# Tag Info

14

For complete layman's terms, an m6 bolt can hold up an elephant. For 10 kg, the bolts won't be the weak link, it will be whatever you are bolting into. Use 3 or 4 and you will probably be fine The fine print: I'm considering an average female Asian elephant (6000 lb or 2700 kg), grade 8 bolts (the expensive "top shelf" bolts) , double shear, and no ...

12

For a layman, you shouldn't use bolts stressed in shear. Bolts seldom hold shear forces. Bolts are used to hold two surfaces together so that the joined components take the shear stress. It is the friction of the two surfaces that support the load. This is the principle that holds wheels on automobile hubs. The friction of hub to the wheel takes the ...

8

The first thing to know is the "grade" of the bolt is available to you. Then use the appropriate table (similar to the one below) from the standard code to determine the loading capacity of the bolt. Note that your application shouldn't involve "bending" and "twisting", for which the table is not appropriate for use. Also, you ...

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 ...

7

In a wing the normal situation is that the aerodynamic force is upwards (resisting gravity). You are right that there is some shear and a fair amount of torsion, but the result is that: the top side of the wing is in compression, while the bottom is in tension. So in essence what happens is the opposite from the picture below (the force is applied upwards, ...

5

So, we apply a point load to the end of the cantilever beam and can see that for external equilibrium, there must be shear reaction at the support. (Leaving aside for now the moment reaction that also exists.) We have an intuitive understanding of this Newtonian action-reaction from our daily lives. So, the next question is, how is the load 'getting' from ...

4

The difference is in the assumptions. The first "shear stress" is simply assuming the stress is uniform over the cross section. As a result, we see a shearing force $P$ shearing a rod of uniform cross section $A$, and due to the assumption or uniform shear stress across the cross section, we have the resulting $\frac{P}{A}$ shear stress. When we abandon ...

4

Shear flow is a quantity which is used to conveniently solve (usually) torsional problems of thin walled beams (it has other applications also). The concept behind it is, that the stress distribution in a wall of a thin-walled beam can be considered constant (while in a circular cross section is proportional to the distance from the shear center. The ...

3

If we cut a section at any point of the beam, an equal and opposite force will be there (shear reaction) to maintain the equilibrium. So if we think it as a differential section, it will be easy to understand. Now imagine this differential section extended to length 'l', each fiber of the beam will have a shear resistance until it becomes equal to the ...

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

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 case), as SolarMike mentioned. This assumption is made easier by the cold work explained by blacksmith37. From an analytical standpoint, all that can be ...

3

Assuming a material like cold rolled steel: The steel will yield and deform slightly at small points of very high stress , equalizing the load/stress. So the bolts will have very similar loads. This plastic deformation and cold work is what makes steel so useful or "forgiving".

3

As mentioned, the formula $\tau_{avg}=\frac{V}{A}$ is called average shear stress. In some calculations it might suffice to calculate $\tau_{avg}$ for a widely used cross-section (let's say a standardized beam,e.g. HEA or IPE in Europe), and apply a safety coefficient to come up for shear flow. The other formula $\tau=\frac{V\cdot Q}{I\cdot t}$ (beam shear ...

3

Source, Eurocodeapplide.com Shear strength of bolts The shear resistance of the bolt per shear plane Fv,Rd is provided in EN1993-1-8 Table 3.4: Fv,Rd = αv ⋅ fub ⋅ A / γM2 where: αv is a coefficient that takes values αv = 0.6 for bolt classes 4.6, 5.6, 8.8 or αv = 0.5 for bolt classes 4.8, 5.8, 6.8 and 10.9. When the shear plane passes through the unthreaded ...

3

IMHO the most important issue, is that the IPE cross-section is considered an open section with respect to torsion. This reduces the resistance to torsion significantly. Figure : Shear stress due to pure torsion of an I beam (source AISC seminars) In essence the effect can be observed easier if you take a straw and cut it open along its length. Then take ...

2

A sketch would be helpful. But if my understanding of what you want to do is correct, you want to cover a 3 by 4 feet sump pit by light material which you could remove in case it is needed. The pvc pipe filled with quickrete or even high strength concrete or any other mix of epoxy and cement is going to fail because it basically hardly can sopprt its own ...

