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

Natural Polymers Rubber is individual natural polymer and was illustrated earlier. Specific trees exude thick liquid that solidified in brittle material termed as rosin. Shell

A thick cylinder 125 mm inside diameter and 250 mm outside diameter is subjected to an internal fluid pressure of 50N/mm 2 . Determine the maximum and minimum intensities of circum

brifely explain 2r-2p invertion


In a double acting vertical steam engine running at speed of 360 r.p.m., cylinder diameter is 25 cm, stroke is 30 cm, diameter of piston rod is about 3.75 cm and length of the conn

Cell Control And Modeling   Various numbers of modeling systems and cell control have been developed to date. Researchers have worked many on three broad regions: scheduling;

Tempering and Stabilization: Tempering of tool steel in heat treatment is again a significant step. The quenching of such steels reasons the existence of untempered martensit

for drawing large size circles diameter greter than 150mm what is attached to compass? \

The joint bevels shall be prepared in accordance with the registered and approved weld procedure. Internal and external surfaces to be welded shall be clean and free from paint, o