A thin sheet made of aluminum alloy with E = 67 GPa and G = 25.6 GPa was used fo

Question

A thin sheet made of aluminum alloy with E = 67 GPa and G = 25.6 GPa was used for two dimensional surface strain measurements. The measured strains are Ex10.5.10 e,,--20-10-5 and ,-240-10-5. 5 Determine from stress and strain a) The principal stresses and strains by solving eigenvalue problem b) The maximum shear stress. Principal directions for stress and strains. What is the relation between principal directions of stress and the principal directions of strains? Why? c) d) Directions of maximum shear stress. Assume 2D states and 2x2 matrices

Explanation / Answer

Principal Strains :

1 =( xx + yy/2) + ( xx + yy/2)2 + xy2

1 =[(10.5 -20.10)/2 + ((10.5 +20.10)/2)2 + 2402] * 10-5

1 = 116.17144 * 10-5

2 =( xx + yy/2) - ( xx + yy/2)2 + (xy /2)2

2 =[(10.5 -20.10)/2 - (10.5 +20.10)/22 + 1202] * 10-5

2 = - 125.77145 * 10-5

Maximum Shear Strain = 240* 10-5

Principal Angle :

Tan 2p = xy/xx - yy =240 / 10.5-(-20.10)

2p = 82.7340 p = 41.3670

Maximum Shear Angle qs1

qs1 = (x-y)2+(xy)2 = (10.5 +20.10)2 + 2402 =88.2 deg

qs2 = -1.82 degrees

Poission Ratio

E = 2G (1+nu) 67 = 2*25.6 (1+nu)

nu = 0.30859

Normal Stresses:

Normal strain = xx = (x/E-vy/E) =( x/67 -0.38 y/67)

Normal strain = yy = (y/E -vx/E) = ( y/67 -0.38 x/67)

On solving above two equaitons we will get

x = 0.1434 Mpa y = -1.424 Mpa

Shear strain xy = xy/G

xy = 6.14678 Mpa

Principal Stress Calculation

1 =( xx + yy/2) + ( xx + yy/2)2 + xy2

1 =(0.1434 -1.424)/2 + ((0.1434 +1.424)/2)2 + 6.146782

1 =5.56 Mpa

2 =( xx + yy/2) - ( xx + yy/2)2 + xy2

2 =(0.1434 -1.424)/2 - ((0.1434 +1.424)/2)2 + 6.146782

2 =-6.84 Mpa

Maximum Shear Stress:

tmax = 1 – 2 /2 = 6.20 Mpa

Principal Angle

tan 2p = txy/ xx - yy = 6.14678 / (0.1434 +1.424)

2p = 82.7340 p = 41.3670 remains same

Maximum Shear Angle qs1

tan 2s = - ( xx - yy /2 txy)) = - (0.1434 +1.424)/2*6.14678

s1 = 86.40 s2 = -3.630

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