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if $ \ r<<R\ $ we can approximate $\text{the area of the cut-off lens as half the area of the small circle}$: Then the approximate answer is $$ J_{circle} = \frac{\pi R^4}{2}$$ $$J_{small circle/2} = …

I would consider placing the shear walls in a way that provided an almost even strength against any earthquake direction. … This means The shear walls are radially almost evenly set around the center of the stiffness of the floor. …

The stresses on the top screws will be tension and shear and shear alone on the bottom. …

Its a bit hard to follow what you are doing. You should use LaTex to write your calculations. The tension on the screw: $100lbs/2=50lbs \quad \text{half wight of the load }$ $50/2=25lbs \ \text{share …

Depending on the mass of the disk and its rotational speed we will have to calculate the procession period and secondary forces caused by that. $$\Omega = \frac{\tau}{ L} = \frac{\tau}{I\omega}$$ $\O …

This is a fix-fix joint beam that will deflect as shown, so by symmetry, we can calculate only the shear and bending moment on half of it as a cantilever beam. …

Common plastic used in grocery shopping has an average shear strength of 200-500 psi. Or check the datasheet! One could multiply this by the length of the cut, or perimeter of the circle. …

Your original stress state is the principal stress state because the shear is zero. …

By definition shear flow on a beam's cross-section is the same on the two sides of any horizontal plane. eg, the top surface is trying to move to the right, the bottom surface is reacting by trying to … move to the left with an equal amount of shear to maintain equilibrium. …

If "y" was a signed value all beams sections would end up having zero shears because the lower half of the section with supposedly a negative shear would cancel the half with positive shear. …

ANSI/AISC 360-16 Specs. pp. 195 has shear in Machin bolts at 60% of Fy. It has solved examples too. … The bolt groups under moment need to be calculated for rotation due to the torque as well as shear. Foundation needs to be designed too. …

The right hand load slopes up from zero at G, to point F reaching loading of w/2, then it slopes down from there to the the point at 4+4=8 ft from G, to reach zero load.

You are right the handle doesn't have any shear. The only member that has shear is member BC. It has a constant shear in the plane XZ of $V=400lbs$ …

Source, Eurocodeapplide.com Shear strength of bolts The shear resistance of the bolt per shear plane Fv,Rd is provided in EN1993-1-8 Table 3.4: Fv,Rd = αv ⋅ fub ⋅ A / γM2 where: αv is a coefficient that … When the shear plane passes through the unthreaded part of the bolt αv = 0.6. …

You should consider sheer density. When the stress increases, the velocity gradient increase. If the distance between velocity contour lines of say 0,1 m/s difference in speed used to be 1mm original …