Find out deflection under the load, Mechanical Engineering

Find out Deflection under the load:

A beam of span 4 m is subject to a point load of 20 kN at 1 m from the left support and a Udl of 10 kN/m over a length of 2 m from the right support.

Find out :

1. Slope at the ends.

2. Slope at the centre.

3. Deflection under the load.

4. Deflection at the centre.

5. Maximum deflection.

Take EI = 20 × 106 N-m2.

Solution

∑ Fy  = 0, so that RA + RB  = 20 + 10 × 2 = 40 kN         --------- (1)

275_Find out Deflection under the load.png

Figure

Taking moments around A,

RB  × 4 = 20 × 1 + 10 × 2 × 3 = 80

RB  = 20 kN                                                      -------- (2)

RA  = 20 kN

M = 20 x - 20 [ x - 1] - 10 [ x - 2] ([ x - 2]/2)

= 20 x - 20 [ x - 1] - 5 [ x - 2]2

EI (d 2 y/ dx2) = M

= 20 x - 20 [ x - 1] - 5 [ x - 2]2       ---------- (4)

 EI (dy / dx )= 10 x2 /3 - (10/3) [ x - 1]2  - (5/3) [ x - 2]3  + C1        ------ (5)

EIy = 10 x3 /3- (10/3) [ x - 1]3  - ( 5/12) [ x - 2]4  + C 1 x + C2  ---------6

at A, x = 0,      y = 0,        C2  = 0

at B, x = 4 m,      y = 0

0 =(10 × 43 )/3- 10 (4 - 1)3 - (5/12)  (4 - 2)4  + C 1 × 4

C1 =- 29.17

EI dy/ dx = 10 x2  - 10 [ x - 1]2  - (5 /3 )[ x - 2]3  - 29.17

 (a)       Slope at A, (x = 0),

θ A = - 29.17 / EI = - 29.17 × 10/(20 × 106)

 = - 1.46 × 10- 3  radians

(b)        Slope at B, (x = 4 m),

EI θB  = 10 × 42  - 10 (4 - 1)2  - 5 (4 - 2)3  - 29.17 = + 27.5

θB = + 1.38 × 10- 3  radians

 (c)       Slope at centre, (x = 2 m),

EI θC  = 10 × 22  - 10 (2 - 1)2  - 29.17

θC  = + 0.04 × 10- 3  radians

Deflection under the load :

EIy = 10 x3 /3- 10 [ x - 1]3  - (5/12)  [ x - 2]4  - 29.17 x

At x = 1 m,

EIy D = (10/3) - 29.17

EIyD  = - 25.84 × 103 × 103/20 × 106

= - 1.29 mm

 (d)      Deflection at the centre :

           x = 2 m

EIy =10 × 23 - (10/3) (2 - 1)3 - 29.17 × 2

yC  = - 1.75 mm

 (e)       Maximum deflection : Let the maximum deflection b/w D and C (x < 2 m).

dy/ dx = 0

10 x2  - 10 ( x - 1)2  - 29.17 = 0

10 x2  - 10 x2  - 10 + 20 x - 29.17 = 0

x = 1.96 m < 2 m

EIy max = (10/3) (1.96)3  - 10 (1.96)3  - 29.17 × 1.96 = - 35

∴ ymax  = - 1.7501 mm

Posted Date: 1/21/2013 5:43:32 AM | Location : United States







Related Discussions:- Find out deflection under the load, Assignment Help, Ask Question on Find out deflection under the load, Get Answer, Expert's Help, Find out deflection under the load Discussions

Write discussion on Find out deflection under the load
Your posts are moderated
Related Questions

Describe vortex flow and derive mathmetical equation Z= w 2 r 2 /2g for forced vertex flow ω= angular velocity z= height of parabola r= radius of parabola

Thermodynamics. Justify that it is the science to calculate energy, energy and entropy. Sol : Thermodynamics is science which deals with conversion of heat into mechanical e

Explain which laws of physics are used to discuss heat loss in a pipe, and briefly explain how the famous equation for the loss of heat in a cylindrical pipe is derived. Explain th

Jewels can only be removed for polishing from either end of the necklace


The basic aim of synthesis in design process is to provide information which can tell you how to improve the design. In general the synthesis process gives many alternatives of

Find out diameter of steel wire: Q: A load of 5 KN is to be raised with help of steel wire. Find out diameter of steel wire, if maximum stress does not to exceed 100 MNm 2

I am attending Thermodynamics class at the university. The professor asked me to write a research on Renewable Energy: Solar Power. The report should be of 10 pages and at least 10

ELECTRIC RESISTANCE WELDING Electric resistance welding is an important and widely applied non fusion welding process. This write up covers the basic principles of various resi