Part 2 of Practice problem-7-32 Discontinuity Functions.
Part 2 of practice problems 7-32 involves drawing the shear-force and Moment diagrams for a given beam acted upon by two vertical loads.
The practice problem is included in Prof. Timothy Philpot’s book An Integrated Learning System, Mechanics of Materials. The same practice problem is P7-56 in the third edition.
We have a given beam spanning 9 feet acted upon by two concentrated loads: P1 = 180 lb, and the second concentrated load is P2 = 450 lb. The two loads act downwards.
Part a) requires using the discontinuity function to write the expression for load w(x), the beam reactions in the expression. For the first part, please refer to the following link.
In part b), w(x) must be integrated twice to determine the V(x) and M(x).
The following slide draws two sketches of V(x) and M(x) based on the information and estimations.
We have a drop in shear value at x=2 ft and x=6 feet.
The shear force diagram shows positive and negative values; the positive values are drawn above the datum line, while the negative ones are below.
The shear force value at A is equal to 290 lbs; the value continues till point b, where there is a drop in shear by a value of 180 lbs. The value of 110 lbs is obtained till point c, where there is a drop of 450 lbs. The shear value equals -340 lbs, which will continue to the end.
All M(x) values are positive; we will draw above the datum line.
Use an Excel sheet and, if then, for the expression for the shear.
In the next slide, I used an Excel sheet. First, I checked the values of Ay and Dy using the sum for vertical loads and the sum of the moments at both A and D using rows from 1 to 8.
I have initiated a tabler for x values and placed x=2 feet twice and x for 6 feet twice to create a step in the Excel sheet.
Remember that we have V(x)= 290*<x-0>^0-180*<x-2>^0-450*<x-6>^0+340<x-9>^0.
I used the if, then, and else functions in the Excel sheet containing three conditions. The first condition is from 0 to <2, followed by x from >=2 to x<6 feet, and the third condition is for x>6 to x<=9
The value of shear is used in cell A15 for =IF(AND(0<=A14,A14<2),290,IF(AND(2<=A14,A14<6),290-180,IF(AND(6<=A14,A14<=9),290-180-450,0))).
A14 here represents the x value. In the first term, we included 290. In the second term, we include 290-180. In the Third term, we include 290-180-450
To create a step at x=2, we consider the term as smaller and equal to 2; we use the A16 cell in the statement as follows:
=IF(AND(0<=A16,A16<=2),290,IF(AND(2<A16,A16<6),290-180,IF(AND(6<=A16,A16<=9),290-180-450,0)))
While in the second term, we use bigger than or equal 2, we use A17 cell in the statement as follows:
=IF(AND(0<=A17,A17<2),290,IF(AND(2<=A17,A17<6),290-180,IF(AND(6<=A17,A17<=9),290-180-450,0)))
In the next slide, we see the equation for a step for the first x=6 value
To create a step at x=6, we consider the term as smaller and equal to 6; we use A21
=IF(AND(0<=A21, A21<2),290, IF(AND(2<=A21, A21<=6),290-180, IF(AND(6<A21, A21<9),290-180-450,0))).
In the second term, we use > or equal 6 in cell A22
=IF(AND(0<=A22,A22<2),290,IF(AND(2<=A22,A22<6),290-180,IF(AND(6<=A22,A22<=9),290-180-450,0)))
Plot the shear force diagram.
In the next slide, please find the plot of shear force created by using a graph with an Excel graph.
Plot the Moment diagram.
In the Last slide, The moment function is :
M(x)= 290*<x-0>^1-180*<x-2>^1-450*<x-6>^1+340<x-9>^1.
I have used the function for cell c14.
=IF(AND(0<=A14,A14<2),290*A14,IF(AND(2<=A14,A14<6),290*A14-180*(A14-2),IF(AND(6<=A14,A14<9),290*A14-180*(A14-2)-450*(A14-6),0))).
The sketch of the moment is drawn using an Excel graph.
The previous post is Part 1 of Practice problem-7-32 Discontinuity Functions.
There is a video for this post; this is the link.
The next post is post 4, Part 1 of Practice problem-7-35 Discontinuity Functions.
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