7

in a cantilever beam the deflection is $$\delta_{max} = \frac {PL^3}{3EI} $$ In this case assuming free sliding between the planks the load P is going to be supported equally between the 3 planks. So the deflection will be $$ \delta_{max} = \frac {(P/3)L^3}{3EI_{\text{single board}}} $$ Because $I$ is proportional to the cube of the board's height (in ...


6

Play around with a simple version of this structure, made from a sheet of paper fixed in a slight curve, and see what happens when you apply a load to the mid point. If the first example, any deflection of the beam will increase its length, which creates more stiffness that is included in the nonlinear analysis but not in the linear one. In the second ...


5

If by axis of bending you mean box functioning as a beam and also the top and walls and bottom of the box are integrated and work like a hollow rectangular beam, then the neutral axis is at the mid height of the box. If the content of the box is liquid, it is evenly distributed along the length of the box and should be considered distributed load. The level ...


4

There are a number of unknowns in the problem statement: The geometry of the bar The location of the supports Whether the beam is simply supported or clamped at these supports Whether the load is a point load or distributed Whether the load is in lbf or lbm What type of stainless steel is being used, etc. However, a rough estimate can be found if we make ...


4

Yes the left column would deflect in the real world. The reasons are: 1) Eccentricity of the connection / column: In reality the load will not be transferred perfectly to the center of the column nor will the column be perfectly straight. This will cause a small amount of bending in the column. In most cases this will be extremely small. However, if the ...


4

No, as the Wikipedia article states, a uniform load's unit is in "force per unit length" ($F/L$), which means it is a load applied along a given distance. So, if you have a 102 g beam, it weighs a total of 1 N. However, that unit is merely "force" ($F$), so we still need to get it into "per unit length" ($1/L$). So, if your beam is 2 m long, ...


4

Prestress creates an upward deflection on the beam due to the combination of elastic strain, creep and shrinkage in the unloaded beam. This negative deflection will bend back to a straight stance or at least minimize much of the deflection of the beam after the loading. The equations of the deflection of a prestressed concrete beam depend on the curvature ...


3

Thanks to @kamran for his answer. I simulated the problem in ANSYS student v19 to verify his approach. In the pictures below, the upper beam is solid, the middle one is split into two segments and the lower one is split into 3 segments. Each segment is allowed to slide in respect to its neighbors. It is clear that the deflection of the 3 segments beam is 9 ...


3

The answer of alephzero is spot on. I just want to mention that the arch structure is particularly sensitive to 2nd order effects when it is shallow (say depth to span ratio ~0.1) . It's not acceptable to model it using 1st order theory. The figure below demostrates the type of behaviour expected (load vs vertical displacement curve) for an arch with pinned ...


3

You are correct that this problem falls within the category of 'large-deflection' problems since the deflection is larger then about $\frac{t}{2}$. To correctly analyse a plate under large-deflections would require solving the Föppl–von Kármán equations, which is generally not possible except for some very specific cases. Numerical solutions to some cases ...


3

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 the beam and $x=L$ at the fixed support, in which case you'd have to change your constraints. What you actually need is: $$EI\dfrac{d^2y}{dx^2} = -PL + Px$$ ...


3

The exact deflection result is a function of the shelving's structural system. In the OP comments you mention that the two options would be either a wooden skin with metal edging or a wooden lattice skinned with wood. Let's start with the wooden skin with metal edging. Structurally, this is equivalent to a slab with multiple supports. The sort of support ...


3

I have no resources to point to in regards to this, but I'm fairly certain your gut-feeling is correct. In fact, end diaphragms exist in bridges primarily to hold the soil on the approach. They can also be useful for temporary lateral stability of the beams, but if that were the only reason, temporary metallic bracing would be much easier and cheaper. ...


3

You need to decide whether the design limitation which determines h is max stress or max deflection. If you decide deflection is more important insert the value for y which is the maximum allowable deflection for this part of your component. If stress is more important (ie you don't care how much it moves but you don't want it to break) then you need to ...


3

I have designed parts which are snap-fit using Polypropylene. They work well, but the best advice I can give make one (or a few) in one or more material and do some life testing to figure out which material performs the best. That's how we determined which material and geometry of the snap-fit design worked the best. Using the expected lifetime of the ...


2

The questions in your title and in your body-text are slightly different. Prestressed girder camber should usually be upwards. This is because one usually applies sufficient prestressing to cancel out (or overcome) all possible dead loads and some fraction of live loads. This means that without live loads the prestressing will probably be larger than the ...


