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42

More compact configuration that fully utilizes the area of the leaf. Avoid stresses concentrated on a single plane that is likely to cause the base material to progressively fail in shear or split. The staggered arrangement is more stable. It provides better strength in resisting the incidental bending resulted from the weight of the door and missing/loose ...


35

Because the screws go into wood and if the screws are in line then the wood will most often split between the screws in the grain direction and then the screws come loose.


17

Staggering the screws will give a better chance for some of them to penetrate into solid grain. and discourage toilet paper perforation pattern. Provide for larger torque resistance. keeping the hinge from develoing a loose play, flapping out of door jamb plane. prhibiting door settling slanted out of its frame.


7

In a wing the normal situation is that the aerodynamic force is upwards (resisting gravity). You are right that there is some shear and a fair amount of torsion, but the result is that: the top side of the wing is in compression, while the bottom is in tension. So in essence what happens is the opposite from the picture below (the force is applied upwards, ...


5

Also someone may have to pick the machine up and carry it. Reducing weight is not just related to the intrinsic function of the machine, its shafts and so on.


5

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 ...


5

The direction of Forces isn't necessarily along the connecting element. If that happens depends a lot on the constraints between the different elements. For example see the following image: In the left column is a "welded" structure, while on the right column is a pin jointed structure (a basic truss) if you like. On the top there are the shapes ...


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

Cost wasn't mentioned yet - Depending on the manufacturer, their processes, the cost of the raw metal, and the accountants it may be that the waste is swept up and reprocessed. For gears made of brass, bronze, aluminium or titanium this is well worth doing. Any other exotic metals depend on the cost per kilo. Copper and Lead would be excellent metals for ...


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 ...


3

Use graphics to visualize the force flow and determine zero-force members. (Draw member forces from joint D)


3

DE and AE are zero are indeed zero. However you need to start from point C to reason about that. If you look at this question, when you have a node without external forces with only two members which are not colinear, the forces on the nodes should be zero. So from this $DC$ and $CB$ are equal to zero. Node D: Now if you take point D and write the equations ...


3

The problem is that the angle between T and the 20 degrees is not only 20 degrees but 20 + 40=60 deg. Look at the triangles that are formed. EDIT>>> A second problem is with the equation for moments $$\tau=T\sin(20)*2a\sin(50)-mg*a\sin(50)=0$$ Although you have included the moment produced by $T_y = T\sin(20+40)$, you need to also include the ...


3

You need to compute the length of the moment arm from $B$ to $\overline{AC}$ That is the height of the triangle, and can be computed directly from tuples as $\frac{\|\overline{AC} \times \overline{AB} \|}{\|\overline{AC}\|}$ So $\overline{AC}\times\overline{AB}= \{72,108,54\}\qquad\|\overline{AC}\times\overline{AB}\|= 140.58\,sqin$ and $\overline{AC}=\{-9,6,...


3

In order for the Resultant force to be vertical you only need the sum of the horizontal components of $F_1$ and $F_2$ to be zero $$F_{1x} - F_{2x} =0 $$ $$F_{1}\cos(70) - F_{2}\cos\theta =0 $$ $$\cos\theta=\frac{F_{1}}{F_{2}}\cos(70) - F_{2} $$ $$\theta=\arccos\left(\frac{F_{1}}{F_{2}}\cos(70) - F_{2}\right) $$ if you substitute you should get $$\theta=50[...


3

Just adding to the rest of the answers, if the door hinge is thick enough to be assumed rigid, then the offset configuration offers better resistance to bending moments in at least two axis. i.e: For the following reference system Case Y-Axis bending moment Z-Axis bending Front View Top View It is noteworthy, that for a pure pullout force (which is ...


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 ...


2

The parallel to f1 is at an angle of 50 degrees with F2. The angle opposite the resultant in the triangle with F2 and the parallel F1 is 130 degrees .


2

I don't have time to type this up nicely - thought it would be quicker to write out by hand but in hindsight my handwriting is terrible. Let me know if you can't read anything!


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 ...


2

The proposed system won't work, as the foot straps are too flexible to prevent the large deflection and rotation of the posts. Instead, you shall embed the posts into the foundation (not shown) below the grade.


2

Wings are designed as a complex structure of spars and ribs, clad with aluminum, titanium, or new composite sheathing, or in the fabric, in the early planes. The Frame has been designed to support all the loads, bending moment of the wings, torsion, shear, compression, and tension. It has been designed to never let the surface buckle because it would be ...


2

The shear stress on hinge screws is close to zero. The door is held by the friction of the plate to the frame, not by the shear strength of the screws. All your calculations ignore that the screws are holding the hinge plate fast to the frame. If you calculated the force holding the plate to the frame you would find that the friction is a very large force....


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

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

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

It's so the screw heads on opposite leaves don't hit each other when the hinge is closed. The heads aren't always 100% flush. Mechanically, this compromises some load cases, but not the limiting ones.


2

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 $\theta = 0$, ...


2

Define: Force away from the joint is positive (+) Draw the FBD as shown on the left, keep in mind that the member end force is the flip of the joint force with the identical magnitude, and the sum of the member end forces must be zero to maintain the static equilibrium of the member.


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