# Tag Info

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If you take a read more carefully the Eurocode, it says "2% annually". That should be translated that in a full year, there is only a 2% chance to get to that wind level. So one simple way to think about this is that there is one chance in 50 years for that failure to occur (which is considered a reasonable scenario. The more correct mathematical ...

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If the joint cracks as shown, you have had a big problem, as it indicates the concrete fails in compression with random failure patterns - the concrete is crumbling, for which the diagonally placed reinforcement solves nothing but hurting the concrete core - by its tendency to straighten and buckle that would create additional stress and splinting the ...

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For hot rolled , the thicker section cools slower so has coarser pearlite and coarser grain size because the grains have more time to grow larger during the slow cool. A secondary factor is the thicker section is worked less / less strain to refine grains. The lower strength affect of the slower cooling thick sections also occurs during normalize and quench ...

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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 ...

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If you are looking for just which option would require the least effort then the answer is the second case (PLAN 2). The help you to understand why is that, do the following mental experiment. Imagine that the support is so high that the fulcrum coincides with the center of the arm. For plan 2 in this scenario, the winch effort to rotate the arm would be ...

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For both methods, you shall first find the tension, T, by taking moments about the reference joint "A" and set sum M = 0. Then you shall find the reactions, V & H, and power demand of the winch. After that, you can determine which method is more desirable, and the reasons.

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The amount of force, F is the tension in the rope which will indicate the torque in the winch. In the top case, it is the lever advantage of the ratio of [CG * mass ] of the antenna to the vertical component of F *the connection of the rope distance multiplied by $$\sqrt{84^2+72^2}*(225/72)$$ $$(360"/2)*84"= 2.1428$$ $$2.1428*105lbs= 225lbs.\text{... 0 Not feasible. Nothing light would be able to support itself. You at least need to be able to withstand wind loading, which at that height will be crazy high. Nothing we have in present day usage. You might be able to build an inflatable that is 50m high, but even that will be a challenge. The only way at present to build such a thing would be a massive ... 2 Using conditions given in the sketch above, let's do a parameter study. M = AsFy(d - a/2) a = Asfy/0.85fc'b, note, since As, b, fc' and fy are constant throughout this exercise, "a" is constant as well. Let's assume a = 50mm. d = 65, M1 = AsFy(65-50/2) = 40Asfy d = 115, M2 = AsFy(115-50/2) = 90Asfy d = 165, M3 = AsFy(165-50/2) = 140Asfy M2/M1 = ... 0 On the figure above and discussion below, M_xx represents the fixed end moment (FEM) when the joints are in locked/clamped/fixed condition. Now, the structure is locked for the calculation of the fixed end moments (FEM). At the rightmost span, M_CD = M_DC = 0, since the span has no load. At the leftmost span, M_AE = 10 kNm, and M_EA = 0, why? (Pay attention ... 0 Its just a matter of using the right formulas and algebraically finding an equation that will leave you with 1 unknown. In this case it is h. 0 Depending on the direction of the pour, the mesh at the leading edge could have been dislocated without noticing. The 40 mm bottom cover thickness is about right for positive reinforcement. If there are two layers of reinforcement, this is nothing to worry about. Otherwise, you might find fine cracks on the top face of the slab area that having a 17mm bottom ... 0 This problem involves the concepts of "conservation of mass" and "conservation of energy". Due to impact, the static deflection has no bearing on this matter. You need to find the force (static + impact) that causes the beam to fail. Assume the beam will fail by flexural (moment), then you can solve this problem by the Energy Method: ... 0 It seems you use the formula for simply supported beam, the cantilever static deflection is as the question already states:$$ \delta_{static} = \frac{mgL^3}{3EI}=\frac{50*1^3}{3*207* 10^{-11}*2.66*10^{-7}}=29.6*10^{-6}m $$Now you can plug this into:$$F_c=(50kg)(1+ \sqrt{1+\frac{2h}{29.6*10^{-6}m}} \quad )$$0 The equations found do match the equations above, they are just written differently but will yield the same outcome. 2 I would say that you can divide them into two groups: At nodes with two member and no external force Node C: CD,CB, Node G: FG, GH Node K: KJ, KL If you write down the equilibrium at node C for x and y you would get:$$x: \quad \sum F_x = 0 \rightarrow CD = 0 y: \quad \sum F_y = 0 \rightarrow CB = 0 $$Same at the other nodes. Note that nodes A, and ... 0 Members KJ and KL by inspection are zero force. there may be more but not immediately evident to mho. 1 Direct answer to your questions: Yes, the opening depth is immaterial in support requirement determination. Round opening is less likely to require support, due to arch action. Wall type is critical when the brick layout pattern affects the available bond between bricks over the opening. If depth of the wall above the opening permits arch action to occur,... 3 Does not apply to this example, but just for completeness: I've also seen CRS as an abbreviation for cold-rolled steel in a material callout in machine drawings, e.g. "1018 CRS". 6 "CRS" is an abreviation for "Centers". So, for example the statement "Deck Joist @ 400mm crs" means "at 400mm centers", i.e. that the centreline of the joists should be 400mm apart. If you were to measure the gap between the joists, it would be less than 400mm, due to the thickness of the joist itself. 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 ... 1 I'm not a geotechnical engineer, but I believe the gravel is there precisely to prevent the soil behind it from getting saturated. That is, if the soil gets saturated, the hydrostatic pressure will squeeze the water out into the gravel, which then drains the water away, thereby reducing or altogether eliminating the pressure on the wall. 3 An extended discussion on internal vs. external forces We usually like to describe hinges as "places where the moment is always zero." But, wait a minute, the moment is always zero anywhere in a stable structure. Don't believe me? Let's take a look at the most trivial structure ever, a simply-supported beam with a uniform load: So, it has a span ... 2 It is important to understand what is an internal hinge, its behavior, and assumptions around it. An internal hinge represents a discontinuity in the beam, it carries a pair of equal but opposite force couple, one to each side of the hinge, carried over from the beam segment it supports/connect to, so sum Fy is zero. Another important assumption is the hinge ... 0 You answered yourself. No matter how much moment is calculated on the left side of the D it has zero effect on the right side of D. The moment does not pass through the hinge and in this case, no shear transfers from the left to the right either because the two supports A and B have taken care of the vertical forces. Regarding image two it's just an easier ... 0 Generally the maximum bending stress \sigma on a beam under bending is given by:$$\sigma_{max} = \frac{M}{I}\cdot y_{max}$$where: I: the second moment of area of the rod (2d, because the internal diameter is by removing two thickness from the extenral )$$I =\frac{\pi}{64} \cdot ( D^4 - (D-2 d)^4)$$y_{max} = \frac{D}{2} in this case M: is the bending ... 0 Given: Total weight = 150 kg, uniformly placed over beam with 0.55m span length. M = wL^2/8 = (150/0.55)(0.55^2)/8 = 10.3 N-m = 0.0103 kN-m For A302, A304, stainless steel tubing, Fy = 207, Fa = 0.6 Fy = 124.2 MPa = 124200 kN/m^2 For D = 10 mm, y = D/2 = 5 mm, I_req = My/Fa = [0.01030.005/124200]*1000^4 = 414.6 mm^4 For round pipe with D = 10 mm, d = 1 mm, I ... 0 If you choose your rods appropriately, a bigger concern is going to be those holes you drill for the rods. For now, I'll leave you with a beam bending approximation approach:$$Stress = \frac{LeverMoment * CentroidDistance}{MomentOfInertia}$$With your tube definitions, you get that the$$CentroidDistance = D/2MomentOfInertia = \frac{\pi}{64} * ( D^4 - ...