2

Full penetration weld would be ok. I am more concerned about that bolts between the beam and the concrete edge. It could be difficult to tight them and inspection would be problematic too. Also you loose a considerable part of the flanges, at the beam edge, and therefore you cause a severe drop of the moment strength. I don't know if this is acceptable for ...

2

The Guide to Design Criteria for Bolted and Riveted Joints 1974 by Fisher and Struik, you can use the approximation: $$S_{Ult-Shear}\sim 0.62\cdot S_{Ult}$$ and from the Distortion Energy Theory: $$S_{YieldShear}\sim 0.58\cdot S_{Yield}$$ The shear force to yield the bolt is: $$F = S_{YieldShear}\cdot A_{bolt}$$ Where $$A_{bolt}$$ is the cross ...

2

For metric bolts the tensile strength can be calculated from the grade of the bolt expressed as A.B (eg 8.8). The first number is the ultimate tensile stress in 100s of Mega Pascals and the number after the point is the yield stress in tenths of the UTS. For example an 8.8 bolt has a UTS of 800 MPa and the yield stress is 8/10 of that (80%). Multiply by ...

2

Yes, shear flow in a cross sectional element will always be in the same direction as the shear force in that same cross sectional element because the shear flow is essentially the distribution of the shear force. Confusion may arise due to the distinction between the "shear force" in an element and the externally applied load that is producing said shear. ...

2

Let's imagine a solid cylinder attached to a fix support at one end and free at other. We twist it at the other end by a torque. It will rotate about it's axis without warping, ( small angles condition). Now we cut the cylinder with laser longitudinally in 4 cuts into 8 triangle sections, radiating from the center of the circle, so if you look at the cross ...

2

The reason is symmetry. For a circular rod under torsion, if you rotate the rod through any angle the warped shape has to look the same, because there is no "special" point around the circumference where you can start measuring angles from. So, if the rod warps, the warped shape can only vary with the radius across the section. Similarly, the warping must ...

2

Assuming that nothing else will give out first, what metric would I use to estimate how much weight it can support before flexing the bolts? Is it "bending stiffness" rather than "shear strength"? The metric that you should be interested in is "bending stiffness" instead of "shear strength". Shear strength has to do with opposing forces that are very close ...

2

You're on the right track, though there looks to be a typo in the last sentence of your question (you wrote "L1" and then typed the length for L2). In short, the "bottom" shear coefficient refers to the span extending toward the left. The "top" shear coefficient refers to the span extending toward the right. Perhaps a brief discussion of how to use ...

2

Maximum shear in a solid cylinder is $$\tau_{max}= 4/3 V_{average}$$ Maximum shear in a hollow cylinder is intuitively zero at the top and bottom of the cylinder and a maximum of 2V/A on the vertical sides. here is a link to analytical calculations: shear in a hollow cylinder. $$\tau_{max}=2V/A$$

2

Wings are designed as a complex structure of spars and ribs, clad with aluminum, titanium, or new composite sheathing, or in the fabric, in the early planes. The Frame has been designed to support all the loads, bending moment of the wings, torsion, shear, compression, and tension. It has been designed to never let the surface buckle because it would be ...

2

In simple tension , the maximum resolved shear stress is 45 degrees to the tensile force . That is why a tensile test bar makes a "cup and cone" fracture face ; the cup edges are 45 degrees to the tensile force . This if for ductile materials that can deform in shear.

2

Concrete crack is essentially caused by the diagonal tension stresses that acting normal to the crack plane. The tension stress is a maximum on a plane that is 45 degree to the analytical axes. You need to review the "Principal Stresses and Maximum Shear Stresses" in the engineer mechanics textbook.

2

The shear stress on hinge screws is close to zero. The door is held by the friction of the plate to the frame, not by the shear strength of the screws. All your calculations ignore that the screws are holding the hinge plate fast to the frame. If you calculated the force holding the plate to the frame you would find that the friction is a very large force....

2

I will start by highlighting the main points (some were also found in other answers). I will additionally provide a short calculation for the tensile strength (and shear strength) very quickly by knowing the type (e.g. M6) and grade (e.g 10.9). and below I will providing a layman's calculation for a metric bold. In the numerical calculation I will use a very ...

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