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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 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, ...


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

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 ...


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 ...


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

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

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

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

The Mohr circle is a tool that helps visualize the stress state in a location in the structure. The way I interpret it is that each point in the Mohr circle represents the stresses at a rotated coordinate system. (For the 2D case) in two of the orientations (the principal directions), there is no shear stress, while in the 45 degrees to those planes the ...


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 ...


1

The graph below provides clear information on stresses variation along a cross-section. Pay attention to the segment "C", which always has the maximum stresses act on it. The magnitude of the maximum shear stress on the Mohr's Circle shall agree with the result from the shear formula, otherwise, it is meaningless. Note: To draw a Mohr's Circle, ...


1

Equation 6.18 is the "design plastic shear resistance", and equation 6.19 is the limiting shear stress for the "design elastic shear resistance. Hope this clears the confusion.


1

$f = \dfrac {VQ}{Ib}$ is the "shear flow" in the flexural beam element caused by the applied load, and its intensity varies along the beam as the internal shear force $V$ varies. The shear stress is always starting from zero at the free surface because shear occurs at the interface of sliding elements as shown below: Mohr's Circle is used to find ...


1

There is one pin in each shaft, so each pin takes the full load but in two places. This is known as double shear. If you put 2 pins for each shaft then there would be 4 sections of pin taking the load so the pins would need re-sizing, otherwise they would be too strong and something else more expensive will break.


1

The only problem I see with your calculation (some might argue against it) is that you are calculating the screws in shear. IMHO, the best practice is design the bolt, so that the axial force of the force is such so that the friction force between the two plates will transfer the torque between the machine elements. So instead of $F_{tau}$, I would calculate ...


1

I would go with M6 or M8 and use several. A variety of values are possible for each size as there are grades of bolts. The material that the rivnut is fitted into may be the limiting factor - body panels are not that strong which is why they have ridges in them especially the supporting ribs. I have often just pried rivnuts out when the bolts have been ...


1

The way I understand it, is that (at least in laminar flow) viscosity is proportional to the "stickiness" between the layers of the fluid. I.e. higher viscosity, means that if a layer starts to move, then it will sweep along more forcefully the adjacent layers. Additionally in the example of the Couette flow, the velocity of the top layer ...


1

This write-up repeats the things been taught in the class. Shear flow $f = \tau t$ = constant For thin wall closed shapes: $f = \tau t = \dfrac {T}{2A_m}$ and $\tau = \dfrac {T}{2tA_m}$ In the formulas, $t$ is the wall thickness, $A_m$ is the area bounded by the centerline of the shape. For example, for a round pipe, $A_m = \pi r^2$, and a hollow rectangle, ...


1

Method 1 - Common Structural Design Approach Assume the door is simply supported by the hinges on the jamb, and solving the reactions on the support hinges. The design shear on the screw is the resultant of T & V, which numerically equals the resultant of C & V. For the purpose of design, you can stop here, since all screws must be the same size for ...


1

Strain is the ratio of change in length of a layer of fiber in the beam with respect to its original length. And it is a function of displacement (elongation/shortening) in direction of the longitudinal axis, x, at time t (for the changes, length and sign, are dynamic in nature).


1

Let's clear the question first: The shear and moment diagrams indicate this is a cantilever beam supported at the left end, and the system is in equilibrium under the applied loads (in green) and the support reactions (in yellow). The range 4<x<7 indicates the internal reactions at a location below the uniform load are in question ("x" is ...


1

IMHO its a combination of many factors. First of all the loading. The loading you are presenting on the top image will create bending load (i.e. normal loads) and shear loads. While the shear loads for a square cross-section can be though of as uniform, then normal loads (which are more significant and generate cracks) are greater away from the neutral axis. ...


1

Buckling occurs because of bending moments which place the upper skin in compression and the lower skin in tension.


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