Deflection at the centre - maximum deflection, Mechanical Engineering

Deflection at the centre - maximum deflection:

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)

1870_Deflection at the centre - maximum deflection.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,

1262_Deflection at the centre - maximum deflection1.png

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 5:36:11 AM | Location : United States







Related Discussions:- Deflection at the centre - maximum deflection, Assignment Help, Ask Question on Deflection at the centre - maximum deflection, Get Answer, Expert's Help, Deflection at the centre - maximum deflection Discussions

Write discussion on Deflection at the centre - maximum deflection
Your posts are moderated
Related Questions
What is The Tunnel Diode and the Current Regulator Diode?

what is the essentials of a good boiler?

Reduced costs result from failure Easier fermenter planning (particularly in light of above) Easier downstream scheduling Off the shelf supplies e.g. motors Routine sup

Question 1: Retail design tends to be rapid turnover business. Your potential client needs a window display for his retail sports wear shop for the summer sales. The assumed bu

Q. What is Stormwater Treatment? Normally, stormwater run-off from process areas needs to be collected and treated before being discharged to land or water. Often, large aerobi

(a) What is the use of ball race drive? (b) How should one install the bottom cone race? (c) What is the use of front fork oil seal driver body and its attachment? (d) Wha

Q. Design Air Coolers for plant layout? As air coolers (sometimes called aerial coolers, or fin-fans) have extensive heat transfer surfaces, they are more vulnerable to failure

Die Press Assembly Drawing The objective of this assignment is to demonstrate the following concepts of design. • Extrude • Revolve • Fillet • Dimensioning • Tolerance • Bas


differentiate between scalar and vector quantities?