1

Deflection of Symmetrical Section Beams

Methods: 1) Successive Integration

2) Singularity Functions (Macauleys Method)

3) Principle of Virtual Work

One of the most important engineering equations which form the basis

for the elastic analysis is

The Engineering Bending Equations

R

E

yI

m

This was derived for symmetrical section beams last semester

Rearranging above equation gives:

R

EIm or

EI

m

R

1

The beam curvature (1/R) is directly proportional to bending moment

m and inversely proportional to the flexural rigidity of the beam EI

2

3

4

5

6

7

SINGULARITY FUNCTIONS

Ex 13.6 shows that the double integration method becomes extremely lengthy when even

relatively small complications such as the lack of symmetry due to an offset load are

introduced. Again the addition of a second concentrated load on the beam of Ex. 13.6 would

result in a total of six equations for slope and deflection producing six arbitrary constants.

Clearly the computation involved in determining these constants would be tedious, even though

a simply supported beam carrying two concentrated loads is a comparatively simple practical

case. An alternative approach is to introduce so-called singularity or half-range functions. Such

functions were first applied to beam deflection problems by Macauley in 1919 and hence the

method is frequently known as Macauleys method.

8

We now introduce a quantity [xa] and define it to be zero if (xa)

9

Numerical Example 1

The beam shown in the following Figure has a flexural rigidity, EI of 1.4 x 104 KNm2.

a) Derive the expression of deflection

b) Determine the position and magnitude of Maximum deflection

A

B

20kN 60kN

2m 2m 3m

30kN

3m

E D C

10

11

12

Numerical Example 2

The beam shown in the following Figure has a flexural rigidity, EI of 425 x 1012 Nmm2.

Calculate the deflection at a point B and at point D using Macauleys method.

RA+RC=80+160+108=320kN

0AM 80 2+160x10+10 8 6-8RC=0; RC=280kN and RA=40kN

Mx = 40x-80[x-2]+280[x-8]-10[x-2]([x-2]/2)= 40x-80[x-2]+280[x-8]-5[x-2]2

BAxxxxxEIy

Axxxxdx

dyEI

xxxxMdx

ydEI X

4333

3222

2

2

2

]2[12

5]8[

3

140]2[

3

40

3

20

]2[3

5]8[140]2[4020

]2[5]8[280]2[8040

y=0 at x=0 B=0

y=0 at x=8 8]28[12

5]28[

3

408

3

200 433 A A=0.83

)83.0]2[12

5]8[

3

140]2[

3

40

3

20(

10425

1 43333

xxxxxy

At B, x=2m; mmy 13.0)283.023

20(

10425

1 33

At D x=10m; mmy 95.7)1083.0812

52

3

1408

3

5010

3

20(

10425

1 42333

A B C D

80kN 160kN

2m 2m 6m

UDL = 10kN/m

y

x

X