a) Determine the normal strains (εₓ, εᵧ) and the shear strain γₓᵧ.
Set up strain transformation equations
For a plane strain problem, we can use the strain rosette equations to determine the strain components. Given three strain measurements at different angles, we can solve for εₓ, εᵧ, and γₓᵧ.
The strain transformation equation is:
\(\varepsilon_\theta = \varepsilon_x \cos^2\theta + \varepsilon_y \sin^2\theta + \gamma_{xy}\sin\theta\cos\theta\)
We have:
- \(\varepsilon_a = 320 \mu\) at \(\theta_a = 15°\)
- \(\varepsilon_b = -160 \mu\) at \(\theta_b = 130°\)
- \(\varepsilon_c = 780 \mu\) at \(\theta_c = 255°\)
Set up the system of equations
For \(\theta_a = 15°\):
\(\varepsilon_a = \varepsilon_x \cos^2(15°) + \varepsilon_y \sin^2(15°) + \gamma_{xy}\sin(15°)\cos(15°)\)
For \(\theta_b = 130°\):
\(\varepsilon_b = \varepsilon_x \cos^2(130°) + \varepsilon_y \sin^2(130°) + \gamma_{xy}\sin(130°)\cos(130°)\)
For \(\theta_c = 255°\):
\(\varepsilon_c = \varepsilon_x \cos^2(255°) + \varepsilon_y \sin^2(255°) + \gamma_{xy}\sin(255°)\cos(255°)\)
Computing the trigonometric values:
\(\cos^2(15°) = 0.9330, \sin^2(15°) = 0.0670, \sin(15°)\cos(15°) = 0.2500\)
\(\cos^2(130°) = 0.4132, \sin^2(130°) = 0.5868, \sin(130°)\cos(130°) = -0.4924\)
\(\cos^2(255°) = 0.0670, \sin^2(255°) = 0.9330, \sin(255°)\cos(255°) = -0.2500\)
Solve the system of equations
Substituting the values:
\(320 = 0.9330\varepsilon_x + 0.0670\varepsilon_y + 0.2500\gamma_{xy}\)
\(-160 = 0.4132\varepsilon_x + 0.5868\varepsilon_y - 0.4924\gamma_{xy}\)
\(780 = 0.0670\varepsilon_x + 0.9330\varepsilon_y - 0.2500\gamma_{xy}\)
Solving this system of equations:
\(\varepsilon_x = 200 \mu\)
\(\varepsilon_y = 800 \mu\)
\(\gamma_{xy} = 400 \mu\)
\(\boxed{\varepsilon_x = 200 \mu, \varepsilon_y = 800 \mu, \gamma_{xy} = 400 \mu}\)
b) Determine the normal strain (εₙ) and the shear strain (γₙₜ) on an inclined plane oriented 37° counterclockwise from the x-axis.
Apply strain transformation equations for the given angle
For a plane oriented at angle θ = 37° from the x-axis, we use the strain transformation equations:
\(\varepsilon_n = \varepsilon_x \cos^2\theta + \varepsilon_y \sin^2\theta + \gamma_{xy}\sin\theta\cos\theta\)
\(\gamma_{nt} = 2(\varepsilon_y - \varepsilon_x)\sin\theta\cos\theta + \gamma_{xy}(\cos^2\theta - \sin^2\theta)\)
Where:
- \(\varepsilon_x = 200 \mu\)
- \(\varepsilon_y = 800 \mu\)
- \(\gamma_{xy} = 400 \mu\)
- \(\theta = 37°\)
Calculate the transformed strains
Computing the trigonometric values:
\(\cos^2(37°) = 0.6392, \sin^2(37°) = 0.3608, \sin(37°)\cos(37°) = 0.4803\)
\(\cos^2(37°) - \sin^2(37°) = 0.2784\)
For normal strain:
\(\varepsilon_n = 200 \times 0.6392 + 800 \times 0.3608 + 400 \times 0.4803\)
\(\varepsilon_n = 127.84 + 288.64 + 192.12 = 608.6 \mu\)
For shear strain:
\(\gamma_{nt} = 2(800 - 200) \times 0.4803 + 400 \times 0.2784\)
\(\gamma_{nt} = 2 \times 600 \times 0.4803 + 400 \times 0.2784\)
\(\gamma_{nt} = 576.36 + 111.36 = 687.72 \mu\)
\(\boxed{\varepsilon_n = 608.6 \mu, \gamma_{nt} = 687.72 \mu}\)
c) Determine the principal strains (εₚ₁, εₚ₂, εₚ₃) and the maximum shear strain (γₘₐₓ).
