Deflection at the centre - simply supported beam, Mechanical Engineering

Deflection at the centre:

A simply supported beam of span 6 m is subjected to Udl of 24 kN/m for a length of 2 m from left support. Discover the deflection at the centre, maximum deflection & slopes at the ends and at the centre. Take EI = 20 × 106 N-m2.

Solution

∑ Fy  = 0, so that RA  + RB  = 24 × 2 = 48 kN          --------- (1)

 

2109_Deflection at the centre - simply supported beam.png

Taking moments around A,

24 × 2 × 1 = RB  × 6

RB  = 8 kN (↑)                     -------- (2)

RA  = 48 - 8 = 40 kN (↑).         ------------(3)

By apply the Udl over the portion DB downwards and upwards,

 

                                 Figure

M = 40 x - 24 x × (x/2) + 24 ( x - 2) ( (x - 2)/2)

Note down that the third term vanishes if x < 2 m.

= 40 x - 12 x2  + 12 ( x - 2)2               ------- (4)

EI d 2 y/ dx2 = 40 x - 12 x 2  + 12 ( x - 2)2          ------- (5)

EI dy / dx = 40 x2/2- 12 x3 /3+ 12 ( x - 2)3/3 + C1

= 20 x2 - 4 x3 + 4 ( x - 2)3 + C1           -------- (6)

EIy = 20 x 2/3 - x4 + (x - 2)4 + C1 x + C2            -------- (7)

Here again note that the third term vanishes for x < 2 m.

at A,      x = 0,    y = 0  ∴ C2  = 0

at B,  x = 6 m,     y = 0         

0 = 20 × 63 /3 - 64  + (6 - 2)4 + C1 × 6

C1 =- 20 × 12 + 36 × 6 - ((16 × 16 )/6)=- 200/3

∴          EI dy/dx = 20 x2  - 4 x3  + 4 ( x - 2)3  - 200/3         -------- (8)

The third term vanishes.

Slope at A, (x = 0),     27

θA  = -200/3EI =- (200 × 103)/ (3 × 20 ×106)

            = -(1/300) rad = - 3.33 × 10- 3  rad

 

Slope at B, (x = 6 m),

EI θ B = 200 × 62  - 4 × 63  + 4 (6 - 2)3  - (200/3)

 θ  = 136/ 3 EI = (136 × 103 )/(3 × 20 ×106)

= + 2.27 × 10- 3  radian

Slope at C, (x = 3 m), i.e. x > 2 m

EI θ C = 20 × 32  - 4 × 33  + 4 (3 - 2)3  - (200/3)

θC = 20 /3 EI = 0.47 × 10- 3  radians

EIy =( 20 x 3/3)- x4  + ( x - 2)4  - (200/3) x                   -------- (9)

Deflection at centre, (x = 3 m),

EIyC = (20/3) × 33  - 34  + (3 - 2)4  - (200 /3)× 3

yC  = - 100 / EI =  - 100 × 103 × 103 / (20 × 106)

= - 5 mm

For maximum deflection,

dy/ dx  = 0

0 = 20 x2  - 4x3  + 4 ( x - 2)3  - (200/3)

= 20 x2  - 4x3  + 4x3  - 32 - 24 x2  + 48 x - (200 /3)

=- 4x2  + 48 x - (296 /3)

∴          x2  - 12x + (74 /3 )= 0

x = 2.63 m , x > 2m

EIy max = (20/3) × 2.633  - 2.634  + (2.63 - 2)4  - (200/3) × 2.63 = - 101.7

∴ ymax  = - 5.087 mm;  - 5.1 mm

Posted Date: 1/21/2013 5:32:03 AM | Location : United States







Related Discussions:- Deflection at the centre - simply supported beam, Assignment Help, Ask Question on Deflection at the centre - simply supported beam, Get Answer, Expert's Help, Deflection at the centre - simply supported beam Discussions

Write discussion on Deflection at the centre - simply supported beam
Your posts are moderated
Related Questions
a steel bar is

how to do work in cam and follower?anser in hindi type

A combined cycle plant is designed to develop 200MW power. The air is obtained by the compressor at 300K and 1 bar pressure. The maximum temperature of the gas turbine cycle is lim

#q(b) Illustrate governing mechanism of Kaplan turbine.uestion..

If Engine Does Not Start in Wet or Damp Weather Causes of Problem Remedy   Insulators are not clean and dry Fuel t

Q. Sewage Treatment? Unless located in a metropolitan area, it is common for plants to have sewage treatment facilities for disposal of sewage and grey water generated in the p

Objectives of the Registration Scheme They are summarized as follows: To enumerate and maintain a roll of small industries to which the package of incentives and suppor

A circular thin steel diaphragm having an effective diameter of 20mm is clamped around its periphery and is subjected to a uniform gas pressure of 180kN/m^2. Calculate the minim

Q. Expected service life of a component in high temperature environment? The expected service life of a component in a high temperature environment is limited by either the mat

the component of computer integrated manufacturing with the diagram