This is Post number 19 of the Steel beam Posts, which includes 1—Solutions for the Block Shear-Coped Beam Problem.
A solved problem 5.13 for Block Shear-Coped Beam-case-1-UBS=1.00.
The solution includes a detailed description of the forces that affect the beam: shear yielding and tensile rupture, Shear rupture, and tension rupture. How do we estimate the LRFD and ASD for block shear?
The next slide image shows part of the shear yielding and tensile rupture estimation, with the values of the shear rupture and tensile rupture.
Solved problem 5-13-List of steel Beam Posts-part-3a
AISC Tables 9-3a,b, and C for Block shear-coped beam.
This is Post number 20 of the Steel beam Posts, which includes: 1- Solutions for Block Shear-Coped Beam Problem for the solved problem 5.13 for Block Shear-Coped Beam-case-1-UBS=1.00. Use tables Aisc Tables 9-3a for tensile rupture for block shear.
Table 9-3b or shear yielding component and AISC Table 9-3C for shear rupture component.
How do we estimate The LRFD and ASD for block shear?
This is a snapshot for Table 9-3a titled block shear-tension rupture component per inch of thickness kip/inch.
AISC Tables 9-3a,b, and C for Block shear-list of Steel Beam Posts-part-3a
This is a snapshot for Table 9-3c titled block shear-shear rupture component per inch of thickness kip/inch.
This is solved problem #10.8 from the Unified Design of Steel Handbook. It requires determining the resistance to block shear of the coped beam W16x40 grade A992. There are two bolt lines, a typical case -2 where UBs=1/2.
This is a snapshot of the shear yielding and tensile rupture calculations versus shear rupture and tensile rupture.
This is the link to post 21-Case 2 for block Shear-Coped Beam Problem.
This post is post Number 19s- for shear of the steel beam, in the list of steel beam Posts Part-3, which includes a Review of shear stresses for steel beams and the relation between horizontal and vertical shear stress.
A-Solved problem- 2-22 for shear stress parts A, B, and C.
First, an Introduction to shear stress for a rectangle; how can we estimate it? How do you get the first moment of area for an I section and derive the shear stress equation? Two posts discuss solved problems 2-22, which include Determining the maximum shearing stress for the following sections when the external shear force equals 75 kips for parts A) & B) and C).
This post includes a solved problem from Prof. Mccormac’s book. A solved problem 10.2- A given steel section W21x 55 with Fy =50 ksi. The loads for the beam are shown in Fig.10.4. It is necessary to check its adequacy in shear.
The second part discusses the Nominal shear strength for steel beams based on CM#14—the relation between Nominal shear and the ratio of hw/tw for different steel grades.
Solved problem 10-2: how do we use Table 3-2 for shear value?
The first part of the post includes a solved problem from Prof. Mccormac’s book. A solved problem 10.2- A given steel section W21x 55 with Fy =50 ksi. The loads for the beam are shown in Fig.10.4, in which I have used Table 3-2 to find the shear value for the solved problem. 10-2.
The second part discusses the Nominal shear strength for steel beams based on CM#15.
Check section M12.5×12.40 for shear and estimate the Nominal shear strength.
Part 2 of Practice problem-7-38 Discontinuity Functions.
Part 2 of practice problems 7-38 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 7-62 in the third edition.
We have a given beam, a given simply supported beam with a span of 20 feet. A downward load of 3000 Lb acts at x=5 feet from the left support.
Simultaneously, one concentrated moment of 8000 lb. ft acts anticlockwise. The downward uniform load of 800 lb/ft has a limited width of 7 feet starting from x=9 ft.
What are the cases that P7-38 covers regarding the discontinuity function in the table?
This problem covers cases 1,2 and 5 from the discontinuity table.
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), V(x) and M(x) must be used to plot shear force and bending moment diagrams.
Use an Excel sheet to find the values of reactions.
This is the Excel sheet I have used: the sum of an Ay +Ey=8600 Lbs, which is 3000 lb+ 800 * 7=8600 lbs.
We get the AY value by taking the moment at E, and the Ay value equals 4750 lbs. From the sum of Y forces, we have Ey=8600-4750=3850 lbs.
Use the if-then function built in Excel for the shear expression for the beam.
Another check to get the Ey is by taking the moment to the left support, which will again give us the same 3850 lbs. These regions are from zero to 5, 5 to 9, 9 to 16, and 16 to 20 feet.
