Solve Prob. 6.85, assuming that the thickness of the beam is reduced to (1)/(4) in.

Name: Problem 86, Chapter 6: Mechanics of Materials
Uploaded: 2020-06-27T19:09:31-08:00
Duration: 5 min 23 s
Channel: Vipender Yadav
Description: Solve Prob. 6.85, assuming that the thickness of the beam is reduced to (1)/(4) in.

Solve Prob. 6.85, assuming that the thickness of the beam is...

Key Concepts

Dimensional Scaling Effects

Understanding how changes in dimensions impact a structure is essential in engineering analysis. When a key parameter like the beam thickness is modified, it scales the moment of inertia and other related factors nonlinearly, affecting the beam’s ability to resist deformation and maintain structural integrity under load.

Cross-Sectional Geometry

Cross-sectional geometry defines the shape and dimensions of a beam’s section, and these features are vital in determining the beam's mechanical properties such as strength and rigidity. A reduction in any key dimension, like thickness, significantly alters the resistance to bending and, consequently, the structural performance under applied loads.

Euler-Bernoulli Beam Theory

This theory provides the fundamental relationship between bending moments, curvature, and deflections in a beam. It simplifies the analysis of how beams bend under loads by assuming that cross-sections remain plane and perpendicular to the deflected neutral axis, which is crucial for calculating deflections and internal stresses in structural elements.

Moment of Inertia

The moment of inertia, also known as the second moment of area, quantifies a cross-section's resistance to bending and is highly sensitive to changes in geometric dimensions. For rectangular or other shaped beams, modifications in thickness will directly affect the moment of inertia, thus influencing the overall stiffness and deflection behavior of the beam.

Transcript

00:01 So here in this problem we have to find the force p to lower the beam in the last problem we have to raise the beam that is problem 8 .52 but in this problem we have to lower the beam and the data is same as of the last problem so me of s it is 0 .3 now first of all we'll be calculating the angle of static friction for steel for the steel surface so it will be fire s will be called to tangent inverse mu s and mu as we know it is 18 point it's 0 .3 and which comes out to be as 16 .7 degree so this is 5 s now next we'll be drawing the free border diagram of the beam and the plate cd so there is a free boarded diagram now i'll be using the equilibrium condition that is sigma f y is equals to to 0 to find the value of the reaction r1 and i'm considering the force in the four direction to be positive and in the download direction to be negative so i can write that r1 cosine 16 .7 5 as we know it's 16 .7 degree minus 18 is equal to 0 from here r1 comes out to be as 18 .79 kips next we'll be drawing the free body diagram of the steel e that is the steel surface e so this is a free boarded diagram now here before solving further we have to calculate this angle and this angle we know that this total angle is 5 and this is 12 degree so this angle would let name this angle to be as theta 2 so theta 2 would be equals to its 5 s is 16 .7 so 16 .7 degree minus 12 degree which comes out to be as 4 degree 4 .7 degree so let me write here that theta 2 is equals to 4 .7 degree so now we can use the equilibrium condition and that is sigma fy is equals to 0 so it's r2 cosine theta 2 so theta 2 is 4 .7 degree minus r1 cosine 5 s 5 as 5 as we know it is 16 .7 degree it will be equals to zero r2 we have to calculate r1 we know so r2 cosine 4 .7 will be equals to r1 we know it is 18 .79 cosine 16 .7 from here r2 will be equals to 18 .05 kips next we can use a equilibrium condition in x direction so sigma f x will be equals to zero in this case i am considering the forces in this direction to be positive so we can write r1 sine 5 s plus r2 sine theta 2 minus p is equal to 0 now we can plug all the values here so r1 we know it is 18 .79 kips times sine 5 s it is 16 .7 degree plus r2 we know it is 18 .05 kips times sine theta 2.

04:00 Sign theta 2 is 4 .7 degree and it will be equals to p.

04:04 So from here p will be equals to 6 .88 kips and the direction of p would be towards this direction, this side...

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