Last Updated on June 5, 2024 by Maged kamel

## Part 2 of Practice problem-7-35 Discontinuity Functions.

Part 2 of practice problems 7-35 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 problem is P 7-59 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.

The following slide draws two sketches of V(x) and M(x) based on the information and estimations.

We have a constant shear value of -5 KN.

At x=0, the moment value equals zero. At x<3 m, the moment value equals -15 KN.m. At x=3m, the moment increases from -15 kn.m to 15 KN.M due to the acting clockwise moment +20 KN.M. The final moment value at x=6m equals zero.

### Use an Excel sheet and, if then, for the expression for the shear.

In the next slide, I initiated a tabler for x values and placed x=3 M twice and x for 6 M twice in the Excel sheet.

Remember that we have V(x)=-5*<x-0>^0+20*<x-3>^-1+5<x-6>^0+10*<x-6>^-1.

I used the if then, else function in the Excel sheet containing two conditions.

The first condition is from 0 to x<3, followed by x from x>3 and less than or equal to 6 M.

The expression for the value of shear is in cell AB175 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 For 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 For 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. We have a constant value of -5KN for the entire span of the beam.

### Plot the Moment diagram.

In the next slide, please find the plot of shear force created by using a graph with an Excel graph.

For the expression of Moment function: 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 to represent the moment value as=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))).

Please refer to the next slide image for more details about the expression for Moment. Column C gives the M value for each x distance.

The sketch of the moment is drawn using an Excel graph.

The previous post is Part 1 of Practice problem-7-35 Discontinuity Functions.

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The next post is post 5, part 1 of Practice problem-7-38 Discontinuity Functions.

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