Last Updated on April 20, 2025 by Maged kamel
In this post, we will continue our discussion of practice problem P7-35. We will use built-in Excel expressions to create expressions for v(x) and M(x) and will be able to draw shear and moment diagrams.
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 at the middle of the span.
Part a) requires using the discontinuity function to write the expression for load w(x) 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), V(x) and M(x) must be used 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. Please refer to the next slide image for more details.
Use an Excel sheet and the statement, if then and, for the expression for the shear. In the next slide, I initiated a table for x values and placed x=3 M twice, since we have a drop at x=3 ft due top the applied moment, 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 B17, =IF(AND(0<=A14,A14<=3),-5,IF(AND(3<=A14,A14<=6),-5,0))
The expression for the value of shear is in cell B18, =IF(AND(0<=A18,A18<3),-5,IF(AND(3<=A18,A18<=6),-5,0))
for x equals 6, the expression for shear in cell B21 is =IF(AND(0<=A21,A21<3),-5,IF(AND(3<=A21,A21<=6),-5,0))
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 -5 KN for the entire span of the beam. It will plotted below the datum line.
Plot the Moment diagram.
The following slide contains the plot of the moment created by using a graph with an Excel graph.
For the expression of Moment function: M(x)=-5*<x-0>^1+20*<x-3>^0+5<x-6>^01+10*<x-6>^0.
 I have used the function for cell c14 for x=0 to represent the moment value as=IF(AND(0<=A14,A14<=3),-5*A14,IF(AND(3<A14,A14<=6),-5*A14+20,0))
 I have used the function for cell c17 for x=3 to represent the moment value before the drop as=IF(AND(0<=A17,A17<=3),-5*A17,IF(AND(3<A17,A17<=6),-5*A17+20,0))
 I have used the function for cell c18 for x=3+ to represent the moment value after the drop as=IF(AND(0<=A18,A18<3),-5*A17,IF(AND(3<=A18,A18<=6),-5*A18+20,0))
 I have used the function for cell c21 for x=6 to represent the moment value at x=6M as=IF(AND(0<=A21,A21<3),-5*A21,IF(AND(3<=A21,A21<=6),-5*A21+20,0))
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 moment value equals zero at x=0 and increases to -15KN.m. An increase to +20KN.M will be noticed at the right side of the same point. The moment at support B equals zero.
The previous post is Part 1 of Practice problem-7-35 Discontinuity Functions.
There is a video for this post; this is the link.
The following post is post 5, part 1 of Practice problem-7-38 Discontinuity Functions.
This is a link to my “