If w=800 lb/ft, determine the absolute maximum shear stress in the beam. The supports at A and B

If w=800 lb/ft, determine the absolute maximum shear stress...

Key Concepts

Simply Supported Beam

A simply supported beam is one that is supported at its ends in such a way that it is free to rotate but not to translate vertically. Recognizing this support condition is important because it governs how the internal forces, such as shear force and bending moment, are distributed along the beam under an applied load.

Cross-sectional Properties

The properties of the beam’s cross-section, such as the area, moment of inertia, and the first moment of area, are fundamental parameters that influence the distribution of stresses including shear stress. These properties are necessary for calculating the exact stress values and identifying the location of maximum stress within the beam.

Shear Stress in Beams

Shear stress in a beam results from internal shear forces acting on a given cross-section. The calculation of shear stress typically involves dividing the shear force by a section property, such as the cross-sectional area or using more advanced formulas (for example, tau = VQ/It) depending on the geometric complexity of the cross-section, making it essential to understand how these stresses are distributed to ensure safe design.

Uniformly Distributed Load (UDL)

A uniformly distributed load is a load that is spread evenly along the length of a beam, meaning each unit length carries the same load intensity. This concept is crucial for determining the reaction forces at supports and for constructing the shear force and bending moment diagrams, which in turn are used in stress calculations in structural members.

Shear Force Diagram

The shear force diagram represents the variation of the internal shear force along the beam’s length. Understanding how to construct and interpret this diagram is key to identifying critical sections, such as where the maximum shear force occurs, which is necessary for subsequently analyzing shear stresses.

Transcript

00:02 Question the shear stress at this specific point will be given by tau max that is equals to v q divided i multiplied by t so for the given geometry which is this this is the given geometry so first of all we will calculate the center of area so using reference point top of the cross section we get y equals to to summation of y i a i divide by summation of a i okay so we can write four multiplied by two multiplied by one plus two multiplied by four multiplied by four multiplied by four divided by two multiplied by four plus two multiplied by four multiplied by one okay so from here we get y equals to two point five inch okay this is the value of y from the top section okay now movement of inertia depends upon the body mass distribution so moment of inertia of the cross section i will be equals to four multiplied by 2 .5 cube multiplied by 1 by 12 plus 4 multiplied by 2 .5 multiplied by 1 .25 square minus 2 multiplied by 0 .5 cube multiplied by 1 by 12 minus 2 multiplied by 0 .5, multiplied by 0 .25 whole square, plus 1 by 12, multiplied by 4 multiplied by 3 .5 cube, plus 4 multiplied by 3 .5, multiplied by 1 .75 square, minus 2 multiplied by 3 .5 cube multiplied by 1 by 12, minus 2 multiplied by 3 .5 cube multiplied by 1 by 1 by 12, minus 2 multiplied by 3 .5 by 3 .5, by 1 .75 square.

02:05 So from here after solving we will get moment of inertia i is equal to 49 .333 inch power 4.

02:17 This is the value of moment of inertia.

02:19 Now since the system is symmetric, we can calculate the reaction forces using equilibrium equation we get summation of f y equals to zero.

02:29 So, now we can draw the shear force diagram for the given condition and the shear force diagram will be look like this.

02:39 This will be the shear force diagram...

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