For the solid beam ABC shown in the figure, the brass rod has a cross-section of 30 x 30 mm with Eb = 85 GPa and the copper rod 20 x 20 mm and Ec = 120 GPa . A) Compute the stresses in the bars. B) Assume the stress in brass = 35 MPa, at what maximum di

A) To compute the stresses in the bars, we can use the equation:

{eq}sigma = \frac{F}{A} {/eq}

where {eq}sigma {/eq} is the stress, {eq}F {/eq} is the force applied, and {eq}A {/eq} is the cross-sectional area of the rod.

For the brass rod, the cross-sectional area is {eq}30times30=900mm^2 {/eq}, and the force applied is {eq}100kN {/eq}. Therefore, the stress in the brass rod is:

{eq}sigma_b = \frac{F}{A} = \frac{100times10^3~N}{900times10^{-6}m^2} = 111.1MPa {/eq}

For the copper rod, the cross-sectional area is {eq}20times20=400mm^2 {/eq}, and the force applied is {eq}100kN {/eq}. Therefore, the stress in the copper rod is:

{eq}sigma_c = \frac{F}{A} = \frac{100times10^3~N}{400times10^{-6}m^2} = 250MPa {/eq}

B) To find the maximum distance from A where the load can be placed, we need to consider the maximum stress that the brass rod can withstand, which is given as {eq}sigma_b=35~MPa {/eq}.

Assuming the load is placed at a distance {eq}x {/eq} from point A, the stress in the brass rod can be calculated using the equation:

{eq}sigma_b = \frac{F_b}{A_b} = \frac{F}{A_b} = \frac{F}{30times10^{-3}times x} {/eq}

where {eq}F_b {/eq} is the force in the brass rod and {eq}A_b {/eq} is the cross-sectional area of the brass rod.

Setting this equal to 35 MPa and solving for {eq}x {/eq}, we get:

{eq}x = \frac{F}{35times10^6times 30times10^{-3}} = \frac{100times10^3N}{35times10^6Patimes 30times10^{-3}m^2} \approx 0.095m {/eq}

Therefore, the maximum distance from point A where the load can be placed is \approx imately 0.095 m or 95 mm.