A uniform beam is 10 m long and is supported at A , at the left end, and at \( \mathrm{D}, 2 \mathrm{~m} \) fron the right end. The beam carries the following loads: - A concentrated load of 30 kN at B , which is 2 m from A . - A concentrated load of 15 kN , at C , which is 3 m to the right of B . - A concentrated load of 40 kN at Eat the right end. Point D is 2 m from E . - Uniformly distributed loads of \( 30 \mathrm{kN} / \mathrm{m} \) are between point A and B and between point C and D. a) Make a neat, labelled sketč of the beam. b) Calculate the reaction forces at A and D .

Sure! Let’s dive into the challenges posed by this beam problem. Imagine the beam as a sturdy plank that is 10 meters long, supported at point A, the left end, and point D, which is 2 meters from the right end (called point E). In total, the beam experiences a mix of concentrated and uniformly distributed loads: - There’s a hefty 30 kN load acting at point B, located 2 meters from A. - Right near B, at point C, you’ll find an additional 15 kN load, 3 meters to the right of B. - At the very end, point E bears a substantial load of 40 kN. - Additionally, there are uniformly distributed loads of 30 kN/m stretching from A to B, and from C to D. Now, regarding the calculations for the reaction forces: To solve for the reactions at A (Ra) and D (Rd), you can start with the static equilibrium equations. First, aggregate the influences of the supports and loads: 1. Sum of vertical forces (ΣFv = 0): \[ Ra + Rd - 30 - 15 - 40 - (30 \times 2) - (30 \times 3) = 0 \] 2. Moment about one of the supports (often convenient to take moments about point A): \[ ΣMoments_A = 0 \Rightarrow Rd \times 8 - 30 \times 2 - 15 \times 5 - 40 \times 10 = 0 \] With straightforward calculations, you'll arrive at the reactions at A and D after evaluating these equations. When you put all that together, it allows you to understand the beam's behavior under load and, ultimately, to keep it from sagging or bending beyond its limits. Happy calculating!