Deflection at the centre - simply supported beam, Mechanical Engineering

Assignment Help:

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


Related Discussions:- Deflection at the centre - simply supported beam

Truss, Truss: What are truss? When can the trusses be rigid trusses? ...

Truss: What are truss? When can the trusses be rigid trusses? State the condition which is followed by simple truss? Sol.: A structure made up of several bars riveted o

Thermodynamics, define the first and second law of thermodynamics.

define the first and second law of thermodynamics.

Advantages of two stroke engine over a four stroke engine, Advantages of tw...

Advantages of two stroke engine over a four stroke engine: The advantages of two stroke engine over a four stroke engine are stated below: 1. A two stroke engine has twice

Determine mass of air needed for complete combustion, How are fuels divided...

How are fuels divided ? Write disadvantages and advantages of each types. (b) A sample of coal supplied to a boiler has the following composition by mass : Carbon = 88%, Hydr

Heat loss in cylindruical pipe, Explain which laws of physics are used to d...

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

Design bulk materials storage, Q. Design Bulk Materials Storage? Consid...

Q. Design Bulk Materials Storage? Consideration shall be given to providing access to safety showers in areas of hazardous bulk material handling and storage. Where bulk handli

Define cases in which automation is done, Define Cases in which Automation ...

Define Cases in which Automation is Done Automation is done in following cases:- a) Loading and unloading of parts. b) Automatic production lines. c) Automatic tool ch

Scrapper -tool and equipment , Scrapper: Scrappers are used to scrap gaske...

Scrapper: Scrappers are used to scrap gasket or other such materials from flat or irregular surface. Scrappers of several sizes are shown in Figure. Figure : Scrapper

What is the net angle of twist at the free end, What is the net angle of tw...

What is the net angle of twist at the free end: The stepped steel shaft illustrated in Figure is subjected to a torque (T) at the free end and a torque (2T) in the opposite di

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