Redo Problen 14.37 , however, with the beams now fixed at the ends as shown in Fig. P.14.38. Fig

Redo Problen 14.37 , however, with the beams now fixed at...

Key Concepts

Euler-Bernoulli Beam Theory

This theory provides the fundamental relationship between bending moments, beam deflections, and material properties. It assumes that plane sections remain plane after bending and that the beam material behaves elastically. The theory is essential for deriving equations that predict the deflection and stress distribution along the beam when subjected to loads.

Boundary Conditions in Structural Analysis

Boundary conditions specify how a structure is supported or restrained, directly influencing the internal force distribution and deflection behavior. For beams, different end supports—such as fixed, simply supported, or free—result in distinct moment, shear, and deflection profiles. Understanding these conditions is crucial for accurately modeling and solving structural problems.

Fixed-End Support Conditions

When a beam is fixed at its ends, both the displacement and rotation are restrained at those ends. This type of support introduces additional moments at the boundaries, which significantly affects the overall bending moment distribution and reduces mid-span deflections compared to simply supported beams. Analyzing fixed-end conditions requires incorporating these restraints in the equilibrium and compatibility equations.

Moment and Shear Force Analysis

Determining the bending moment and shear force distributions along a beam is key to assessing its structural behavior under load. Techniques such as static equilibrium methods, differential equation integration, and diagram plotting are employed to evaluate the internal forces. This analysis helps in designing beams to safely withstand applied loads by ensuring that stresses remain within allowable limits.

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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