# Tag Info

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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 ...

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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 \... 4 Some other answer have touched upon this, but I think it needs to be made explicit: Your mistake is in thinking that civil engineering is about making the lightest structure possible. It's not. Instead, it's about making the most cost-effective structure possible. Give me a material that weighs a billion tons per cubic centimeter but is cheap enough to let ... 4 Shear deformation causes the cross-section to twist, thus the plane no longer remains plane, which is one of the cornerstones of the Bernoulli Beam Theory. Also, the large deflection invalidates the solution of differential equations of deflection for a beam with small angles of rotation, which was solved with some simplifications. When the slopes and ... 4 Apart from the other answers another reason why the Bernoulli-Euler beam theory falls apart at large deformation is due to the approximation about the radius of curvature. The Bernoulli-Euler beam theory is usually encountered in the following equation form: $$EI w''(x) = M(x)\qquad\text{ or } \qquad w''(x) = \frac{M(x)}{EI }$$ where: E: elastic modulus I:... 3 This is too difficult as a starting problem. There are many things not considered/ not correct. I will start with the most elementary, and move progressively (IMHO) to the more advanced subjects. Lateral forces on the tarp. The following is an part of your image you are assuming that because the load is 300 kg (and therefore the load is 3000 N), that ... 3 If you know how to calculate the reactions (R, V & M) of a simply supported beam with a single concentrate load, you can repeat that calculation for every single load in a multi-load situation, then super-impose the results at the points of interest. The graph below shows the process of a beam with two concentrate loads$P_A$&$P_B$, and how to ... 3 The most important advantage of reinforced concrete over pure steel structure is availability. Many countries/regions on earth rely on imports for general goods production needs, the cost of steel can run prohibitively high compared to concrete. The next advantage is weight. In certain types of structure weight counts, such as underwater powerhouse. However, ... 3 I will make a slight add to the answers given here already, as follows. Take the case of a bridge. When loaded, some parts of it will be in compression, other parts will be in tension. The designer's job is to manage the stress levels in the most cost-effective manner. If a certain amount of concrete is capable of handling the compressive stresses less ... 3 First of all, whenever I see stresses so localised near the boundary, I always think that there is something wrong. IMHO, the max stress at corner is an artifact, probably due to overconstraining the model. I.e. I expect you have fixed the entire face, and as a result there is an additional (non-real) component of the stress. There may be other factor ... 3 Yes, I think you can say so if you reverse its concept exactly the way it is concerned. The explanation (of the concept) below is quite straightforward and precise: Saint-Venant's Principle simply states that the stress measured at any point on an axially loaded cross section is uniform given that the measured location is far enough away from the point of ... 3 You seem to be mixing up a few concepts. As others have mentioned, bending moment is independent of a structure's cross-sectional dimensions. After all, bending moment is simply the sum of$F_i\ell_i$, where$F_i$represents the different forces applied on a structure and$\ell_i$is the perpendicular distance from the force to the point of study. That is it,... 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 The hinged region at B is not a support, actually. Thats just a connection (or to be more precise, a pinned connection). A pinned support is something different; a pinned support is something what you see on the right end C. The support itself is connected to the ground (for example), and cannot translate in any of the directions (which means that the right ... 2 A partial listing ( in addition to the previous answer) ; for reinforced concrete only simple generic shapes are needed ( bars) while steel only needs "I" , "H" beams, gussets, fasteners, welding . Careful inventory control of the previous steel items. Coating of these shapes to reduce rust during construction. Inspection to verify the ... 2 Usually continuity is to be maintained between all joints - wall/slab/beam, and side wall/side wall, or sidewalls/column, to provide water tightness. So, yes, moments due to the pool water (horizontally and vertically) need to be considered. 2 To follow up on answer by NMech: I suspect if this were built, it would be necessary to have tension adjustment, due to elasticity of the tarp. Then, the sag and any 3D "catenary type" shape (a gnarly math problem of its own, and that link doesn't include there being more load in the center in our case), is simply a byproduct of however much the ... 2 The tarp is too flexible to resist the horizontal forces induced by the anticipated 5cm deflection, thus the cable is required to be able to hold the post in position as shown in the sketch below. Without the cable, the horizontal force produces a moment (torque) about the support point "A", and the fun system will simply fall flat due to ... 2 I think your big misunderstanding is in this paragraph: [...] if I were to scan, starting from the leftmost side towards the right. I would initially see a compressive force onto the beam from R, then, I would see compressive forces (by using the section method at point 1). However once I am coming up to the 20kN force, I would, from my perspective, see a ... 2 This sketch is traced off from yours, you may update any missing/incorrect details. However, my impression is you were calculating the forces on plane a-a with the force applied as shown. For such a case, the bolt acts as a shear lug/pin and subjects to shear force only, no bending will occur or only with a negligible amount of moment resulted from the force ... 2 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 ... 2 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], ... 2 The forces on section b-b are calculated below.$\sum F_J = 0, R_I= +8.33$Draw section diagram with member forces indicated as shown in graph below. Write the external equilibrium equations for the section:$\sum F_X = 0, F_{BA} + 0.8F_{EG} + F_{HG} = 0$-----(1)$\sum F_Y = 0, F_{BD} + 0.6F_{EG} - 8.33 = 0$-----(2)$\sum M_I = 0, 6F_{BA} + 8F_{BD} + ...

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Increasing the size of the gusset plate is a bit deceiving first thought. But a disproportionally large gusset can cause stress concentration and even section failure right at the point of termination of the gusset. There are strong joint connections that are designed to be tough and ductile at the same time, this is one example. The stiffeners make the ...

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The graph below shows a fixed-end frame subjected to a concentrated load in the mid-span of the horizontal member and the resulting deflected shape. Let's draw tangent lines on the deflected shapes at joints B, we note that both the curved segments of BF & BE are deflected away from the respective tangent lines with the angle of rotation $\theta = 0$, ...

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Consider a point along the length of the beam at coordinate $(u_0, z)$ let's call the x of the point U and the Y of the neutral axis w. After deformation: $$U_{x,y}= (U_{0(x)} - z*sin(x))$$, and $$W_{x y}=W_{(x)}+z*cos\theta-\theta$$ refer to the diagram. $\theta$ is the rotation of the neutral axis. Under the assumption of small deformation we have, \$Sin(\...

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The problem is that when you reduce the width but keep the area constant you are increasing one moment of area but you reduce the other. Although there is a benefit in bending the same section with buckle (the term in structural engineering is lateral torsional buckling). Figure: Lateral torsional buckling in a cantilever beam L. Dahmani, S. Drizi If you ...

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On an appropriate scale many composite materials (even unidirectional with the appropriate layup or fabric) can be considered homogeneous and (quasi-) isotropic materials. E.g.: in the case of beams with short randomly oriented fibres should be adequately modelled using the assumption of isotropy and homogeneity. The exception in that case, would be if the ...

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As addressed before, you should formulate the problem with the x-axis pointing from the free end to the fixed end, which is the conventional method, and was used by the wiki example as well. On wiki example: I've carefully reviewed the wiki example, the solution seems incorrect. However, I've failed to pinpoint the root problems, as the steps jump around ...

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Like you said at point B there will be a reaction force from the left. However, if you look that system in isolation, that reaction needs to be supported at the end. And that force at B creates a reaction moment on B. Since the reaction at B will be equal to P/2 (just take the FBD of the right part and calculate the reactions), then the bending moment at ...

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