Things topple when their center of mass moves past their supports. Think of a square box: if you push it over slightly and let go, it'll fall back onto its original position. But if you push it past ...

Once again, the superposition principle is our friend. It basically means we can look at each load in isolation, find the individual results, and then add them all together to get the final result. ...

Your bending moment diagram is very wrong. It is continuous even though there are concentrated bending moments (which create discontinuities), and you have $F$ generating positive bending moment when ...

You are correct that symmetry demands that rotation at midspan be zero. However, as you also mentioned, there will obviously be deflection at midspan as well. You therefore cannot simplify the ...

YES This is known as the superposition principle. For beams composed of linear-elastic materials where all loads are constant and independent from changes in beam geometry (i.e. not rain or snow ...

As mentioned by other answers, when dealing with a statically determinate structure, the stiffness of each element is irrelevant when calculating the bending moment, but a key variable when ...

The equation you're talking about is the following: $$\sigma = \dfrac{My}{I}$$ The reason this equation only considers bending moment even though "moment is caused [by] shear stress", is because, ...

Imagine if AB didn't exist, so all we have is BD with the pinned support at D and the force at C. In this case, BD is hypostatic and becomes a mechanism, rotating around D. Obviously, we know that ...

In these cases, it is always worth thinking about how the stress "flows" through the element. In this case, that's quite simple: The stress near the center of the element is basically unaffected by ...

The issue is simply cost. The loads in structures such as benches (which you describe in comments under the question) are very low and such structures often have huge safety margins. These margins are ...

As others have mentioned, the load will be distributed between the bolts. How they will be distributed is more complicated, but it is usually assumed to be an equal division between the bolts (in this ...

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

Yes. In a perfectly elastoplastic material (where the stress-strain relationship is perfectly linear until the yield point and is then constant and equal to the yield point), the material will ...

The depth of a member is merely equal to its total height (or whichever total dimension is parallel to the loading direction).

This is also considered a simply-supported beam. If you want to be more descriptive, you can call it a simply-supported beam with overhangs (the cantilevers beyond the main span). Basically, any beam ...

In theory, a splice behaves exactly like a single continuous bar, so there should be no effect as to where you actually do the splice. However, in the real world it depends. If you're dealing with a ...

Whether a structure is determinate or indeterminate is just a description of how easy or hard it is to solve. For a 2D planar structure such as a beam or frame, there are only three global equations ...

Ok. Yeah. Your question seems to be why a fixed support has zero rotation ($\theta=0$). The answer is simple: that's the definition of a fixed support. Here is a table representing the constraints ...

The reason is that both the horizontal and the vertical component of $N_B$ generate a moment around $A$. Since the moment due to a force is equal to the product of the force and the perpendicular ...

The fundamental beam equation is $$\dfrac{\text{d}^2}{\text{d}x^2}\left(EI\dfrac{\text{d}^2w}{\text{d}x^2}\right) = q$$ Which basically translates to "the fourth derivative of the deflection ...

As @AndyT stated, you need to add the horizontal reaction at A. You start by finding the reactions. Your global equilibrium equations are \begin{align} \sum F_x &= R_{A,x} + F = 0 \\ \... View answer 3 votes Questions 1 + 3: Your first question is unanswerable. Any design is theoretically safe. Just create a massively thick slab with high-strength concrete and massive reinforcement and you can do just ... View answer 3 votes As @JMac pointed out in a comment, the cut in the beam causes the following cross-section to occur (exaggerating the size of the cut): Originally, the shear stress can flow from the top to the bottom ... View answer Accepted answer 3 votes Both answers are incorrect. Your answer to (a) would imply that the bending moment is equal to zero at x=0 and increases linearly until x=L. This is only true if you put x=0 at the free end of ... View answer Accepted answer 3 votes Before discussing any such matters, we must define our internal force sign conventions. It is common to define them thusly: So, tension is positive, compression is negative. Shear forces are positive ... View answer Accepted answer 3 votes In structural analysis, one represents a real beam (or column, or any other "unidimensional" element) is represented by one or more bars. A bar is defined as a connection between two nodes. If you ... View answer Accepted answer 3 votes This answer will assume the applied load is not enough to damage your structure (the stresses generated are lower than the elastic yield stress). I am also going to ignore the distinction between ... View answer Accepted answer 3 votes The density of structural steel is usually adopted as 7850 kg/m3 (or thereabouts). According to this source, the cross-sectional area of an IPE 300 section is 53.8 cm2. The linear mass of ... View answer Accepted answer 3 votes The deflection of a simply-supported beam under a concentrated load at midspan, regardless of the cross-section, is equal to:\delta = \dfrac{PL^3}{48EI} So the only thing that changes in your ...