The cross section of an extruded beam is a hollow square of side a=3 in. and thickness t=0.25 in

The cross section of an extruded beam is a hollow square of...

Key Concepts

Shear Stress in Beams

Shear stress refers to the internal forces per unit area that develop in a beam’s cross-section when it is subjected to a shear load. In structural analysis, the distribution of shear stress helps determine how different parts of the beam will perform under applied forces. The most common expression, ? = VQ/It, indicates that shear stress depends on the applied shear force (V), the first moment of area (Q) of the portion of the cross-section, the total moment of inertia (I) of the cross-section, and the local thickness (t).

Shear Flow

Shear flow is a measure of the rate of transverse force transfer along or between sections of a beam, defined as q = ?t. It is particularly useful in the analysis of thin-walled sections where shear stresses are not uniformly distributed. Understanding shear flow helps in determining how shear forces are transmitted through the structure, especially along walls or flanges, and is critical for ensuring the continuity and integrity of the cross-section.

First Moment of Area

The first moment of area, often symbolized as Q, is calculated by integrating the area above (or below) a given line multiplied by its distance from the neutral axis. It plays an essential role in shear stress calculations because it quantifies the contribution of different segments of the cross-section to the overall distribution of shear forces. Its value is key to accurately determining the maximum shearing stress in non-uniform cross-sectional shapes.

Second Moment of Area (Moment of Inertia)

The second moment of area, or moment of inertia (I), measures a cross-section's resistance to bending under load. It captures the spread of the cross-section's area relative to a given axis and is a fundamental parameter in both bending and shear analysis. In the context of shear stress distribution, it normalizes the contribution of the first moment of area, ensuring that the beam’s stiffness and geometric properties are taken into account.

Analysis of Hollow Sections

Hollow sections, like thin-walled extruded beams, exhibit non-uniform stress distributions due to their geometry. The analysis of these sections involves understanding how the material is distributed around the perimeter, which affects both the moment of inertia and the flow of shear forces. Analyzing hollow sections requires careful consideration of local thickness variations and the individual contributions of each segment to the overall structural performance.

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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