lecture6 shear force and bending moment in beams

7/30/2019 Lecture6 Shear Force and Bending Moment in Beams

1/23


2/23

Unit 2- Stresses in Beams

Lecture -1 Review of shear force and bendingmoment diagram

Lecture -2 Bending stresses in beams Lecture -3 Shear stresses in beams Lecture -4- Deflection in beams Lecture -5 Torsion in solid and hollow shafts.

Topics Covered


3/23

Why study stresses in

beams


4/23

What are beams A structural member which is long when compared

with its lateral dimensions, subjected to transverse

forces so applied as to induce bending of the

member in an axial plane, is called a beam.


5/23

Objective When a beam is loaded by forces or couples,

stresses and strains are created throughout the

interior of the beam.

To determine these stresses and strains, theinternal forces and internal couples that act on the

cross sections of the beam must be found.


6/23

Beam Types Types of beams- depending on how they are

supported.


7/23

Load Types on Beams Types of loads on beam

Concentrated or point load

Uniformly distributed load

Uniformly varying load

Concentrated Moment


8/23

P

Happy Beam is +VE

+VE (POSITIVE)

M M

Q Q

Sign Convention for

forces and moments


9/23

-VE (POSITIVE)

M M

Q Q

Sign Convention for

forces and moments

Sad Beam is -VE

P


10/23

Sign Convention for

forces and moments Positive directions are denoted by an internal shear

force that causes clockwise rotation of the member

on which it acts, and an internal moment thatcauses compression, or pushing on the upper arm

of the member.

Loads that are opposite to these are considerednegative.


11/23

SHEAR FORCES AND BENDING

MOMENTS

The resultant of the stresses must be such as tomaintain the equilibrium of the free body.

The resultant of the stresses acting on the crosssection can be reduced to a shear force and a

bending moment.

The stress resultants in statically determinatebeams can be calculated from equations ofequilibrium.


12/23

Shear Force and Bending

Moment in a Beam


13/23

Shear Force and Bending

Moment Shear Force: is the algebraic sum of the vertical

forces acting to the left or right of the cut section

Bending Moment: is the algebraic sum of themoment of the forces to the left or to the right of thesection taken about the section


14/23

SF and BM formulas

B

L

x

A

W

SFW

WxL BM

Fx= Shear force at X

Mx= Bending Moment at X

Fx=+W

Mx=-Wxat x=0=> Mx=0at x=L=> Mx=-WL

Cantilever with point load


15/23

SF and BM formulas

B

L

x

A

wLBM



Fx=+wx

at x=0 Fx=0at x=L Fx=wL

Mx=-(total load on right portion)*Distance of C.G of right portionMx=-(wx).x/2=-wx

2/2

at x=0=> Mx=0

at x=L=> Mx=-wl2/2

w Per unit

length

wL2/2

Cantilever with uniform distributed load


16/23

SF and BM formulas

B

L

x

A



at x=0 Fx=0

at x=L Fx=wL/2

Mx=-(total load for length x)*Distance of load from X

at x=0=> Mx=0

at x=L=> Mx=-wl2/6

wL/2

Cantilever with gradually varying load

w wx/L

C

Fx =wx

2

2L

Mx=

wx3

6L

Parabola

Cubic


17/23

SF and BM formulas

B

L

x

A

W

SFW/2

WL/4BM



Fx=+W/2 (SF between A & C)Resultant force on the left portion

Simply supported with point load

RA=

W

2

RB=

W

2

C

SF W/2

BaselineA C

B

BCB

W

2W

=

W

2

Constant force

between B to C


18/23

SF and BM formulas

B

L

x

A

W

SFW/2

WL/4BM



at A x=0=> MA=0

at C x=L/2=>

Simply supported with point load

RA=

W

2

RB=

W

2

C

SF W/2

BaselineA C

B

BCB

Mx= R

Ax =

W

2x

MC=

W

2L

2

for section

between A & C

Mx= R

Ax W x

L

2

=

W

2x Wx +W

L

2

= W

2x +W

L

2

for section

between C & B

MB=

WL

2

W

2

L = 0


19/23

SF and BM formulas

B

L

x

A

wL/2BM



w Per unit length

wL2/2

Simply supported with uniform distributed load

CRA RB

wL/2

C BA

wL2

8

RA

= RB

=

wL

2

Fx= R

Aw.x =

wL

2w.x

x = 0 FA

=

wL

2w.0

2=

wL

2

x =L

2 F

C=

wL

2wL

2= 0

x = L FB

=

wL

2wL =

wL

2


20/23

SF and BM formulas

B

L

x

A

wL/2BM



w Per unit length

wL2/2

Simply supported with uniform distributed load

CRA RB

wL/2

C BA

wL2

8

Mx= R

Ax w.x

x

2

=

wL

2x

w.x2

2

x = 0 MA

=

wL

2.0

w.0

2= 0

x =L

2 Mc =

wL

2.L

2w

2

L

2

2

=

wL2

4wL

2

8=

wL2

8

x = L MB

=

wL

2L

w

2L

2= 0


21/23

Load

P Constant Linear

Shear

Constant Linear Parabolic

Moment

Linear Parabolic Cubic

SF and BM diagram


22/23

Load

0 0 Constant

Shear

Constant Constant Linear

Moment

Linear Linear Parabolic

M

SF and BM diagram


23/23

Relation between load, shear

force and bending moment

B

L

x

AC

1

1 2

2

dx

F+dFF

w/m run

dF

dx= w

dM

dx= F

The rate of change of shear force is equalto the rate of loading

The rate of change of bending moment

is equal to the shear force at the section

M+dMM

lecture6 shear force and bending moment in beams

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