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5

For a rectangular simply supported plate with length $a$ and width $b$ (Roark's Formulas for stress and strain): $$\sigma_{max} = \sigma_b= \frac{\beta qb^2} {t^2} $$ $\beta$ ranges from 0.287 for a=b to 0.75 for a very long and narrow beam. 0.75 would make it equal to a beam. At a=5b ,$\beta$ is 0.74. Approximately at $a> 5b$ the plate behaves very ...


5

The derivation for bending stress is depended on the assumption that the strain distribution across the thickness is linear. i.e. $$\epsilon(z) = a_1 z + a_0$$ where: $a_0$ $a_1$ are coefficients of the slope. Because in the linear region the constitutive equation is $\sigma(z) = E\cdot \epsilon(z)$, the development of stress at distance z is proportional ...


4

Because we're not interested in the moment of inertia or the "c" ordinate of a particular area element of the beam for their own sake - we're interested in them as intermediate steps on the way to computing the bending moment and/or the axial load. That involves multiplying by axial longitudinal stress and integrating. If you take the origin ...


4

Yes, the equations are valid for both imperial and metric systems. The most important thing is to use consistent units of a system throughout the calculations. Consistency of units matters as most mistakes are caused by mixing the imperial and metric units without conducting correct unit conversions (such as 1 in = 25.4 mm, 1MPa = 145 psi), or forgetting to ...


3

In setting small-angle deformations as a limit we set the limit for strains in all configurations to be elastic and totally reversible. Meaning if the beam deflects under the load or twists under the torque, etc, it will recover its original shape fully after removing the load. As long as the strains and stresses are within this envelope, you can add or ...


3

The superposition principle is valid if the assumptions it makes are valid. A commonly used set of valid assumptions are small displacements and strains and linear elastic material behavior. For large displacements, large strains, nonlinear elasticity, plastic deformation, etc it is usually not valid.


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 ...


3

First of all the bending moment, according to the usual convention (and the one you are presenting here) is positive. However, the problem originates when you add the area of the shear force. (Probably without realising it), what you are doing is you integrate the shear force over dx $\int_0^L V(x)dx$. I am using the mathematical notation because it is ...


2

If you are using the term "elastic" to means something flexible and not necessarily stretchy, consider that leather is a good choice. Leather can be bonded to wood using mechanical means (bolts, furring strips, etc) as well as with various adhesives. Using mechanical fasteners allows for easy replacement as required. I've constructed leather hinges ...


2

The bending stiffness will be determined by the second moment of area ($I$). The formula you provide $\int\int r^2 da$ is for the Polar Moment of area ($J_p$), and is valid for torsional problems. Apart from little issue you are on the right track. Assuming that: x is the horizontal axis y is the vertical axis then you are after $I_{xx}$. Additionall, I'm ...


2

The following graph shows you the properties of Polypropylene at high strain rates As you see here the modulus increases with increasing displacement rate (its not entirely equal to strain rate but its closely related). The effect is due to this viscous damping. Polymers usually are more strongly affected, in terms of modulus and ultimate tensile strength. ...


2

If you are an engineering student, you understand that beam deflects under load, and the longer the more flexible than the shorter beam, thus deflect more. The answer to this problem can be worked out from the concept of "beam on elastic (deflectable/settled) support" For your case, with the identical uniform load on both beams, it is obvious that ...


1

If you are only interested in approximation, the inner tube can be considered as a clamped cantilever after getting in contact with the walls of the outer tube, thus the applied load is resisted by the forces at the points in contact. If you want to go one step further, the clamped inner tube is resembling a beam simply supported on two hinges with an ...


1

The maximum shear stress due to vertical load is $$\tau_{max} = \frac{VQ_{max}}{I_c b}=\frac{4V}{3A}=\frac{4*3kN}{3\pi25^2}$$ And ot happens at the horizintal axis passing through the centroid. The max shear stress due to torsion happens at the surface of the bar. $$\tau_{max} = \frac{2t_{max}}{\pi r^3}=\frac{2kN*1000mm/m}{\pi *25^3}$$ So the maximum shear ...


1

You are not alone, many people are confused about the moment of inertia and polar moment of inertia of the circular shape, $I, I_p$, and their uses. Let's review two equations: $\sigma_b = \dfrac {My}{I}$, where $\sigma_b$ is the normal stress induced by the bending moment $M$, and $I$ is the moment of inertia. For circle, $I = \dfrac {\pi d^4}{64}$, where $...


1

Assuming x is the horizontal axis parallel to P =15kN (+x to the right, -x to the left) y is the vertical axis on the plane (+y upwards, -y downwards) z is the axis perpendicular to the x-y (coming out of the image). Also Q= 3kN is the load at the end of the beam $M_t=1 [kNm]$ is the torsional load Then there are three stresses at each point A, B an ...


1

What's an equation? In life there are two types of equations: theoretical equations are obtained from first principles: make some assumptions and then play around with variables until you get a useful equation empirical equations are obtained by experiment and then finding an equation that adequately describes the observed behavior. Theoretical equations ...


1

SI In the SI, you are correct that the idea is that you don't need to modify the equations. You can use any units in the equation and perform the conversions in the calculation. Typically, what one (usually a beginner) would do is convert to the basic SI units and then perform the calculation. For example a steel (E=200GPA) cantilever beam with L = 2[m], ...


1

There is no error in your diagram. if this is the moment diagram the short beam moment diagram should be a superposition of a triangle due the concentrated load from the long beam and the parabola due to the uniform load on the beam. And it is. unless your long beam is much stiffer than the shot beam which is highly unusual. The long beam moment diagram is ...


1

With the selection of constraints you have selected it is as if the longer beam (C and D) is resting (or is being supported by) the shorter beam (A and B). What happens is that while the load is distributed among the individual beams, the bending moments and the lengths are different. Therefore independently the longer beam (assuming they have the same ...


1

Depends on base metal, filler metal ( if any ) and weld process and technique. Welds may be weaker or stronger ; more or less ductile than the base metal. No problem with a properly made weld in the right base metal.


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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

TL;DR: If your goal is to find likely points of failure, take the absolute value It may seem I'm repeating what other have said. I'm merely trying to expand and I'll be trying to give another angle on this. Hopefully, some might find it helpful. How we perceive Shear forces is different than tensile/compressive forces. Tensile and compressive forces produce ...


1

Short answer Get the absolute value, so the maximum is 60. Longer answer What defines whether a shear force is positive or negative? Is it something intrinsic to the force or the applied loading? Well, no. If it were, then indeed, the maximum force would be +40. But that'd mean that pushing down on a symmetric beam would break it at a different force than if ...


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