# Tag Info

6

I think you might have a misconception regarding to how far the pressure from the fasteners extends. One subject you might want to have a look into is "bolt joint stiffness". The most popular is the "Rotscher’s pressure-cone method". Essentially there is a pressure cone which radiates outwards with an pressure cone angle a. According to ...

6

The Load-Displacement (or Load Extension) and stress strain diagrams are two diagrams identical in shape. See below. The main visible difference is the values on axis (which are at first glance neglected). So, it is natural when you first encounter them to question why do you want to learn about a stress and strain diagram which has obscure quantities (as ...

5

Material Properties For the linear-elastic analysis you describe, the key pieces of material information are the elastic modulus (E) to calculate deflection and the yield stress (Fy) to check if the material remains elastic under the given loading. For steel, E is commonly around 200 GPa. The Fy can vary significantly depending on the grade of steel, so you'...

4

First of all, one small but important note: The relationship between shear yield stress $S_{sy}$ and the (tensile) yield stress $S_y$ is dependent on the failure theory. Von Mises: $S_{sy} = 0.577 S_y\approx 0.6 S_y$ Tresca: $S_{sy} = 0.5 S_y$ I.e. the Tresca is a more conservative criterion.. That is probably the reason that it is preferred for materials ...

4

What is really proportional to the distance from the neutral axis (let's denote it $z$) is the strain. For pure bending and a symmetric cross-section the equation for strain is given by $$\epsilon(z) = \frac{z}{\rho}$$ where: $\rho$ or $R$ is the radius of curvature of the beam Because in the linear region the constitutive equation is $\sigma(z) = E\cdot \... 3 From what I see, in this picture from the Shigley book, if I really had to, I would extrapolate to 0.5 (or more precisely interpolate between the limit cases). The limit cases for very thick plate (d/h=0) and very thin plate ($d/h\approx \infty$) have values for d/w close to 0.7. (I would hazard a guess that these solutions are based on analytical ... 2 Although the deflections equation @kamran reported, are indeed what most textbooks give, the problem is that they only hold true for relatively small deflections. I also agreewith @alephzero that there are a few errors in the calculations. e.g. I calculate$Ix = 0.0491 cm^4$. By my calculations, you would need a steel rod with almost 11cm diameter to support ... 2 Assuming your sketch has been drawn to scale, it won't be easy to make a hand calculation of the location of the shear center. The challenge is not the use of two different materials but that the usual assumption of a thin-walled cross section won't be very accurate. If you need the accurate location of the shear center, you will pretty much have to use a ... 2 Actually, the pressure inside the vessel is uniform and constant everywhere. What happens is that geometry results in reduced stresses in one direction and not in the other. There may be some confusion here with respect to stress and pressure being the same thing. Although they share a few similarities ($\frac{F}{A}$and units$[Pa] or [psi]$, they have a ... 2 Since you are essentially using infinitesimal changes, then higher order differences can be neglected. I.e. following from your equation $${\Delta A \over A} = {{0.25\pi(d+\Delta d)^2 - 0.25\pi d^2} \over {0.25\pi d^2}} = {{2d \Delta d} \over d^2} + {{\Delta d^2} \over {d^2}} = -2 \nu { \Delta L \over L} + \big( \nu {\Delta L \over L} \big)^2$$ because$\...

2

Your best bet is to rig the planer with a couple of carpenter's horses or something on the sides. However, if we want to go your way need to check the following. Overturning moment on the fully extended setting with 44" cantilevered rollers. Torque on the joint between the two drawer slides. Let's say your machine and its base weigh 40# and the base ...

2

There are at least two contributing factors to that: a) development of grains (the following is mainly an excerpt from this question ) and this Usually thinner steels exhibit higher yield points (see cold roll sheet catalog page.8) and ultimate tensile strengths (see Steel construction) because, sheets of steel, that come out of rolling processes (especially ...

1

I don't have idea about the material you are mentioning, so no comment there. Instead, I suggest to provide bracings and additional support beams, using shelve angles, as shown below. Also, adding 1/2" - 3/4" plywood board can stiffen the shelve bed. Note, you shall confirm the shelve is rated for your application.

1

Deformations and stresses are different things. Stresses and strains are a function of the derivative of deformations, not the values. Consider a rod in tension, fixed at one end with a tension force applied to the other. The normal stress in every part of the rod is force over area. This is just as true for the fixed face as the other faces. This has ...

1

It's straightforward to model very slight deformation of the sphere (i.e., slightly flattening its surface). It's also pretty straightforward to model the case when the sides are already fully flattened and meeting at the corners. It's trickier to bridge the gap. The small-deformation case has an exact solution for a single side (Case 2 here): the ...

