30.03.2023 - 19:32

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

Question:

For the solid beam ABC shown in the figure, the brass rod has a cross-section of 30 x 30 mm with {eq}E_b = 85 ~GPa {/eq} and the copper rod 20 x 20 mm and {eq}E_c = 120 ~GPa {/eq}.

A) Compute the stresses in the bars.

B) Assume the stress in brass = 35 MPa, at what maximum distance from A can the 100 kN load be placed?

Answers (1)
  • enCore
    April 5, 2023 в 19:54

    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.

Do you know the answer?
Not sure about the answer?
Find the right answer to the question 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 by subject Physics, and if there is no answer or no one has given the right answer, then use the search and try to find the answer among similar questions.
Search for other answers
New questions in the category: Physics
Authorization
*
*

Password generation