I"ve been attempting this fundamental shear force diagram trouble for several days, but can"t it seems to be ~ to get the exactly result. I"m do the efforts to calculate the shear pressure diagram in regards to \$x\$, but I"m unsure around the intensity \$w(x)\$ the the triangular load distribution between \$0m le x lt 3m\$. Ns am able to calculation the correct an outcome for the last section \$3m lt x le 6m\$, therefore I"m a tiny confused as to what the exactly intensity that the triangular load distribution is and also how to calculate the exactly shear force using the exactly intensity \$w(x)\$?

Below I"ve enclosed the problem and calculated the assistance reactions, which are \$A_y=15kN\$ and \$B_y=15kN\$.

You are watching: Triangular distributed load shear and moment diagram

Now, I"ve attached my free body chart of the very first section between \$0m le xlt 3m \$ and also indicated the confident sign convention because that this beam.

I climate proceeded to discover the shear force in regards to \$x\$ together follows:

\$sum F_y=0:\$

\$\$15-w(x)·x·frac12 - v_1 = 0 quad (eq 1)\$\$

Where \$w(x)·x·frac12\$ is the area of the triangular pack distribution.

This is wherein I gain confused. My knowledge of triangular load circulation in terms of the intensity \$w(x)\$ is that:

\$\$w(x)=fracw_0xL\$\$

Where \$w_0 = 10\$ and also \$L=3\$ because that this problem.

But substituting this values into the strongness \$w(x)\$ and back into \$(eq 1)\$ gets me the wrong an outcome of:\$\$v_1=15-frac53 x^2\$\$

After analysis multiple textbooks and watching numerous videos, ns finally uncovered out that if the maximum pack of a triangular load circulation is at the initial point \$x=0\$ then the complying with formula should be applied:

\$\$w(x)=fracw_0xL-w_0\$\$

I now know this a bit, but I am wondering where I might get a an excellent explanation regarding why?

I"m struggling to find a good explanation as almost every example I"ve found in textbooks/videos usage triangular fill distributions that boost from the initial point and not decrease.

However, ~ utilising this formula, ns still obtain the dorn solution. My functioning out is as follows:

\$sum F_y=0:\$

\$\$15-Bigl(w(x)·x·frac12 Bigr) - v_1 = 0\$\$\$\$15-iggl(Bigl(frac10x3-10Bigr)·x·frac12 iggr)- v_1 = 0\$\$\$\$15-iggl(Bigl(frac10x6-frac102Bigr)·xiggr) - v_1 = 0\$\$\$\$15-iggl(Bigl(frac5x3-5Bigr)·x iggr)- v_1 = 0\$\$\$\$15-frac5x^23+5x - v_1 = 0\$\$\$\$Rightarrow v_1=15-frac5x^23+5x\$\$

The actual equipment is:\$\$v_1=15+frac5x^23-10x\$\$

So I"m not sure whether I"m utilizing the exactly intensity \$w(x)\$ and/or whether the triangle area has been appropriately calculated utilizing this soot \$w(x)\$.

For the second section \$3mle xlt6m\$ i am may be to calculation the exactly shear pressure in terms of \$x\$, this systems is:

\$\$v_2=-15-frac5x^23+10x\$\$

Plotting a chart of the correct shear pressures \$v_1\$ and \$v_2\$ in terms of \$x\$ watch the following:

For her reference, this problem (F11.6) can be uncovered in chapter 11 the Statics and Mechanics of materials (4th Ed. SI edition) through Hibbeler.

I"d evaluate if someone might explain intensity loads for situations similar to over and wherein I went wrong in my calculations.

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Thank you.

Edit:

After analysis a couple of examples, I uncovered that if i calculate the shear force from the left finish I am able to acquire the correct shear force using my initial strongness \$w(x)=fracw_0xL\$ and not the latter intensity \$w(x)=fracw_0xL-w_0\$.

However I"m unsure why i can"t calculate this indigenous the best end? walk it have actually something to perform with the left support \$A_y=15kN\$ developing a discontinuity? If i calculate indigenous the left finish am ns correct in an altering the section"s variety to \$0m lt x le 3m\$ to not include the left assistance \$A_y\$?

My functioning out is together follows:

\$sum F_y=0:\$

\$\$-Bigl(w(x)·x·frac12 Bigr) + v_1 = 0\$\$\$\$-iggl(Bigl(frac103(3-x)Bigr)·(3-x)·frac12 iggr)+ v_1 = 0\$\$\$\$-iggl(Bigl(10-frac10x3Bigr)·(3-x)·frac12 iggr)+ v_1 = 0\$\$\$\$-iggl(igl(30-10x-10x+frac10x^23igr)·frac12 iggr)+ v_1 = 0\$\$\$\$-iggl(igl(30-20x+frac10x^23igr)·frac12 iggr)+ v_1 = 0\$\$\$\$-igl(15-10x+frac10x^26igr)+ v_1 = 0\$\$\$\$-15+10x-frac5x^23+ v_1 = 0\$\$