- #1

WillemBouwer

- 81

- 1

## Homework Statement

Designing a safety pin for gate valve, picture is attached! Material: EN24 (λ=680MPa and E= 207 GPa)

Total distance between supports = 98 mm! Force applied by gate is 64705 N! Question: Minimum diameter of pin for 1) No shearing , 2) a maximum deflection of 1 mm?

## Homework Equations

For circular sections: λ=(4/3)*(V/A)

Where λ= Maximum Shear(MPa) V = applied force (N) and A = Cross-sectional area of the pin (mm^2)

Normal shear at point right next to the supports is: σ = M*c/I

Where σ= Maximum Normal Shear(MPa) M = Moment created by V (N.m) c = distance from center to outside of pin:radius (mm) I = Moment of Inertia (mm^4) = ∏*r^4/8

Maximum deflection y=(V*l^3)/(48EI) Fairly well known equation...

**3.The attempt at a solution**

1) Safety factor = 1.5

Hence λ=(4/3)*(V/A)

But Vmax = 64705/2 = 32352.5 N because there is a counter force at other support side

680/1.5=(4/3)*(32352.5/(∏d^2/4)

giving d = 11 mm

BUT PROLBLEM σ = M*c/I gives

680 = (32352.5*49)*r/(∏*r^4/8)

giving r = 18.106 mm, d = ±36mm which I personally think is way to big! Please help clarifing this! Sorry I'm Afrikaans not the best speller!

2) y=(V*l^3)/(48EI)

y=(64705*(0.098^3)/(48*207*10^9*((∏*r^4)/8)

r = 11 mm which gives d = 22 mm

Seeing that it is a safety pin and will have to be removed I think a 1 mm deflection over 98 mm span is not unexceptable, but now I don't know whether it will shear because of the problem listed in the 1st question... Any help will be much appreciated... Thanks