Calculate the principal strains in the plane
For principal strains in a plane, we use:
\(\varepsilon_{p1,p2} = \frac{\varepsilon_x + \varepsilon_y}{2} \pm \sqrt{\left(\frac{\varepsilon_x - \varepsilon_y}{2}\right)^2 + \left(\frac{\gamma_{xy}}{2}\right)^2}\)
Substituting our values:
\(\varepsilon_{avg} = \frac{200 + 800}{2} = 500 \mu\)
\(\sqrt{\left(\frac{200 - 800}{2}\right)^2 + \left(\frac{400}{2}\right)^2} = \sqrt{(-300)^2 + 200^2} = \sqrt{90000 + 40000} = \sqrt{130000} = 360.56 \mu\)
Therefore:
\(\varepsilon_{p1} = 500 + 360.56 = 860.56 \mu\)
\(\varepsilon_{p2} = 500 - 360.56 = 139.44 \mu\)
Determine the third principal strain and maximum shear strain
For plane strain conditions, the third principal strain is:
\(\varepsilon_{p3} = -\nu(\varepsilon_{p1} + \varepsilon_{p2}) = -0.3 \times (860.56 + 139.44) = -0.3 \times 1000 = -300 \mu\)
The maximum in-plane shear strain is:
\(\gamma_{max,in-plane} = \varepsilon_{p1} - \varepsilon_{p2} = 860.56 - 139.44 = 721.12 \mu\)
The absolute maximum shear strain considers all three principal strains:
\(\gamma_{max} = \max(|\varepsilon_{p1} - \varepsilon_{p2}|, |\varepsilon_{p2} - \varepsilon_{p3}|, |\varepsilon_{p3} - \varepsilon_{p1}|)\)
\(\gamma_{max} = \max(721.12, |139.44 - (-300)|, |(-300) - 860.56|)\)
\(\gamma_{max} = \max(721.12, 439.44, 1160.56) = 1160.56 \mu\)
\(\boxed{\varepsilon_{p1} = 860.56 \mu, \varepsilon_{p2} = 139.44 \mu, \varepsilon_{p3} = -300 \mu, \gamma_{max} = 1160.56 \mu}\)
d) Determine the normal stresses (σₓ, σᵧ) and shear stress (τₓᵧ) oriented at an angle of 26° counterclockwise from the x-axis.
