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
(a) Why do we change the engine oil at periodic intervals? (b) What are the recommended instructions to change engine oil? (c) How will you measure the engine oil level?

Q. What do you mean by extrusion? • Extrusion is a process used to create objects of a fixed cross-sectional profile. A material is pushed or drawn through a die of the desir

Question: The figure below shows a packed bed in a 0.1 m diameter tube.  The particles are cylindrical in shape with surface volume mean diameter of 300 µm. The density and ma

The Ordinate Dimension Command We can use The Ordinate command to annotate co-ordinate points with X or Y values. This might be useful for setting-out on site plans.

Classificatio n of I.C. Engines: IC engines are classified as follows: 1 . Nature of thermodynamic cycles as:  Otto cycle engine, Diesel cycle engine and Dual combus

Explain Principal Process for Abrasive Jet Cutting Machines Abrasive particles are fed from the hopper in to the mixing chamber by regulating device as shown in figure.

Define the nature of force in a member of truss: How can you define the nature of force in a member of truss? Sol.: We know that whenever force is applied on a cross se

Derivation of formula of thermal stress

The layout of a transmission shaft carrying two pulley C and B and supported on bearings A and D is shown in figure. Power is supplied to the shaft by means of vertical belt on pul

An unsymmetrical I-Section as shown in figure below. If maximum bending stress is not extend more than 40MPa in the section than find out the moment which it can bear