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
Con e of Friction: When the body is having impending motion   in direction of P , the frictional force will be limiting friction and the resultant reaction R wil

You need to sketch shear force and bending moment diagrams and estimate maximum bending and shear stresses acting on the beam using your knowledge in beam analysis.  Please note

What is wave? How many type of the waves?

Q. Cold service insulation on pipe? Factory fabricated sectional pipe insulation shall be applied in staggered joint construction. On multiple layer construction, all joints sh

why compression molding process is impractable for thermoplastics?

Explain Atmospheric Pressure (P atm ) Atmospheric Pressure is the pressure exerted by atmospheric air on any surface. It is calculated by a barometer. The standard values of it

What is the importance of the concept of continuum in engineering thermodynamics. Sol: Tough the simplification of matter into atoms, electrons, molecules, and so on, is very c

Loa d Diagram and BM D fro m the Given SF D: Q: Shear force diagram of simply supported beam is given in the figure below.  Calculate the support reactions of beam and

Q.   Derive the expression of torque developed in closely excited magnetic system. Clearly explain then assumption made.   Sol. Double - Excited System   A doubly - ex

Explain Index Jig? 1. This type of jig, as the name suggests, incorporate a suitable indexing mechanism, by means of which the work piece can be indexed to required posit