The first column shows the x distance in feet, the second the shear Value in lbs, and the last the moment value in Lb.ft.
We have a drop at X equal to 5 feet.
We have a drop in shear by Value of 3000 lbs and a decline of moment value by 8000 lb.Ft. Also, again, there is a change in the slope at X = 9 ft—a drop of 800 lb/ft due to the distributed load.
How do we express the value of the shear and the value of the moment at the point of x=5? Please find the expression for shear.
There are two terms: one term for x<=5; in this term, the shear value equals 4750 lbs. Meanwhile, the second term is for X>5 and less than and equal to 9; the shear value is 4750-.3000.
The third term is for X>9 and less than and equal to 16; the shear value is 4750-3000-800*(A19-9).A19 is the cell for x value where x is 5 ft.
The fourth term is for X>16 and less than and equal to 20; the shear value is 4750-3000-800*(A19-9) + 800*(A19-16). A19 is the cell for the x value, where x is 5 ft.
I have added two cells for x=5. To create a step in the Excel shear at x=5, the upper cell B19 is for x<=5 in the expression. In the second cell, B20 is the expression for x<5 with no equal sign.
For x=5 and x=5 plus, I put two values for 5, five here and five here. And in the expression.
We have A19 > or equal zero. For A19 < or equal 5, putting the equal sign here and Less than five is essential.
The following slide image shows the plot for the shear force diagram created using an Excel graph. The x-axis represents the distance in Feet, while the y-axis represents the corresponding shear Value.
Above the datum will be considered for the positive values, while below the datum represents the shear values with negative signs. We have a shear value of 4750 Lb, from x=0 to x=5 feet.
Due to the concentrated load, we have a drop of 3000 lbs. So we have 4750-3000, and it will be 1750. Then comes the uniformly distributed load of 800 going downwards. We have a slope or drop of 800 lbs per foot.
At X equal to 9 feet, We have 1750 going down with the slope of 800 lb per foot. So, to get the point where the shear equals 0, we can divide 1750 / 800 and get 2.1875 ft.
We add this Value to 9, and the final distance is 11.1875 ft. This is the distance to the point of zero shear.
Please scroll to review all images in sequence.
Shear expression for x=5-P7-38.
Shear expression for x=9 ft by using if then function.
Part 2 of Practice problem-7-38-The shear diagram by using an excel sheet.
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Use the if-then function built in Excel to get the expression for Moment.
In the next slide image, there is an expression for M(x) using the if-then function in Excel. This is for the value of C19. We have a drop at the moment value at x=5 Ft. This is the first expression of<=5.
The sketch of the moment is drawn using an Excel graph.
We used this bending moment diagram, which we drew using the Excel sheet.
We started the moment at zero and then moved until X was equal to 5. At x=5, we have 23,750 lb .ft; due to the anticlockwise concentrated moment, there is a drop at X equal to 5, which equals 8000 LB .foot.
The Value of the moment after the drop equals 15,750 LB feet.
It starts from this point. We have a slope until the maximum moment, which is 24,664 Lb.ft. This point is at x=11.1875 Ft.
Please scroll to review all images in sequence.
The moment expression in excel using if then function.
Plot for Moment diagram P7-38.
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We have come to the end of our discussion about 5a-Part 2 of Practice problem-7-38 Discontinuity Functions.
Thanks a lot.
The previous post is Part 1 of Practice problem-7-38 Discontinuity Functions.
Part 1 of Practice problem-7-38 Discontinuity Functions.
We will solve practice problem 7-38, 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 7-62 in the third edition.
The practice problem is included in Prof. Timothy Philpot’s book, An Integrated Learning System, Mechanics of Materials.
We have a given beam, a given simply supported beam with a span of 20 feet.
A downward load of 3000 Lb acts at x=5 feet from the left support.
Simultaneously, one concentrated moment of 8000 lb.ft acts anticlockwise. The downward uniform load of 800 lb/ft has a limited width of 7 feet starting from x=9 ft.
This problem covers cases 1,2 and 5 from the discontinuity table.
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) is to be integrated twice to determine V(x) and M(x).
Part C needed V (x) and M(x) to plot shear force and bending moment diagrams.
There is also a limited width downward uniform load of 800 Lb/ft with a length of 7 feet starting from x=9 ft.
