mechanics of solids (me f211) - bits pilani

BITS Pilani K K Birla Goa Campus VIKAS CHAUDHARI

MECHANICS OF SOLIDS

(ME F211)

Vikas Chaudhari BITS Pilani, K K Birla Goa Campus

Mechanics of Solids

Chapter-8

Deflections due to Bending



Contents:

The moment curvature relation

Integration of moment curvature relation

Principle of Superposition

Load- deflection differential equation

Energy methods




Deformation of an element of a beam subjected to bending moments Mb

The relation between curvature of neutral axis and bending moment is given by E is modulus of elasticity and Izz is moment

of inertia. Longitudinal dimension of the beam will be

in x direction. Bending will take place in xy plane about z

axis. Instead of symbol Izz for moment of inertia,

we use the abbreviation I.

0

1lim b

szz

Md

s ds EI




Geometry of the neutral axis of a beam bent in the xy plane

Deflection of neutral axis from the knowledge of its curvature Slope of the neutral axis

Differential with arc length s.

tandv

dx

22

2

22

2

sec

cos

d v dx d

dx ds ds

d d v dx

ds dx ds




Geometry of the neutral axis of a beam bent in the xy plane

From Figure

The curvature is

1 2

2

1cos

1

dx

dsdv dx

2 2

3 22

1

d d v dx

dsdv dx




Above equation is nonlinear differential equation for determination of

v as a function of x2, if Mb is known.

When the slope angle is small, then dv/dx is small compared to unity

Equation 1 is called moment curvature relation

The term EI is referred as flexural rigidity or bending modulus.

2

2

d d v

ds dx

2 2

3 22

1

b b

zz

M Md d v dx

ds EI EIdv dx

2

2 ------- 1bMd v

dx EI



Integration of the moment curvature relation

Integration of moment curvature relation leads to the correct

deflection curve.

However suitable boundary conditions should be chosen to determine

the integration constants.

Figure shows the suitable boundary condition encountered in various

supports.



Problem: The simply supported beam of uniform cross section shown in figure is subjected to a concentrated load W. It is desired to obtained maximum slope and maximum deflection.

L = 3.70m a = b = 1.85m W = 1.8kN E = 11GPa I = 3.33 107mm4



Solution:

Singularity function method is used to find bending moment.

The maximum deflection is the point at which the slope is zero.

Therefore maximum deflection will be at x = L/2.

Maximum slope will be at x = 0 or at x = L.

Maximum Deflection vmax will be Maximum slope max will be

3

max 25.19

48x L

WLv v mm

EI

2

max

0

0.0042 0.2416

o

x

dv WLrad

dx EI



Problem: A uniform cantilever beam has bending modulus EI and length L. It is built in at A and subjected to a concentrated force P and moment M applied at B as shown in figure. Find deflection and slope at point B.



Solution:

Deflection at point B

will be

Deflection at point B

will be

3 2

3 2B x L

PL MLv

EI EI

2

2B

x L

dv PL ML

dx EI EI



Superposition

The total deflection is the sum of deflections due to individual

load (Mb)

The deflections in the standard cases are given in table 8.1. The

solution of the original problem then takes the form of a

superposition of these solutions.

Deflection of a beam is linearly proportional to the applied load

The linearity between curvature and deflection is based on

assumption that

Deflections are small

material is linearly elastic



Problem: The cantilever beam shown in figure carries a concentrated load P and end moment Mo applied at B as shown in figure. Find deflection at point C in terms of the constant bending modulus EI.



Solution:



The Load-Deflection Differential Equation

An alternative method to solve beam deflection problem.

The differential equations for force and moment equilibrium are

Therefore,

Using above equation and moment curvature relation, we obtained a single differential equation relating transvers load-intensity function q and transverse deflection v.

0 and 0bdMdVq V

dx dx

2

2

bd Mq

dx

2 2

2 2

d d vEI q

dx dx



Boundary Conditions

Figure shows the suitable boundary conditions corresponds to four

types of supports.



Problem

The beam shown in figure is built-in

at A and D and has an offset arm

welded to the beam at the point B

with a load W attached to the arm at

C. It is required to find the deflection

of the beam at the point B.



Solution

Load intensity

Boundary conditions

Load-deflection differential equation

2 13 3

3

WLq x L W x L

0 and 0 at 0 and dv

v x Ldx

4

4 2 13 3

3

d v LEI W x L x L

dx



Solution

By integrating previous equation

Boundary conditions gives four integration constants

22

1

1 2 3

33

3 2 2

x Ldv W L xx L c c x c

dx EI

33 2

2

1 2 3 4

33

6 6 6 2

x LW L x xv x L c c c x c

EI

1 2 3 4

8 4; ; 0

27 27c c L c c



Solution

By inserting boundary conditions in deflection equation

Deflection at point B, by setting x = L/3

2 3 3 29 9 43 3 2

27 2 2 3

Wv L x L x L x Lx

EI

3

3

14

2187B x L

WLv

EI



Castigliano’s Method to find the deflections Strain energy due to transverse loads

If total elastic energy in a system is expressed in terms of external loads Pi, the corresponding in-line deflections i are given by partial derivatives

2

2 22

2

2

1

2 2

1

2 2

2

xx x

b b

L A

b

L

U dxdydz dxdydzE

M y MU dxdydz dx y dydz

E I EI

MU dx

EI

i

i

U

P



Castigliano’s Method to find the deflections Important

If, we may wish to know deflection at a point where external

force is zero .

In such case a fictitious force Q is to be considered at that point.

Deflection at that point in the direction of Q is given by U/Q

and setting Q = 0.

Similarly slope is given by U

M



Castigliano’s Method to find the deflections Simplified equation

Deflection

Slope

0 0

2

2

L L

b b b bi

i i i

M M M MUdx dx

P EI P EI P

0 0

2

2

L L

b b b bi

i i i

M M M MUdx dx

M EI M EI M



Problem

Using Castigliano’s method, determine the slope and deflection at

point B (loading diagram is shown in figure below).



Solution

Bending moment = and

Deflection at point B will be

After solving above equation

0 0

2

2

L L

b b b bB

i

M M M MUdx dx

P EI P EI P

bM P L x M bML x

P

3 2

3 2B

PL ML

EI EI



Solution


Slope at point B will be

After solving above equation

bM P L x M 1bM

M

2

2B

PL ML

EI EI

0 0

2

2

L L

b b b bB

M M M MUdx dx

M EI M EI M



Problem

Using Castigliano’s method, determine the reaction at point A.

Take a = b = L/2



Solution

We know that deflection at point A is zero. Let’s find reaction at A.


Deflection at point A will be

1

b AM R x P x a b

A

Mx

R

0

0

L

b bA

A

M Mdx

EI R

2

0

10 ( )

2

L

A A

LR x Px x dx

EI



Solution

3

00

03 2 2

L L

A

x L dx LR P x x x

dx

3 350

3 48

AR L PL

5

16AR P



Problem

Using Castigliano’s method, determine horizontal deflection at point

A (consider deflection due to only bending moments).

313

192A

PL

EI ANS.



References

1. Introduction to Mechanics of Solids by S. H. Crandall et al

(In SI units), McGraw-Hill

mechanics of solids (me f211) - bits pilani

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