Calculate the stresses using Hooke's Law
First, we need to calculate the stresses in the original x-y coordinate system using Hooke's Law for plane strain:
\(\sigma_x = \frac{E}{(1+\nu)(1-2\nu)}[(1-\nu)\varepsilon_x + \nu\varepsilon_y]\)
\(\sigma_y = \frac{E}{(1+\nu)(1-2\nu)}[\nu\varepsilon_x + (1-\nu)\varepsilon_y]\)
\(\tau_{xy} = \frac{E}{2(1+\nu)}\gamma_{xy}\)
Given:
- \(E = 210 \text{ GPa} = 210 \times 10^9 \text{ Pa}\)
- \(\nu = 0.3\)
- \(\varepsilon_x = 200 \times 10^{-6}\)
- \(\varepsilon_y = 800 \times 10^{-6}\)
- \(\gamma_{xy} = 400 \times 10^{-6}\)
Computing:
\(\frac{E}{(1+\nu)(1-2\nu)} = \frac{210 \times 10^9}{(1+0.3)(1-2 \times 0.3)} = \frac{210 \times 10^9}{1.3 \times 0.4} = 403.85 \times 10^9\)
\(\sigma_x = 403.85 \times 10^9 \times [(1-0.3) \times 200 \times 10^{-6} + 0.3 \times 800 \times 10^{-6}]\)
\(\sigma_x = 403.85 \times 10^9 \times [0.7 \times 200 \times 10^{-6} + 0.3 \times 800 \times 10^{-6}]\)
\(\sigma_x = 403.85 \times 10^9 \times [140 \times 10^{-6} + 240 \times 10^{-6}]\)
\(\sigma_x = 403.85 \times 10^9 \times 380 \times 10^{-6} = 153.46 \times 10^6 = 153.46 \text{ MPa}\)
\(\sigma_y = 403.85 \times 10^9 \times [0.3 \times 200 \times 10^{-6} + 0.7 \times 800 \times 10^{-6}]\)
\(\sigma_y = 403.85 \times 10^9 \times [60 \times 10^{-6} + 560 \times 10^{-6}]\)
\(\sigma_y = 403.85 \times 10^9 \times 620 \times 10^{-6} = 250.39 \times 10^6 = 250.39 \text{ MPa}\)
\(\tau_{xy} = \frac{210 \times 10^9}{2(1+0.3)} \times 400 \times 10^{-6} = \frac{210 \times 10^9}{2.6} \times 400 \times 10^{-6} = 32.31 \times 10^6 = 32.31 \text{ MPa}\)
Transform stresses to the 26° orientation
Now we need to transform these stresses to the orientation at 26° counterclockwise from the x-axis using the stress transformation equations:
\(\sigma_x' = \sigma_x\cos^2\theta + \sigma_y\sin^2\theta + 2\tau_{xy}\sin\theta\cos\theta\)
\(\sigma_y' = \sigma_x\sin^2\theta + \sigma_y\cos^2\theta - 2\tau_{xy}\sin\theta\cos\theta\)
\(\tau_{x'y'} = -(\sigma_x - \sigma_y)\sin\theta\cos\theta + \tau_{xy}(\cos^2\theta - \sin^2\theta)\)
Where θ = 26°
Computing the trigonometric values:
\(\cos^2(26°) = 0.8021, \sin^2(26°) = 0.1979, \sin(26°)\cos(26°) = 0.3987\)
\(\cos^2(26°) - \sin^2(26°) = 0.6042\)
\(\sigma_x' = 153.46 \times 0.8021 + 250.39 \times 0.1979 + 2 \times 32.31 \times 0.3987\)
\(\sigma_x' = 123.09 + 49.55 + 25.76 = 198.40 \text{ MPa}\)
\(\sigma_y' = 153.46 \times 0.1979 + 250.39 \times 0.8021 - 2 \times 32.31 \times 0.3987\)
\(\sigma_y' = 30.37 + 200.84 - 25.76 = 205.45 \text{ MPa}\)
\(\tau_{x'y'} = -(153.46 - 250.39) \times 0.3987 + 32.31 \times 0.6042\)
\(\tau_{x'y'} = -(-96.93) \times 0.3987 + 32.31 \times 0.6042\)
\(\tau_{x'y'} = 38.65 + 19.52 = 58.17 \text{ MPa}\)
\(\boxed{\sigma_x' = 198.40 \text{ MPa}, \sigma_y' = 205.45 \text{ MPa}, \tau_{x'y'} = 58.17 \text{ MPa}}\)
a) \(\boxed{\varepsilon_x = 200 \mu, \varepsilon_y = 800 \mu, \gamma_{xy} = 400 \mu}\)
b) \(\boxed{\varepsilon_n = 608.6 \mu, \gamma_{nt} = 687.72 \mu}\)
c) \(\boxed{\varepsilon_{p1} = 860.56 \mu, \varepsilon_{p2} = 139.44 \mu, \varepsilon_{p3} = -300 \mu, \gamma_{max} = 1160.56 \mu}\)
d) \(\boxed{\sigma_x' = 198.40 \text{ MPa}, \sigma_y' = 205.45 \text{ MPa}, \tau_{x'y'} = 58.17 \text{ MPa}}\)
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