This problem covers cases Number 1,2, and 5 from the discontinuity table. Please refer to the next slide image for details on cases 1 and 2.
cases 1 and 2 for w(x),V(x) and M(x) expressions for discontinuity functions.
Please refer to the next slide image for details on case 5.
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.
Find the beam reactions for Practice problem 7-38.
First, we draw the Free-body diagram for the beam, which includes two reaction forces. Ax and Ay at support A and one upward reaction force.
EY at support E is the ruler support. The total uniform load value is 800 multiplied by 7 equals 5600 lbs.
We have Ax and Ay for reactions; on the other support, we have a vertical load, EY.
This is a concentrated downward load of 3000 lbs and the concentrated moment of 8000 LB feet, acting anticlockwise and is a total load of the uniform load, as shown.
We have a positive direction from left to right for horizontal forces, and from below to up, it is considered positive for vertical loads.
While for moments, the anticlock direction is the positive direction.
If we start with the sum of X equal to 0, we cannot find any horizontal forces acting on the beam if we investigate it. So we could say Ax=0. We consider the sum of ∑Y=0. We can assume Ay + Ey Equals (3000 lbs + 5600 lbs). Then, Ay and Ey will equal 3000 + 5600 = 8600 lbs. The last equation of equilibrium is the sum of ∑M at E=0. for all the forces. Considering the anticlockwise direction is positive, the AY multiplied by its arm will create a moment in that direction.
Since it’s a moment in the clockwise direction, we’ll consider it negative. The Ay value is +4750 lbs, meaning the assumed direction is correct. The vertical reaction Ey equals 3850 Lbs.
Part 1 of Practice problem-7-38-Reactions values.
Write the two expressions for w(x) and V(x).
We will move to the next slide. In this practice problem, we have 3 cases, 1-2 and 5 in the discontinuity function tables.
Case one is the case of concentrated moment, which we have anticlockwise of 8000 lb.Ft.
Case number two is the concentrated load, including the reaction at A, the focused load acting after 5 feet, and the other at E.
While Case 5 is the partially loaded uniform load, we have partially loaded uniform load acting. After 9 feet from the left, support and limited width, which is 7 feet.
We have several regions. We have one region from zero to five feet, followed by another area.
From 5 feet to 9 feet, then from 9 feet to 16 feet, and later from 16 to 20 feet.
We start to write the load intensity W(x) equals. The first reaction is 4750 lbs, acting upward in the positive direction.
We’ll write 4750 multiplied by(x-0)^-1 and deduct the downward load, writing -3000 lbs *(x-5)^-1. We will extend the distributed load of 800 lb/ft to the end and add an upward distributed load of 800lb/ft with a length of 4 feet.
For the extended load. We write -800 * (x-9)^X ^0 +800* (x-16)^0.
Finally, the reaction of the right support, which is 3850 * (x-20)^-1, will be ignored in the Calculation since this will start to work for more than 20 feet.
We integrate once to get the expression for v(x). The full detailed expression is shown in the next slide image.
Part 1 of Practice problem-7-38-W(x) and V(x) expressions.
The moment expression M(x).
In the next slide, we continue to integrate, and we have the expression of M(x) equals 4750 * (x-0)^1- 3000 *(x-5) ^1, -8000 * (x-5)^0- (800 / 2) *(x-9)^2+800 / 2 * (x-16)^2+3850 * (x-20)^1. The last term will be ignored. Thus, we have completed parts A and B.
Part 1 of Practice problem-7-38-Moment expression.
The shear values are based on V(x) from x=0 to x=11 Ft.
Part C will use V(x) and M(x) to draw and plot shear-force and bending moment diagrams.
In the next slide, we’ll introduce the expression of V(x) and start to get the values Of the shear at various points.
We start with V(x=0). We consider it as 0 plus or a small amount more than 0. We have the V(x=0 plus), which will be 4750 multiplied by.
Positive Value is to the power of 1; we consider it one. So V of X equal to 0 would be equal to 4750 lbs.
V of X of 3=4750 lbs. V of X of 5 equals 4750 * 5 raised to the power of 0=4750 lbs.
V of X of 5 equals 4750 * 5 raised to the power of 0=4750 lbs.
V(X) of 7, 4750 * 7 raised to the power of 0= 4750 – 3000 by 2^0=1750 lbs.
