# Shear force and bending diagram un-uniformly distributed load

I already got great help on this topic, but I'd like to make sure that I get it.

For example, take a beam of length 6 meters, left side fixed support and right side rolled support. There is a load of 30 kN-m that starts at 3 meters and ends at 6 meters with a magnitude of 60 kN-m.

The function of the load is easy, f(x)=10x. The centroid is given by $$\frac{\int_3^6 10x² dx}{\int_3^6 10x dx} = 4.67m$$

And the force translated into a point load is $$\int_3^6 10x dx = 135 kN$$

So the sum of forces as seen from a is 0, and can be used to calculate Mb $$\sum Ma =0 = 4.67*135 -6*Mb \Rightarrow Mb = 105 kN \Rightarrow Ma=35kN$$

Then to find the shear forces I integrate the equation from 3 to x taking into account Ma $$\int_3^x 10x dx -Ma = 5x² -75$$

And to find the Bending moments I just integrate again, this time for the whole equation

$$\int_0^x (5x²-75 )dx = \frac{5x³}{3} -75x$$

Alas this does not give the complete answer, I have many uncertainties, for the shear forces I notice that it only gives the correct answer for x between 3 and 6; for x under 3 it becomes Ma, and I imagine if the length of the beam would be greater than the distributed load's length, it would be Mb for x>6. If I would add a point load of 3kN at l=2, do I need to replace Ma with Ma'- 3kN in the shear force equation? What would happen with a point load after 6 meters (eg lenght of beam = 9 meters, point load at 7m)?

As for the bending moments, I kinda get the right answer for x between 3 and 6, but it is 90 kN too much, which is the BM at the start of the distributed load. For x=2 for example I get completely wrong answers, and I need to revert to calculating BM by using trigonometry to calculate the actual area under the shear force curve.

I mean I can solve them, but I don't get the why - why can't I use the BM function to calculate all of x, considering the boundary condition should have been solved in the first integral (from 3 to 6 ). Why do I need to substract the BM at the beginning of the distributed load,...

I would greatly appreciate if someone would take the time to enlighten me - as it is I'm not really confident for the exam (I can't afford to fail)

I am a bit confused here, so I went under the assumption you have a statically undetermined beam with a fixed support (moment resisting) on the left and a roller support (non-moment resisting) on the right. In either case, this method still applied, but the constants become significantly easier in the case where it is a pinned support (non-moment resisting, but fixed in x-axis) on the left side. I'll describe the more complicated case and you can see how it can simplify to the less complicated case.

Seeing how you want a method that gives the complete picture for all of x, I will suggest what I personally use, and what books such as Roark's formulas for stresses and strains uses, the Macaulay Brackets. These are defined as follows:

$$\langle x-a \rangle^n = \begin{cases} 0, & x < a \\ (x-a)^n, & x \ge a. \end{cases}(n \ge 0)$$

Essentially, you can ignore what is in the bracket when x is less than a, and then start from 0 when x is greater than a. It is even defined at n=0 to be equal to the heaviside step function. This forms Macaulay's method.

$$w(x) = 30 \frac{kN}{m} \langle x-3 \rangle^0 + 10 \frac{kN}{m^2} \langle x-3 \rangle^1$$

This looks a lot more complicated then it did before. But the same theory applies:

The centroid of the load is still at $\frac{\int w(x)*x dx}{\int w(x) dx}$ and has a value equal to $\int w(x) dx$. We then need to work on the remaining equations. The reaction forces at each end are - actually unknown if this is a statically indeterminate beam as described above.

We can determine the shear and bending moment in the beam throughout the section simply by integrating w(x):

$$V(x) = 30\langle x-3 \rangle^1 + 5\langle x-3 \rangle^2 + F_a$$ $$M(x) = 15\langle x-3 \rangle^2 + \frac{5}{3}\langle x-3 \rangle^3 + F_a*x + M_a$$

We then can divide by $EI$, the beam property, and continue to integrate to determine the slope and deflection:

$$EI\theta(x) = 5\langle x-3 \rangle^3 + \frac{5}{12}\langle x-3 \rangle^4 + \frac{F_a}{2}x^2 + M_ax + C_1$$ $$EIy(x) = \frac{5}{4}\langle x-3 \rangle^4 + \frac{1}{12}\langle x-3 \rangle^5 + \frac{F_a}{6}x^3 + \frac{M_a}{2}x^2 + C_1x + C_2$$

We see immediately that $C_2 = 0$ because the deflection at a is 0. We then see that $C_1 = 0$ because the slope at a is 0. Finally, we can solve for the reactions. We can use the fact that the moment at b is 0 on M(x) to get a single equation:

$$M(6) = 0 = 15\langle 6-3 \rangle^2 + \frac{5}{3}\langle 6-3 \rangle^3 + F_a*6 + M_a$$ $$0 = 180 + 6*F_a + M_a$$

We then use the deflection equation for the last equation:

$$EIy(6) = 0 = \frac{5}{4}\langle 6-3 \rangle^4 + \frac{1}{12}\langle 6-3 \rangle^5 + \frac{F_a}{6}6^3 + \frac{M_a}{2}6^2$$

$$0 = 101.25 + 20.25 + 36F_a + 18M_a$$ $$0 = 121.5 + 36F_a + 18M_a$$

This can then easily be resolved into $F_a = -43.3125$, $M_a = 79.875$. $F_b$ follows shortly thereafter.

• That was quite thorough, thank you for taking the time ! Aug 30 '18 at 10:51