Find out deflection under the load, Mechanical Engineering

Assignment Help:

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


Related Discussions:- Find out deflection under the load

Classification of system - thermodynamics, Classification of system - Therm...

Classification of system - Thermodynamics: (1) Kitchen refrigerator, (2) Ceiling fan (3) Thermometer in mouth (4) Air compressor (5) Pressure Cooker (6) Carburetor (7) R

What is passive fire protection, Q. What is Passive Fire Protection? Pa...

Q. What is Passive Fire Protection? Passive fire protection is defined as any fire protection system that, by its nature, plays an inactive role in protecting personnel and pro

Industrial jobs, i wanna know is production job is best or marketing after ...

i wanna know is production job is best or marketing after doing btech

Generation of steam at constant pressure - thermodynamics, Generation of st...

Generation of steam at constant pressure: Steam is pure substance. Like any other pure substance it can also be converted into any of the three states, that is, solid, liquid

Concept of continuum - thermodynamics, Concept of continuum - thermodynamic...

Concept of continuum - thermodynamics: Sol: Even the simplification of matter into molecules, atoms, electrons, etc. is too complex a picture for many problems of thermodyna

Advantage of electro chemical machining, Advantage of Electro Chemical Mach...

Advantage of Electro Chemical Machining 1. As the tool does not come in contact with the work there for tool wear is practically nil. 2. The metal removal is faster rate.

Write Your Message!

Captcha
Free Assignment Quote

Assured A++ Grade

Get guaranteed satisfaction & time on delivery in every assignment order you paid with us! We ensure premium quality solution document along with free turntin report!

All rights reserved! Copyrights ©2019-2020 ExpertsMind IT Educational Pvt Ltd