2

Consider a horizontal beam of length $L$ along the x-axis with vertical displacement $v$. Also consider that is has an applied load $P$ at $\frac{L}{2}$: $$\dfrac{d^2v(x)}{dx^2} = \dfrac{M(x)}{EI}$$ Start with finding the moment in terms of x. You should know that the moment reaction on the left end is $-PL/8$ and the force reaction is $P/2$ so for $ 0 \le ...


2

An exact theoretical analysis of this is rather involved so I will take an approximate approach. We can idealise the beam as a single degree of freedom (SDOF) dynamic system. The theory behind this can be found in a variety of dynamics textbooks. For example: Biggs - Introduction to Structural Dynamics (1964). Because the beam has uniform mass and ...


2

For a cantilever with a point load at the free end, the deflection at that free end is: $$\dfrac{WL^3}{3EI}$$ Handbook of Structural Steelwork - 3rd edition where $W$ = point load $L$ = beam length $E$ = Young's modulus $I$ = second moment of area The Young's modulus depends on the material, see Wikipedia for various examples. The second moment of area ...


2

Note: I'll be using the following notation: $q_1, q_2$ are the distributed loads (where you use $w$, which I reserve for deflections) $L = c + 2(a+b)$, the total length of the span $\ell = a + b$ I'll also be making use of the structure and loading's symmetry, and adopting the following model: First let's get the reactions. Since the structure and the ...


2

Between disc 1 & 2, there is a reaction force on disc 1 which is not going to act on the edge rather at the distance of 2nd disc edges. This means you need to have the model of 1st disc bending developed for 2 cases, Pressure drop load acting at the radius of 2nd disc For the 2nd disc, you have 2 cases reaction load from the 1st disc at the edge of ...


2

The principle behind that table is that you will be doing a strength calculation in any case and need the table for a guideline of whether deflections or bending strength will be the design driver. High strength reinforcement does not contribute any more to the stiffness than low strength reinforcement, as the elasticity modulus of steel is the same for ...


2

... the spider is the more accurate method as multiple SPCs make the model overly stiff ... That depends what you are trying to do. Remember that (1) the RBE2 does not have to connect to all 6 DOFs at the end of each "leg", and (2) not all 6 DOFs at the "independent" grid (GN in the documentation) have to be constrained. If either of those conditions is ...


2

Concrete is strong in compression but weak in tension. In tension, it will crack at a much lower stress than the allowable compression stress. A reinforced concrete beam without any prestressing will generally have compression in concrete and tension in the rebars. If there are any significant tension stresses in the concrete, the concrete will crack and ...


2

In US, The American Society of Civil Engineers (ASCE) Structural Engineering Institute (SEI) publishes Minimum Design Loads for Buildings and Other Structures (ASCE/SEI 7-10). Section 12.8.3 Vertical Distribution of Seismic Forces, details the vertical distribution of seismic forces. It roughly applies the total design base shear of the building in a ...


2

There are two aspects to this problem that are each significantly more complicated than what you seem to have tried so far: The angle of 45 degrees by which the cross section is changing dimension is just too much for the usual approximations to make sense. At such a large angle, plane section will not remain plane and the standard beam equations no longer ...


1

The Radius of curvature on a beam bent under moment is $1/r = m/EI $ If we use this just as a rough model the m affecting each disk gradually becomes smaller as we go up and radius of curvature bigger. So it is safe to assume only the outer edge of each top disk touches the lower disk.


1

10kg of cover on roughly 10kg worth of pipe? Never mind moist environment where rust will be a serious problem... never mind if you intend it to roll by crank (like in the PDF attached), in the beginning you'll be rolling about 10cm of the cover per revolution. Go with a much larger diameter PVC pipe - 10cm or more. Rust-free, can be easily worked with ...


1

mg4w's answer has some great examples of how your modelling assumptions aren't true in reality. But even ignoring all of those there will still be deflection in the left hand column. You may not find any deflection in your analysis, see for example Sam's answer. But this is because you're doing an over-simplistic analysis. We often assume that we have ...


1

1) Theoretically this is true, there would not be any horizontal deflection as the fixed supported (left column) would offer complete rigidity and the hinge connection at node 2 would not transfer any bending. There would also be no horizontal force from the top member as the force is completely vertical: Source: SkyCiv Frame Analysis Software However, in ...


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