRelations Among Load, Shear, and Bending Moment

xwV

xwVVVFy

0:0

D

C

x

xCD dxwVV

wdxdV

• Relationship between load and shear:

221

02

:0

xwxVM

xxwxVMMMMC

D

C

x

D Cx

dM Vdx

M M V dx

• Relationship between shear and bending moment:


Draw the shear and bending moment diagrams for the beam and loading shown.

SOLUTION:

• Taking the entire beam as a free body, determine the reactions at A and D.

• Apply the relationship between shear and load to develop the shear diagram.

• Apply the relationship between bending moment and shear to develop the bending moment diagram.


5- 13

SOLUTION:

Taking the entire beam as a free body, determine the reactions at A and D.

kN2.81

kN8.52kN6.115kN54kN900

0FkN6.115

m4.8kN8.52m2.4kN54m8.1kN90m2.70

0

y

y

y

A

AA

DD

M

Apply the relationship between shear and load to develop the shear diagram.

dxwdVwdxdV

- zero slope between concentrated loads

- linear variation over uniform load segment


- bending moment at A and E is zero

- total of all bending moment changes across the beam should be zero

- net change in bending moment is equal to areas under shear distribution segments

- bending moment variation between D and E is quadratic

- bending moment variation between A, B, C and D is linear

dxVdMVdx

dM

Apply the relationship between bending moment and shear to develop the bending moment diagram.


Draw the shear and bending moment diagrams for the beam and loading shown.

SOLUTION:

• Taking the entire beam as a free body, determine the reactions at C.




SOLUTION:

• Taking the entire beam as a free body, determine the reactions at C.

330

0

021

021

021

021

aLawMMaLawM

awRRawF

CCC

CCy

Results from integration of the load and shear distributions should be equivalent.


curveloadunderareaawV

axxwdx

axwVV

B

aa

AB

021

0

20

00 2

1

- No change in shear between B and C.- Compatible with free body analysis



203

10

320

0

20 622

awM

axxwdx

axxwMM

B

aa

AB

323 0

061

021

021

aLwaaLawM

aLawdxawMM

C

L

aCB

Results at C are compatible with free-body analysis

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