I'm trying to solve the question attached below (please note the handwritten values are ones that I calculated/not part of question) I am assuming that to find the load $q$ I need to get the bending stress (which from my calculation was $45\text{ N/mm}^2$) and substitute into equation of Bending stress $= \dfrac{My}{I}$ to solve for M which should be equated to $\dfrac{ql^2}{4}$ then find $l$, however I think I messed up in the first part of the question because I'm not taking the area of part A into consideration, can you please advise me on how to approach depth-varying beams?

enter image description here


1 Answer 1


As always, the first thing you need to do is calculate your properties. \begin{align} A_A &= 150\cdot10 + 6\cdot(200-10) = 2640\text{ mm}^2 \\ \overline{y}_A &= \dfrac{150\cdot10\cdot195 + 6\cdot190\cdot95}{2640} = 151.8\text{ mm} \\ I_A &= \dfrac{150\cdot10^3}{12}+150\cdot10\cdot(151.8-195)^2+\dfrac{6\cdot190^3}{12}+6\cdot190\cdot(151.8-95)^2 \\ I_A &=9919274\text{ mm}^4 \\ A_B &= 150\cdot10 + 6\cdot(100-10) = 2040\text{ mm}^2 \\ \overline{y}_B &= \dfrac{150\cdot10\cdot95 + 6\cdot90\cdot45}{2040} = 81.76\text{ mm} \\ I_B &= \dfrac{150\cdot10^3}{12}+150\cdot10\cdot(81.76-95)^2+\dfrac{6\cdot90^3}{12}+6\cdot90\cdot(81.76-45)^2 \\ I_B &=1369647\text{ mm}^4 \end{align}

Now, you need to observe the three sources of stress in your structure:

  1. The axial load at point B;
  2. The moment due to the eccentricity of the load at point B;
  3. The moment due to the load $q$ to be determined.

So, what you need to do is calculate the stresses due to the known sources (#1 and #2) to see what slack you have for the load $q$. Strictly speaking you should check points A and B, but I'll only do A here.

So, the axial load at point B is equal to $70\cdot2040=142800\text{ N}$. This immediately gives us a uniform stress of $$\sigma_1 = \dfrac{142800}{2640} = 54.09\text{ N/mm}^2$$

However, the resultant force of the load at point B is located at the centroid of the section at point B, which is different from the centroid at point A. This causes a moment, which is equal to \begin{align} M &= 142800\left((200-151.8)-(100-81.76)\right) = 4278288\text{ Nmm} \\ \therefore \sigma_2 &= \dfrac{My}{I} = \dfrac{4278288\cdot(200-151.8)}{9919274} =20.79\text{ N/mm}^2 \end{align} Note the stress is being calculated in the top fiber, since that is where the tensile stresses occur.

So, just these loads give you already $\sigma = 54.09+20.79=74.88\text{ N/mm}^2$, meaning you are allowed an increment of $\sigma_3 = 115-74.88 = 40.12\text{ N/mm}^2$.

You are correct that you need to back-calculate the load via stress, so: \begin{align} M &= \dfrac{qL^2}{2} \\ \sigma_3 &= \dfrac{My}{I} \\ \therefore \sigma_3 &= \dfrac{qL^2y}{2I} \\ \therefore q &= \dfrac{2I\sigma_3}{yL^2} = \dfrac{2\cdot9919274\cdot40.12}{(200-151.8)\cdot2500^2} = 2.64\text{ N/mm} \end{align}

I may have bundled the calculations, but that's the general gist of it.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.