A plate is under biaxial loading as shown in the Figure belo

A plate is under biaxial loading as shown in the Figure below. Given two choices of plate material: A514 Steel: E_xt = 29 times 10^6 psi sigma_yt = 100 ksi rho_at = 0.284 lb/in^3 6061-T6 Aluminum: E_At = 10 times 10^6 psi sigma_y At = 37 ksi rho_At = 0.098 lb/in^3 a) Design the thickness, t, of the plate to prevent yielding for each of the two choices of material if a hole is not present. b) Design the thickness of the plate for each of the two choices of material if a hole is present. Use the creteria that the von Mises stress must be less than sigma_yp/2. Assume the infinite width plate analysis. c) Would the plate with a bowl at the minimum thickness be lighter if it was made of aluminum or steel?

Solution

solution:

here by maximum shear stress theorey

Sx-Sy/2=Syt/2

1) for steel without hole

Sx=5000/t

Sy=-3333.33/t

hence from above relation

t=.0833 in

with hole

Sx=5128.2/t

Sy=-3389.83/t

hence thickness is

t=.08518 in

3) for alluminimum

without hole

t=.2252 in

with hole

t=.2352 in

4)here mass of material with hole is given as for steel

Vs=W*H*t=30*20*t=30*20*.08518=51.108 in^3

mass of steel=density*vs=.284*51.108=14.51 lb

5)here mass of material with hole is given as for alluminium

Vs=W*H*t=30*20*t=30*20*.2302=138.12 in^3 in^3

mass of Al=density*vs=.098*132.18=13.53 lb

hence alluminium is lighter than steel for given case