Last Updated on May 30, 2024 by Maged kamel
Part 1 of Practice problem-7-35 Discontinuity Functions.
We will solve practice problem 7-35, which involve developing load function W(x), shear function V(x), and moment function M(x) using the discontinuity function. The same problem is P 7-59 in the third edition.
The practice problem is included in Prof. Timothy Philpot’s book An Integrated Learning System, Mechanics of Materials. The same practice problem is on pages 7-56 in the third edition.
We have a given cantilever beam with a span of 6 m acted upon by one downward concentrated load of 5 KN at point A and one concentrated moment of 20 KN-M acting clockwise.
Part a) requires using the discontinuity function to write the expression for load w(x) in part a) and to include the beam reaction in the expression.
In part b), w(x) must be integrated twice to determine the V(x) and M(x).
In part c), it is necessary to use V(x) and M(x) to plot shear force and bending moment diagrams.
What is the case in the discontinuity table that covers our Practice problem 7-35?
In the next slide image, we can find the first and the second cases, where the beam is acted upon by a vertical load and a moment covering our practice problem 7-35
Find the beam reactions for Practice problem 7-35.
First, we draw the free body diagram for the beam, which includes two reaction forces and a moment for the fixed support C: Cx, Cy, and Mc.
We are using the three equations of equilibrium: ∑X=0, ∑Y=0, and ∑M=0.
Following the standard rule for directions. For horizontal loads, the positive direction is to the right. For vertical loads, the positive direction is to the upward direction. For moments, the positive direction is the anticlockwise direction.
The horizontal reaction component is at Cx=0 since there are no acting horizontal loads.
From ∑ Y=0, Cy =5 KN acting upward. From ∑M=0. Take a moment for all forces about the right support C, multiplying each force and reaction by the arm length to support D and considering the directions of moment rotations. +5*6-20-Mc=0.
From ∑M=0. Take a moment for all forces about the right support C, multiplying each force and reaction by the arm length to support D and considering the directions of moment rotations. +5*6-20-Mc=0. Mc value equals 10 KN.M; since it is positive, it means as assumed in direction.
Write the three expressions for w(x), V(x), and M(x).
In the next slide, we are ready to write the expression for W(x) for the given beam; these are cases 1 & 2 from the discontinuity table discussed earlier. Consider the left point A as your 0,0 point. The horizontal axis X passes by the(0,0) point. The Y-axis will be pointing upwards.
Following the positive direction, the upward direction for positive loads is essential. The Load of 5 KN Ay is downward, so it is negative, acting at x=0, so we write -5*<x-0>^-1 since we are dealing with load intensity.
After three meters comes a clockwise moment of 20 KN.M KN, so the expression is +20*<x-3>^-2. At x=6m, there is an Upward reaction load of 5 KN, its expression is +5*<x-6>^-1, and a moment of 20 KN.M, its expression is +0<x-3>^-1.
Finally, we have Cy of +5 KN acts at x=6 m and a moment of +10 KN.M. The W(x)=-5*<x-0>^-1+20*<x-3>^-2+5<x-6>^-1+10*<x-6>^-2.
We do not consider the last two reactions in our calculation since they represent a step function for distance >6 m as we will see next.
The next step is integrating one to get the V(x) for shear.
V(x)=-5*<x-0>^0+20*<x-3>^-1+5<x-6>^0+10*<x-6>^-1. We have integrated by adding only n+1, and we do not have divisions by n+1. Again integrate to get the value of M(x)=-5*<x-0>^1+20*<x-3>^0+5<x-6>^01+10*<x-6>^0. Then, parts A and B were completed.
The shear values are based on V(x) from x=0 to x=5 m.
In the next slide, we write the V(x) expression. V(x)=-5*<x-0>^0+20*<x-3>^-1+5<x-6>^0+10*<x-6>^-1.
If x <3 the term *<x-3>^-1 is zero. If x>3, it will be positive.
The term <x-6>^0, if x is less than 6, is zero, while if x >6, the term is positive. The moment term does not affect shear. For x=0, V(0)=-5*1+0+0+0=-5 KN. For x=2, V(2)=-5*1+0+0+0=-5 KN. For x=3,
V(3)=-5*1+0+0+0=-5 KN. For x=4, V(4)=-5*1+0+0+0=-5 KN. For x=5, V(5)=-5*1+0+0+0=-5 KN.
For x=6, V(6)=-5*1+0+0+0=-5 KN.
Values for M(x) for x=0 to x=6m.
In the next slide, we write the expression for M(x). M(x)=-5*<x-0>^1+20*<x-3>^0+5<x-6>^01+10*<x-6>^0. For x=0, M(x=0 )=-5*(0)=0 KN.M.
For x=2 M, M(x=2)=-5*(2-0)^1=-10 KN.M. For x=3 M, M(x=3)=-5*(3-0)^1=-15 KN.M.
For x=3 plus M(x=3 plus)=-5*(3-0)^1+20*(+ve)^0=-15+20=+5 KN.M.
For x=4 M, M(x=4)=-5*(4-0)^1+20*(+ve)^0=-20+20= 0 KN.M.
For x=5 M, M(x=5)=-5*(5-0)^1+20*(5-3)^0=-25+20=-5 KN.M.
For x=6 M, M(x=6)=-5*(6-0)^1+20*(6-3)^0=-30+20=-10 KN.M.
This is the end of Part 1 of Practice problem-7-35 Discontinuity Functions. Please refer to the second part of this post.
The previous post is the 3a-Part 2 of Practice problem-7-32 Discontinuity Functions.
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