The beam is made from a polymer and is subjected to a shear of V=7 kip. Determine the maximum sh

The beam is made from a polymer and is subjected to a shear...

Key Concepts

Moment of Inertia (I)

The moment of inertia of a beam’s cross section is a measure of its resistance to bending and is determined by the geometry of the cross section. It plays a critical role in both bending and shear analyses. In the context of shear, the moment of inertia normalizes the effect of the distribution of area (captured by Q) across the entire section, influencing the overall shear stress distribution.

Numerical Evaluation and Graphical Representation

For practical engineering applications, it is often necessary to evaluate the shear stress at discrete intervals across the beam’s depth to analyze and visualize the variation in stress. This step-by-step evaluation at specified intervals allows engineers to generate a plot of the shear-stress distribution, ensuring that no part of the beam exceeds the material's allowable stress limits.

First Moment of Area (Q)

The first moment of area, Q, is a geometric property that quantifies the distribution of an area relative to a given axis. It is defined as the integral of the area times the distance from the centroid of that area element. Q is key in shear stress calculations because it reflects how much of the cross-sectional area is located at a given distance from the neutral axis, which directly affects the magnitude of the shear stress.

Shear Formula (tau = VQ/Ib)

The shear stress at any point in the cross section of a beam is calculated using the formula tau = VQ/Ib, where V represents the applied shear force, Q is the first moment of area above (or below) the point in question, I is the moment of inertia of the entire cross section, and b is the width of the section at the point where the stress is being calculated. This formulation provides insight into how the geometry of the cross section influences the stress distribution.

Shear Stress Distribution

Shear stress distribution in a beam describes how shear force is spread across the cross-sectional area. The distribution is typically non-uniform, reaching a maximum value at a certain point (often in the centroid region) and tapering off toward the edges. Understanding this distribution is crucial for ensuring that the material and cross-sectional design can safely withstand the applied loads without failure.

Transcript

00:09 This question we have moment of inertia i equals to 56 inch square okay so the shear stress is given by tau equals to v q by i t it can be written as v y bar y bar dash a dash divided by i t where t is the thickness which is equals to 1 okay and the a a dash is the area above the area above the n okay so at y equals to 0 we can write tau y equals to 0 equals to substituting the values we get 7 2 multiplied by 1 multiplied by 1 plus 4 into 1 into 2 plus 1 by 2 plus 1 by 2 divided by movement of inertia is 56 and t is 1.

01:17 So we get 1 .5 ksi.

01:19 At y equals to 0, this value is maximum.

01:23 So now moving to y equals to 0 .5 inch, we get tau y point y equals to 0 .5 inch.

01:36 We get 7, 2 minus 1 by 2, multiplied by 1 multiplied by 1 by 1 by 2 plus 2 minus 1 by 2 divided by 2 plus 4 multiplied by 1 multiplied by 25 divided by 56 multiplied by 1 so from here we get value equals to 1 .484 k s i okay so now solving at y equals to 1 we get y equals to 1 inch we get tau y equals to 1 inch equals to 7 1 multiplied by 1 multiplied by 1 plus 1 by 2 plus 4 multiplied by 1 multiplied by 1 multiplied by 1 divided by 56 multiplied by 1 and this value comes out to be 1 .438 k s i so now solving at y equals to 1 .5 inch we get tau one y equals to 1 .5 inch that is equals to 7 multiplied by 1 by 2 made by 1 by 2 plus 1 by 2 plus 1 by 2 plus 1 by 2 56 multiplied by 1 this value comes out to be 1 .359 ksi okay similarly solving at y equals to 2 inch so we get y equals tau at y equals to 2 as 7 multiplied by 4 mudd by 1 mudd plaid by 2 plus 1 by 2 divided by 56 muddclad by 1 this value comes out to be 1 .25 k si and now y equals to 2 plus okay this value is at 2 minus so this value comes out to be 7 into 4 into 1 multiplied by 2 plus 1 by 2 divided by 56 multiplied by 4 here t is equal to 4 4 full flange width okay so this value comes out to be 0 .313 k s i now now, at y equals to 2 .5, tau y equals to 2 .5 inch, we get 7 multiplied by 4, multiplied by 1 by 2, 2 plus 3 by 4 divided by 56 multiplied by 4...

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