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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 ...


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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 ...


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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 ...


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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 ...


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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".


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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 ...


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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 ...


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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 ...


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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 ...


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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. ...


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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 ...


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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 ...


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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 ...


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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 ...


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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 ...


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For a C channel with the following dimensions: b = width of the flange measured to the center of the web, inches or mm c = Distance to the shear center, inches or mm h = Height of section measured to the center of flanges, inches or mm I = The second moment of area, in^4 or mm^4 t = flange thickness, inches or mm $ c=\frac{b^2h^2t}{4I_x}$ If we apply ...


1

What you are referring to is the shear stress in Mohr circle at a plane rotated counterclockwise 45 degrees from the horizontal $ \sigma_n $ axis. It means if we consider a differential small cube of the rod under just plane compression, normal stress, at a plane with an angle of 45 the average $ \sigma_n = \frac{ (o+\sigma_{n\ max})}{2 }= \tau_{xy\ max} $. ...


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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 ...


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Right, or just as a double check, we say: The reaction at leftmost support plus the lower shear at the 2nd support must add up to total load on the span, meaning (1)*5*8.9kN. $$0.34 + 0.66 = 1 \quad ,Check. $$ Edit So,it is 29.37kN at the bottom of the 2nd support and the top shear is Vtop= 8.9*7.5*0.484=32.307.


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It makes sense to stream line the shape of connection at the end so as to eliminate the part that is not in the path of stress for two reasons: A - In a situation of dynamic loading such as a truss in a bridge under impact of vibration of passing traffic the part which is not sharing the stress, does not strain with the rest of the structure either, and in ...


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I use to build a very simple model first, then try to expand it. Here i try to find the minimum, force require to cut the wire, by using Tresca criterion: Yielding starts when the maximum shear stress in the material $\tau_{max}$ equals the maximum shear stress at yielding in a simple tension test $\tau_x$. You can verify the final result with the Mohr's ...


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Although I agree with @kamran, I have another way of thinking about it The deflection of the structure is $$\delta=\frac{P\cdot L}{3\cdot E\cdot I}$$ The only difference in this problem between then bonded and the decoupled is the I. Assuming b: breadth, and h: height For the coupled: $I_{coupled}= \frac{b\cdot h^3}{12}$ For the decoupled, the things ...


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I am a bit confused here, so I went under the assumption you have a statically undetermined beam with a fixed support (moment resisting) on the left and a roller support (non-moment resisting) on the right. In either case, this method still applied, but the constants become significantly easier in the case where it is a pinned support (non-moment resisting, ...


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From the image you should see that the pin on the left has two "lines" of shear stress while the one on the right only one, bending is of course an issue.


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This is very similar to the punching shear in column caps. One would calculate the shear at a distance d/2 away from the rod, with d being the thickness of flange. Hwever if your guid rail or slidin track has prticular constrains you need to consider the impact of that too. I would apply ultimate stress method and just divid the factored load by the ...


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Force is normal to plane in which cross section area is d*t on both sides thus 2dt. It is full force because steel strut caries only axial force and full force is transmitted to the pin and force is normal to this cross section plane. This is similar to calculation of stress in cylindrical vessel where the area used to calculate stress is the cross section ...


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One other possibility is to mount the motor / pulley in a drill stand and drill a "keyway" axially between the shaft and pulley then put a pin in that hole - 2mm dia but 10mm long which would work well... You may have to do the first pass with a similar metal to the shaft : a nut and then drill again with the pulley as the drill may wonder off as the pulley ...


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Propably the confusion comes from different conventions used for the internal forces. If we make a cut at any section of a member, two internal actions are defined, at opposite directions, each applied to one of the two faces of the cut. Consider as an example the following cantilever beam problem, with two cuts: The shear force, and the shear stresses have ...


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What you're looking for is commonly referred to as the shear area of the section (see here for a list of shear areas). Consider a solid rectangle with depth 20mm and width 10mm. From this calculator we can see that the section properties are: $I = 6667 \text{ mm}^4$ $Q = 500 \text{ mm}^3$ $t = 10 \text{ mm}$ So the shear area is: $A_{s} =\dfrac{It}{Q}=\...


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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 ...


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