Define the stress and strength? A material has yield strength 100 kpsi. A cantilever beam has length 10 in and a load of 100 Lbf is applied at the free end. The beam cross section is rectangular 2""x5'. Is the beam design acceptable or not for a factor of safety 2?

Stress is a force that acts on a unit area of a material. The strength of a material is how much stress it can bear without permanently deforming or breaking.

Is the beam design acceptable for a SF of 2? YES

Explanation:

Your factor of safety is 2, this means your stress allowed is:

σall = YS/FS = 100kpsi/2 = 50kpsi

Where:

σall = > Stress allowed YS = > Yield Strength FS = > Factor of safety

Now we are going to calculate the shear stress and bending stresses of the proposed scenario. If the calculated stresses are less than the allowed stress, that means the design is adequate for a factor of safety of 2.

First off we calculate the reaction force on your beam. And for this you do sum of forces in the Y direction and equal to 0 because your system is in equilibrium:

ΣFy = 0 - 100 + Ry = 0 thus, Ry = 100 lbf

Knowing this reaction force you can already calculate the shear stress on the cantilever beam:

τ = F/A τ = 100lbf / (2in*5in) τ = 10 psi

Now, you do a sum of moments at the fixed end of your cantilever beam, so you can cancel off any bending moment associated with the reaction forces on the fixed end, and again equal to 0 because your system is in equilibrium.

ΣM = 0 - 100lbf*10in + M = 0 M = 1000 lbf-in

Knowing the maximum bending moment you can now calculate your bending stress as follows:

σ = M*c/Ix

Where:

σ = > Bending Stress M = > Bending Moment c = > Distance from the centroid of your beam geometry to the outermost fiber. Ix = > Second moment area of inertia

Out of the 3 values needed, we already know M. But we still need to figure out c and Ix. Getting c is very straight forward, since you have a rectangle with base (b) 2 and height (h) 5, you know the centroid is right at the center of the rectangle, meaning that the distance from the centroid to the outermost fibre would be 5in/2=2.5in

To calculate the moment of Inertia, you need to use the formula for the second moment of Inertia of a rectangle and knowing that you will use Ix since you are bending over the x axis:

Ix = (b*h^3) / 12 = (2in*5in^3) / 12 = 20.83 in4

Now you can use this numbers in your bending stress formula:

σ = M*c/Ix σ = 1000 lbf-in * 2.5in / 20.83 in4 σ = 120 psi

The shear stress is 10psi and the bending stress is 120psi, this means you are way below the stress allowed which is 50,000 psi, thus the beam design is acceptable. You could actually use a different geometry to optimize your design.