# Tag Info

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If you consider only the static forces then indeed the thickness might seem over-engineered. However, engine blocks are not statically loaded. They operate in the range of a few hundred to a few thousand rpm (Revolution Per Minute), so there are dynamic considerations here. Fatigue When materials are subjected to cyclic loading they exhibit a reduction in ...

16

You need to consider that the complete engine block has to withstand the reciprocating forces generated by each of the pistons and con rods moving as well as the rotational forces from the rotating crankshaft. The 1 litre 4 cylinder engine used in the Hillman imp was known for twisting under load especially once it was tuned as it was an aluminium block. One ...

12

The vibration, loading, and fatigue aspects have already been addressed, but a wide range of operating temperatures is another factor. A typical consumer engine can be deployed in anything from say -50F to 120F (-45C to 50C), and some blocks will crack when operated or even stored at those extremes. An engine operating at those extremes will experience ...

11

For example, take a look at the following static systems. Assume they have the same length and the same (constant) cross-section. Thus an equal allowed bending moment $M_u$. The first system is statically determinate, as it is supported only by simple supports. The maximum moment developing within the beam is $M=\frac{QL}{4}$, thus the load under which the ...

11

In itself putting paper inside a glass will make very little difference. Glass is a brittle material and tends to fail by shock and point loading. Its static tensile strength is actually pretty good. What will help is packing paper between the glasses as it will help prevent them coming into contact with each other and dampen vibration and impact forces. ...

11

The angle is arbitrary. One can decide which angle is more practical. However, it comes at a cost: the more the angle from the vertical the more pulling and pushing force on the anchors or escrows. Let's say your box is 100lbs such that its center of the gravity is one foot away from the wall. We check the pullout force on the top anchor/s which happens ...

9

As mentioned in the linked text and in @grfrazee's answer, the secret is influence lines. Or, more generically, influence surfaces. For starters, let's stick to influence lines, since they are far easier to describe. An influence line is a diagram for a given point on an object composed of unidimensional beam elements. It describes the internal force that ...

9

For a simple visual demonstration, take one of the spans in your example. If it is fully hinged, then each span can be represented as a simply-supported beam. If the beam is continuous, then each span can be represented as fixed-and-pinned (or at least pinned and partially fixed). The only difference between these two models is the moment reaction at the ...

9

Things scale, but not always in a nice simple way. For example consider the units of stiffness of part of the structure are $E$ (Young's Modulus) times some length, but the units of weight are $\rho$ (density) times some length cubed. So a scaled-down model structure will deform much less under its own weight than the original. (This explains why elephants ...

9

You are worrying about the numbers and forgetting what the numbers mean. This is for the PE exam, so this is a very important topic to make clear in your mind. You solved the problem. You came up with an answer for what is required. You now have two options: Choose a small footing that your own solution just proved is too small for the requirements. Choose ...

8

Say we have a slab supported on a beam. The centroid of the slab and the centroid of the beam are not coincident. Fortunately, in FEM software packages the geometric centroid of the element can be offset from the nodes that define the element. The sketch below shows a case where the shells have been offset such that the nodes are at the bottom face and the ...

8

Well, such an equilibrium (be it of moment, or shear or axial force) is necessary for any static system, and can be trivially demonstrated with Newton's second law. For forces, that is the classic $F=ma$ (force equals mass times acceleration), while for moments it is equal to $M = I\alpha$, where $M$ is moment (or torque), $I$ is the rotational inertia (also ...

8

Filling inside the cup with paper is just to lock in the wrapping of the same paper around the glass. It also offers coverage and protection of the rim of the glass. Because of the integrated, all-around coverage the whole package remains intact in the box with no chance of clinking against each other and it renders strength to the box as well. It is even ...

8

As your construction is 3D printed, increasing the strength at the point of current breaking will more likely transfer the damage to a new location. The obvious and possibly impractical solution is to solve the jamming problem, rather than to try to power through it. If it's not practical to prevent the jamming, consider to convert the jamming related ...

7

The short answer to your question, How can we compute the Equivalent Static Lateral Forces, the overturning moment and other quantities in an irregular building/structural discretized by finite element? is, simply, you can't. ASCE7-10 speaks directly to this in Section 12.6 when it describes the conditions under which Equivalent Lateral Force (ELF) can ...

