- #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