1

If I understand your question correctly your qualm is about $\tau$ being negative. I will use the same image as karman because it is a very good one: you will notice that shear stress is defined as: $$\tau = \mu\frac{d u}{dy}$$ Now notice in your image: that for increasing y, the velocity increases up to the middle of the pipe. (That is because the fluid ...

1

If I get your point correctly, It is only a sign convention. say in a pipe with the positive flow gradient we consider a plane C parallel to the pipe axis. The shear stress above the Yc plane is considered positive and the shear stress below that same plane is considered negative. both of these stresses are due to $\tau \$ which is positive. and the sum of ...

1

I am 68 kilograms and you could never convince me to sit on this! let's say the table is 90cm high and the legs can extend out, by just eyeballing and correct me please, 50cm. and your lege ar 2cm pipe and the screws are on a 3cm diameter circle. One can not carefully place the load at the center of the table every time, also because of the fact the legs are ...

1

You have to remember that you can control independently (or with a fixed gear ratio), the rotational velocity for each roll. So, in order to decrease the tension, what you do is you decrease the rolling speed progressively in the output zone. A very similar concept which is under the same restrictions, is the belt pulley system. Below it shows the tension ...

1

If we consider the pins on the top and bottom brackets free to rotate the system will buckle randomly either way, to right or left. With the first buckling column changing the geometry of the system and sparing the other one from buckling. Unless the width, $\theta \geq L/10$, or whatever short column index for this material, the $K=1$. However, The ...

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In the old days we used the formula- Stress = PD/2t , and the hoop stress was twice the axial /longitudinal stress . Pressure vessel heads were usually half the thickness of the walls because of the lower stress relative to the walls.

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Without having access to AISC 360-16, but from experience from other structural codes, I'll hazard a guess. you can check for allowable stresses with respect to: yielding ultimate failure fatigue torsional contact pressure dynamic loading ... Also, I am not certain, the context of the word 'element', because that could include buckling, etc.

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Your static deflection is way too low. let's consider a 1-inch length of your angle for a basic check. Assuming steel E=29000ksi $$I=1*0.38^3/12=0.00457 in^4$$ $\delta = \frac{wL^3}{(3EI)} = \frac{11760*1^3}{3EI}=0.0295 in$ Therefore $$p = p_s\left(1+\sqrt{1+\frac{2h}{y}}\right) =11760(1+\sqrt{1+\frac{2*0.5}{0.0295}}=11760(1+6.899)=92786 psi \ \text{no ... 1 Titanium modulus ranges from 15 to 18 million psi depending on alloy. Ti grades 1 to 4 are unalloyed with different levels of impurities . Grade 2 is by far the most common. Grade 5 is the most common alloy , 6 Al : 4 V. Grade 5 has a "martensite" transformation like steel ,about the only metal that does, so it can be hardened by quenching and ... 1 Using the properties you mentioned, the equation for bending stress is$$ \sigma = \frac{M}{I_{xx}}\cdot y_{max} where : $M=F\frac{l}{2}=0.5kNm$ (that's a lot) $I_{xx}$= 1800mm4 $y_{max} = \frac{thickness}{2}=3[mm]$ The results is a whopping 833.[MPa], so it will fail. Assuming you use a solid beam (same material), you can estimate the minimum ...

1

As a body, say a bar, is being subjected to a load, say a tensile load, there is an extensional strain induced due to the load. This changes the cross-section of the bar due to Poisson effect. Now, stress induced in the bar is dependent on its cross sectional area (as stress = $\frac{\text{Tensile load}}{\text{Cross-sectional area}}$). Engineering stress is ...

1

Diving deep, given the correct material/component, I was think if it is possible to use a strong external fast changing EM field (electro-magnetic) to control the shape of melting metal and keep it suspended/floating in space, and cool down using water or a high speed jet of liquid air. this will result in a material as hard as rupert's drop with no tail (no ...

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Logically I would use 4 bolts. Since you expect a moment to develop in the y axis and x axis you ought to have bolts suffice to resist those forces with a safety margin of 1.15 of the maximum allowable force. An additional 2 bolts through the vertical face would be a good idea by extending the lip upwards to the top of the bearing case that these new bolts ...

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In general the idea of bolt connections is that there should be no stress on the holes themselves. The purpose of bolts is to push the surfaces together so that the friction between the surfaces takes on the stress. So its important that the torque on the bolts is correct and the primary force direction isnt upwards in this case. (If it is bolt from sides) ...

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A one-meter cantilever shaft supported by a single ball bearing is not going to work regardless of the bearing's specs. All cantilever shafts and rods when spinning go into an increasingly unbalanced whiplash motion bending the shaft like a quadratic graph rotating about the X-axis with the centripetal force constantly increasing, leading to breaking the ...

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