7

I believe you have chosen a poor reference. Indeed, that link has absolutely nothing to do with structural analysis, but rather a specific implementation in steel structures. A pinned support is a boundary condition which restricts all displacements but allows the structure to rotate. A hinge (more formally, an internal hinge), on the ...

7

Those aren't "smoke" stacks but cooling towers designed to let hot air rise and expand optimally. These things are typically 200 m high. Static pressure of 200 m of water is pretty high. They were never required to be water tight so they will leak. In fact towers are open at the bottom to let air in and rise inside. So to start you would need to seal that ...

7

There are several general reasons for conducting practical tests in addition to theoretical analysis. to verify that the theoretical approach and associated assumptions are valid. Pretty much all loading models involve some simplifications and assumptions. for quality assurance : practical tests can pick up manufacturing/material defects, design ...

6

Firstly, the shear center is the point at which an applied load produces no torsion on the section. For a singly or doubly symmetric section, the shear center will lie on the axis of symmetry. For the unequal leg angle shown here, we can say by inspection that the shear center must be at the intersection of the two legs because the shear flow will produce no ...

6

Summary: 1) The answer to this question is difficult. You would need to know how austenite and ferrite behave in relation to what you are doing to them. You would also need to know their compositions, temperature field, etc. The results here could vary significantly depending on the specific parameters and how they change with time and with each other. 2) ...

6

What chapter/section of AISC codifies this kind of problem? If you're looking for a specific code provision for this exact type of weld, you're out of luck. However, it is possible to solve this problem using basic mechanics of materials principles. For this case, I'm going to use elastic design. You could also use the instantaneous center of rotation (...

6

How should we obtain $F_x$, given that it is lateral static force, but we are doing dynamic analysis? $F_x$ is the sum of the lateral force for a story. If you've performed a dynamic analysis, this is computed (usually) by multiplying your nodal masses by their corresponding nodal accelerations. You should be able to easily output both these from your ...

6

This means that the member is considered to be fully loaded for a failure mode which is being analyzed by that stress calculation. Some things to (probably) keep in mind include: Some conservatism likely exists in the analysis/design calculation. A member reported to be at 100% capacity is (probably) not actually at 100% capacity. This conservatism could ...

6

There is no maximum angle. Theoretically a shelf could be supported solely via cantilever (a support coming straight out from the wall. If the angle in use is the angle from the wall to the support, the compression stress on the support will be the load divided by the cosine of the angle. At straight up and down, the angle is zero, and the cosine is 1, so ...

5

This looks like a Collar Beam Roof (or sometimes a Collar Tie Roof). It is often used in lightly-loaded residential construction. One popular use is to form Cathedral Ceilings. Strictly speaking this is not a truss since the structure relies primarily on bending of the rafters to support the roof loads. By definition trusses are assumed to consist entirely ...

5

Yes this is correct - if you want to eliminate the moments in the columns then a pinned connection between the beam and columns will achieve this. To demonstrate this I have recreated your structure with some imaginary point loads and dimensions in a structural analysis software. The columns are fully fixed at the base. The diagrams on the left the images ...

5

As @pauloz1890 has already stated in his answer, if you add a hinge to a beam-column connection, you will eliminate the bending moment on your column (assuming no horizontal loads are applied). That being said, your rendering shows a frame composed of concrete beams and columns. The joints between the beams and columns are clearly monolithic. If this is the ...

5

I see another answer has given input on the strength side of things, where the economics isn't as clear, so I will add that one of the crucial aspects that usually make continuous beams economical is deflection. For a given load the deflection of a continuous beam is smaller (usually much smaller) than what it would be for a simply supported beam of the ...

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They way to answer this question is transform the dynamics model into modal coordinates, and see what happens to the force term. Suppose we can describe the stiffness and mass properties of the structure as matrices $\mathbf K$ and $\mathbf M$, and its displacement as a vector $\mathbf x$, in a physical coordinate system, and we apply a vector of forces ...

5

Your equation for bending is wrong. Force x distance is appropriate for a cantilever. For a point load applied to a "table" you are either talking simply supported (in which case Mmax = Force x span / 4) or fixed-ended (in which case Mmax = Force x span / 8). Note that you have not provided enough info to show whether a calc this simple is appropriate or not....

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