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Following the points alephzero made, assuming we have a box with a sturdy geometry, 50 kg divided between the 100 cm of the edge of a square bottom (25cmx25cm) that are hanging from vetical walls of the box is way bellow the ripping resistance of the sandwich cardboard. Of course using good filler material helps a lot, be it old clothes or bubble plastic ...

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Depending on the design, these folded boxes are good for weights between 60 and 300 pounds. The box is "locked in place" because paper or cardboard can not be stretched. The only way you can easy way to change its shape is to bend it, and you the only way you can bend a piece of paper is to shape it into a cylinder. But you can't "bend" ...

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Just a few notes here: With or without gravelly backfill, the retaining wall must be designed for the maximum groundwater anticipated for the site. Your geotechnical engineer shall provide this information. The 300mm clean free draining material is the minimum requirement, it is recommended to backfill the entire cut (excavation) with well graded gravelly ...

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First thing first, you need to get hand on the ASME Boiler and Pressure Vessel Code (BPVC Code). Starting division 8, section 1, for safety and quality of your pressure vessel. After checking on all relevant provisions/standards, you can use the equations to derive the required wall thickness for your application. Notation σ_H = hoop stress, psi or MPa D = ...

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That is certainly the hard way to do it. You could go to a local welding supplier and buy or lease a standard steel pressure bottle / tank ; good for probably 3 X your maximum pressure (designed and built in conformance to ASME Sec. 8 , Div.1 ). Stainless steel is certainly not needed for air or most gasses. You don't need a turbine to get the cooling ...

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The scaling for a simple case is easy to calculate. Assume a spherical pressure vessel of radius $r$ and wall thickness $t$ with internal pressure $P$. The stress in the walls scales as $$\frac{Pr}{t}.$$ So, if you double $r$, for the same quantity of gas you reduce $P$ by a factor of 8 (since the internal volume scales as $r^3$), which reduces $t$ by a ...

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It’s interesting that you know the price. Is there a supplier - if so, contact customers for their experience. It’s my opinion that an un-reinforced post will break during lifting due to the bending moment tension within the concrete. I wouldn’t expect corrosion of the reinforcement to be a problem. The steel should be at the center and covered by concrete. ...

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First of all, there will be two deflection equations, because the stiffness is not uniform. I'm getting the impression you somehow tried to use a single equation for the whole beam, which would obviously fail. The basic equation for cantilever beam you posted seems correct. Now for the left part you can use that as is, substituting the total length (L1+L2) ...

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Put a modest floor layer on the bottom (another piece or two of cardboard), and the failure location (with careful handling, no bouncing resulting in concentrated load) becomes the bends at the edges, so I would guess the max load should be proportional to the perimeter.

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Your question lacks many essential specifics required for a meaningful comparison. However, in the general sense, while the larger box have more volume/room, thus may hold more weight, but a smaller box will be more effective, and stronger, compared to the former. The reason is the smaller box posses the shorter span length, for everything else holding ...

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A riser is a static reserve volume of metal available to flow into the casting to compensate for shrinkage as the liquid freezes. Risers are relatively large because they should not freeze until after the casting has frozen. Risers may be "hot topped", exothermic material is put on to provide extra heat to keep the riser liquid as long as possible. ...

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