If the bar assembly is made of a material having a yield stress of σY = 45 ksi , determine the minimum required dimensions h1 and h2. Apply a factor of safety F. S. = 1.5 against yielding. Each bar has a thickness of 0.5 in.

The bar assembly is missing, so i have attached it.

Minimum required dimensions are; h1 = 2 inches and h2 = 4 inches

Explanation:

We are given;

Yield stress; σY = 45 ksi

Factor of safety; F.S = 1.5

Applying conditions of Equilibrium Σx = 0 and Σy = 0,

At segment AB in the diagram with Σx = 0 , we have;

-30 + N_ab = 0

N_ab = 30 kip

The cross sectional area of AB will be;

A_ab = (h1 × 0.5) in²

Now, since we have a factor of safety as 1.5,our allowable normal yield stress will be calculated from;

σ_allow = σY/F.S

So, σ_allow = 45/1.5

σ_allow = 30 ksi

Now, normal allowable yield stress is also calculated as;

σ_allow = N_ab/A_ab

So plugging in the relevant values, we can find h1.

So,

30 = 30/0.5h1

h1 = 30/(30 × 0.5)

h1 = 2 inches

Similarly, we can do the same for BC.

The cross sectional area of BC will be;

A_bc = (h2 × 0.5) in²

At segment BC in the diagram with Σx = 0 , we have;

-30 - 15 - 15 + N_bc = 0

N_bc = 60 kip

σ_allow = N_bc/A_bc

So,

30 = 60/0.5h2

h2 = 60/(0.5 × 30)

h2 = 4 inches

Minimum required dimensions are; h1 = 2 inches and h2 = 4 inches