V(X) of 9, 4750 * 9 raised to the power of 0= 4750 – 3000 by 4^0=1750 lbs.
V(X) of 10, 4750 * 10 raised to the power of 0= 4750 – 3000 by 5^0-800*1=950 lbs.
V(X) of 11, 4750 * 11 raised to the power of 0= 4750 – 3000 by 6^0-800*1=150 lbs.
We will notice that there is an 800-lb drop. We are interested in the maximum moment or zero shear point. We assume that this point will occur between 9 and 16. So, we will write V(X ) =Zero.
The equation will be 4750*1-3000-800(x-9)=0. We have 800 * (x-9) = 4750 -3000, which gives 1750. We adjust the term and get the maximum value of X, which is 11.1875 feet.
This X is > than 9, but smaller than 16 feet.
Part 1 of Practice problem-7-38-Shear values from x=0 to x=11 FT.
The shear values are based on V(x) from x=12 to x=20 Ft.
We want to get the values from X equal to 12 to X equal to 20. V(X) of 12, 4750 * 12^0= 4750 – 3000 by 7^0-800*3=-650 lbs. V(X) of 15, 4750 * 15^0= 4750 – 3000 by 10^0-800*6=-3050 lbs.
V(X) of 16, 4750 * 16^0= 4750 – 3000 by 11^0-800*7=-3850 lbs. V(X) of 18, 4750 * 18^0= 4750 – 3000 by 13^0-800*9+800*2=-3850 lbs.
V(X) of 20, 4750 * 20^0= 4750 – 3000 by 15^0-800*11+800*4=-3850 lbs. This is precisely the numerical Value of the reaction at support E, which we have already calculated.
Part 1 of Practice problem-7-38-Shear values from x=12 to x=20 FT.
Values for M(x) for x=0 to x=11 Feet.
In the next slide, we want to get the expression of All the values for the moment at several points starting from zero until X is equal to 11. M(X) =4750*<x-0>^1- 3000*<x-5>^1-8000*<x-5>^0 -400*<x-9>^2+400*<x-16>^2+38/50*<x-20>^0.The last term will be ignored.
The Value of M(X) of 0 equals 4750*0=0 lb.Ft. The Value of M(X) of 3 equals 4750*3=14250 lb.Ft.
The Value of M(X) of 5 equals 4750*5=23750 lb.Ft. The Value of M(X) of 7 equals 4750*7-3000*2-8000*1=19250 lb.Ft. The Value of M(X) of 9 equals 4750*9-3000*4-8000*1=22750 lb.Ft. The Value of M(X) of 11 equals 4750*11-3000*6-8000*1-400*2^2=24750 lb.Ft.
Part 1 of Practice problem-7-38-Moment values from x=0 to x=11 FT.
Values for M(x) for x=12 to x=20 Feet.
In the next slide, we estimate the moment value for the points from X=12 feet to X=20 feet.
The first region is from zero to 5, the second region moves from 5 feet to 9 feet, and then we start from 9 to 16. The Value of M(X) of 12 equals 4750*12-3000*7-8000*1-400*3^2=24400 lb.Ft. The Value of M(X) of 14 equals 4750*14-3000*9-8000*1-400*5^2=21500 lb.Ft.
The Value of M(X) of 16 equals 4750*16-3000*11-8000*1-400*7^2=15400 lb.Ft. All values so far as positive will continue to be positive. The Value of M(X) of 18 equals 4750*18-3000*13-8000*1-400*9^2+400*2^2=7700 lb.Ft.
The Value of M(X) of 20 equals 4750*20-3000*15-8000*1-400*11^2+400*4^2=0 lb.Ft.
Part 1 of Practice problem-7-38-Moment values from x=12 to x=20 FT.
This is the end of Part 1 of Practice problem-7-38 Discontinuity Functions. Please refer to the second part of this post.
The previous post is the 4a-Part 2 of Practice problem-7-35 Discontinuity Functions.
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), 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.Please refer to the next slide image for more details.
Part 2 of Practice problem-7-35 Discontinuity Functions-Sketch for Shear force and Moment
Use an Excel sheet and use 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 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
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. it will plotted below the datum line.
Plot the Moment diagram.
Please find the plot of the moment created by using a graph with an Excel graph in the next slide.
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 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.
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), V(x) and M(x) must be used 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.
Part 1 of Practice problem-7-35 Discontinuity Functions
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 distances>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.
