A compression bar, having a square cross section of side length b, must support a compressive load of magnitude W, uniformly distributed over the bar cross section. The bar is made of two segments connected by a glued joint along plane pq, which is at an angle α to the vertical
Consider the rotated coordinate frame (x1,y1) defined, as indicated in the figure, with the coordinate axis y1 normal to the plane of the joint.
Remember that, in the stress-transformation relationships, the angle θ from the x-axis to the x1-axis is defined positive counterclock-wise so, for this problem, θ=−(π2−α). Also keep in mind the positive convention for σy1 and τx1y1
Use the stress-transformation relationships for the plane stress state in the x−y plane to obtain symbolic expressions for the normal and shear stress components acting on the joint, σy1 and τx1y1, respectively. Express your answer in terms of W, b, and α.
In your symbolic expression, type α as "alpha".
σy1= ?
τx1y1= ?
Now take the magnitude of the compressive load to be W=1000N and the angle of the joint to be α=35∘.
The bar is constructed of a polymer for which the maximum allowable magnitude of stress in compression is σmax=1.1MPa. Also, the joint will fail under excessive levels of shear stress: take the maximum allowed magnitude for the shear stress along the joint to be τmax=0.5MPa.
If you want to avoid both failure of the polymer in compression and failure of the joint in shear, determine the numerical value, in units of cm, to the third decimal digit, of the minimum side length b of the bar:
bmin= ?
To solve this problem, we need to use the stress-transformation relationships for the plane stress state in the x-y plane.
1. Normal Stress, σy1:
The normal stress, σy1, can be calculated using the equation:
σy1 = σx * cos^2(θ) + σy * sin^2(θ) + 2 * τxy * sin(θ) * cos(θ)
2. Shear Stress, τx1y1:
The shear stress, τx1y1, can be calculated using the equation:
τx1y1 = -σx * sin(θ) * cos(θ) + σy * sin(θ) * cos(θ) + τxy * (cos^2(θ) - sin^2(θ))
Now, substituting the values given in the problem, we have:
W = 1000 N (compressive load)
α = 35 degrees (angle of the joint)
σmax = 1.1 MPa (maximum allowable stress in compression)
τmax = 0.5 MPa (maximum allowable shear stress along the joint)
By substituting these values into the stress equations, we can solve for the minimum side length, bmin of the bar.
To calculate bmin:
1. Calculate the normal stress, σy1 using the equation:
σy1 = σx * cos^2(θ) + σy * sin^2(θ) + 2 * τxy * sin(θ) * cos(θ)
2. Calculate the shear stress, τx1y1 using the equation:
τx1y1 = -σx * sin(θ) * cos(θ) + σy * sin(θ) * cos(θ) + τxy * (cos^2(θ) - sin^2(θ))
3. Set the normal stress σy1 to the maximum allowable stress in compression, σmax:
σmax = σy1
4. Set the shear stress τx1y1 to the maximum allowable shear stress along the joint, τmax:
τmax = τx1y1
5. Solve the equations for σmax and τmax and substitute the given values:
1.1 MPa = σx * cos^2(θ) + σy * sin^2(θ) + 2 * τxy * sin(θ) * cos(θ)
0.5 MPa = -σx * sin(θ) * cos(θ) + σy * sin(θ) * cos(θ) + τxy * (cos^2(θ) - sin^2(θ))
6. Use trigonometric identities to simplify the equations and solve for σx, σy, and τxy.
7. Finally, use the calculated values of σx, σy, and τxy along with the given compressive load W = 1000 N to determine the minimum side length bmin of the bar using the equation:
bmin = Sqrt(W / (σx * b))
By following these steps, you will be able to numerically calculate the minimum side length bmin in units of cm to the third decimal digit.