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Defining $F_1$ as the triangular load, we have $$F_1 = -\dfrac{4b}{2} = -2b$$ Defining $F_2$ as the uniform load, we have $$F_2 = 2.5(a+b)$$ As you stated, $F_1+F_2 = 0 \therefore b = -5a$. This r …

The answer can be found with the same process used in the previous question. Defining $F_1$ as the downwards load, we have $$F_1 = -\dfrac{4b}{2} = -2b$$ Defining $F_2$ as the upwards load, we have …

$\newcommand{\a}[1]{\langle#1\rangle}$The reason is simple: the singularity-function form of the shear and bending moment equations already perfectly describes the results without the need for integra …

Your result is correct: member 4 is what's known as a zero-force member. This result is inescapable given it's definition as a rod which can only take in axial loads. Since it's on a diagonal, any ax …

The misunderstanding here is quite simple: there is no support at B. You are absolutely correct that there's clearly something there, as indicated by the grey circle. However, it isn't a support. The …

Bending moments due to forces and distances are found by the product of the force and the perpendicular distance of its projection to the point of study. You have the following structure (ignore the …

Let's start by transforming the load $G$ into the resultant forces applied to the structure. At point $R$, we have a vertical component equal to $G$ and an inclined component $G$. Since the inclined c …

In this situations, it's worth remembering Hooke's Law, which tells us that internal forces and deformations are related. A beam can deform either because a load is directly applied to it or because …

It's given by the question. If you pay attention to the original image, you'll see that the internal angle ($180−\alpha−\beta$) has the classic right-angle symbol. Source

To answer your question, you just need to break the inclined force into its component X and Y forces: $$\begin{alignat}{4} f_x &= \dfrac{20}{\sqrt{1^2+1.5^2}}&&=11.09\text{ kN} \\ f_y &= -1.5f_x&&=-1 …

You say you've already identified the zero-force members, so I'll skip part (a) of the question. The structure then becomes (deleting all the zero-force members except for DE which is needed for struc …

The first thing to note is the fact that, without some sort of bracing within BCEF, the structure is unstable. As-is, that square is completely flexible, and can flatten completely: CE can move down-a …

None of these elements are two-force members Before anything, it's worth remembering what two-force members are: bars which only have two equal-and-opposite colinear forces applied, one at each extre …

Your general idea is correct. Your fingers apply a force which can be translated to a moment at the rotation axis. The reaction forces applied by the object to be cut must generate an equal moment at …

For starters, as @alephzero mentioned in a comment, in the real world there are no rigid bodies, so this question is entirely theoretical. No matter how massive and rigid the beam, some infinitesimal …