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Force, P = 1000 lbs Beam Span, L = 20' Max. Moment, M = PL/4 = 1000(20)/4 = 5000 lb-ft Shear, V = P/2 = 1000/2 = 500 lbs According to the table below, both of your choices are capable of handling the load. LVL 2650Fb-1.9E Technical Guide 24F Douglas-Fir Design Table

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If this is a slab-on-grade problem, open the program and from the main page, select "Define" tap, and find " Define Spring Constants" in the sub-manual to select the type of spring and provide the required data. I suggest reading the ETABS User Manual first to learn how to model a slab-on-grade structure.

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I would use a 1/8 x 4 x 10-inch plate on each side with a proper pre-drilled hole on it for your eye bolt with four 1/4 inch bolts on each corner. A swing not supported by a plate will eventually split the beam. Hanging them to the bottom of the beam has its own risks of failure.

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The deflection of E depends on the L length of the wires. if L is long enough to allow the softer wire to expand ultimately longer than the distance of the beam AB it will never receive half of P and will be just riding along. Most of the load will be carried by the stiffer wire and beam AB will rotate 90 degrees. let's call the stiffness of the wires K1 and ...

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Weight. As turbines get larger, it's the point of diminishing returns it takes a steady and constant wind velocity in order to accomplish the same goal. The bigger the machines get...........the efficiency rate declines. It's all based on the Cube square law. When an object undergoes a proportional increase in size, its new surface area is proportional to ...

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Basically AC and BD can be thought of as two springs. Because the equivalent "spring constant" is given by: $$k_{eq}= \frac{EA}{L}$$ you need to calculate the "spring constant" for copper and aluminium. If the spring constant is the same, then the elongation will be the same. If one of the is larger (e.g. $k_c> k_{al}$), then the ...

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They won't deform by the same amount. Assuming that the horizontally beams are deformable bodies (i.e. not rigid), you can calculate the forces acting on each of the corners by simple statics. Aluminum is experiencing P/2 load on each end, and so is Copper. A difference in Elastic Modulus exist between these two materials, so for the same equal force on both ...

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You made a mistake in solving the reaction at joint "A". See calc below: $\sum M_G = 0$ $R_A = \dfrac{22.31*8}{12} = 14.873$ kN Solve internal member force using the method of section: Since there is only one unknown in the vertical direction, so we can solve the member force $F_{BC}$ directly by $\sum F_X = 0$ $\sum F_X = 0$ $-F_{BC}cos 30^{o}$ +...

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I/L is relative stiffness. For the same flexural member material and cross sectional characteristics, the longer the flexural member, the more it displaces/deflects. Therefore, I/L provides a relative stiffness value depending on member's length.

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Let's try this way: $k = \dfrac{F}{\Delta}$, and $\Delta = \dfrac{FL}{EA}$. Rearrange the terms, it becomes $\dfrac{F}{\Delta} = \dfrac{EA}{L} = k$ Note the terms $F$ (internal force/reaction of a member in an assembly) and $\Delta$ are both "unknowns" before solving the problem. The member stiffness is therefore characterized by the terms on the ...

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If this is a roof hatch, it shall have its own frame to support the steel panel. You shouldn't have the problem figuring out the required bolts and bolt size by modeling a flexible beam between the hinge supports. The important thing is not only knowing the person will not fall through, but the deflection of the plate also counts heavily on human perception ...

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If your framing is lumber, it offers a bit of elasticity for your sheet metal to deflect a little, say 1/8" which is a relaxing deformation. Then by checking Roark's formulas for stress and strain 7th ed. on flat plates restrained on 4 sides pp 514 $$\sigma_{center of long side}=\frac{\beta_1q*b^2}{t^2}$$ $$\beta_1=0.3834$$ . Basically, your stresses ...

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The invariants affect the shape of the yield surface. The von Mises condition assumes that the yield surface remains cylindrical in principal stress space. If you want pressure-dependence (the circular cylinder becomes a circular cone), then you add the first invariant into the mix. If the yield surface varies depending on whether you are in pure triaxial ...

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TL;DR: you need to prioritize which behaviour is important and scale the load to investigate this behavior. It is very difficult to scale the load and expect to obtain the full behaviour of the structure. IMHO it is very difficult to scale the load and expect to obtain the full behaviour of the structure. Different loading conditions have different ...

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I am going to provide the hints for you to solve this problem. Hint 1: Since the moment diagram has a singular point (change shapes) at the internal supports B & C, so you need to consider each segment (A-B, B-C & C-D) separately in writing the moment equations. Hint 2: However, due to symmetry in both beam geometry and loadings, you only need to ...

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An easy way is to think of it as two separate beams. The center beam with L= 7meter will have half of its total 7meters loading as reaction on supports B and C and maximum moment $$M_{center segment}= \frac{\omega, (2236.13kgm) 7^2} {8}$$ Eaxh of the two side legs can be considered a cantilever beam but their end momend, because of symetry will in effect ...

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1.) "... if we stretch it too far, or keep it under tension for too long without firing, it would deform and then not work anymore, am I right? In a general sense, your thoughts are fine, but let's be more specific about the different scenarios: (1a.) - If the spring is stretched beyond its breaking strength, it breaks and would no longer work. (1b.) - ...

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According to this article: A Phased Approach to Optimized Robotic Assembly for the 777X drilling multi-material composites does require some special trickery, especially when automated assembly is part of the process: There were lessons learned relating to process capabilities whereby technologies were added or modified to prevent the need to return to hole ...

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Just to be clear, there are external forces at work here! When you are extending the spring (or whatever), the box doesn't move downwards due to the restoring force of the ground it's sitting on. If you tried to make the box move while it was floating in zero-gee, conservation of momentum would make that impossible.

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Question 1: If we don't pull the spring hard/far enough, when we release it it wouldn't provide much force. But if we stretch it too far, or keep it under tension for too long without firing, it would deform and then not work anymore, am I right? You are right. The effect you are describing is called creep. most materials suffer from it. However, in this ...

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Depending on how high you want it to go, you may not even need a spring, just a solenoid with sufficient travel and weight to it. The force generated will depend on $F=ma$ with the $m$ of the thing moving and how fast it starts and stops moving, but this is probably difficult to determine. It would be better to run some tests.

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The paper quoted is Open Access, and gives the reason in the opening paragraphs: for most stack machining operations settled in aerospace applications, liquid cutting fluids are prohibited due to quality and automation requirements They in turn cite Rivero A, Aramendi G, Herranz S, de Lacalle LL (2006): An experimental investigation of the effect of ...

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