Question 1.3.2: Derive the loading, shear-force, and bending-moment relation...

Using Table 3–1 and q\left(x\right)   for the loading function, we find

q=R_1\left\langle x\right\rangle^ {-1}− 200\left\langle x − 4\right\rangle^{-1} – 100\left\langle x − 10\right\rangle^{-1} − R_2\left\langle x − 20\right\rangle^{-1}        (1)
Integrating successively gives

V=\int{q dx } =R_1\left\langle x\right\rangle^ {0}− 200\left\langle x − 4\right\rangle^{0} – 100\left\langle x − 10\right\rangle^{0} − R_2\left\langle x − 20\right\rangle^{0}      (2)

M=\int{V dx } =R_1\left\langle x\right\rangle^ {1}− 200\left\langle x − 4\right\rangle^{1} – 100\left\langle x − 10\right\rangle^{1} − R_2\left\langle x − 20\right\rangle^{1}        (3)

Note that V = M = 0 at x = \bar{0} ?.
The reactions R_1 \ and R_2 \   can be found by taking a summation of moments and forces as usual, or they can be found by noting that the shear force and bending moment must be zero everywhere except in the region 0 \leq x \leq 20 \ in.

This means that Eq. (2)
Figure 3–5
should give V = 0 at \ x slightly larger than 20 in.

Thus R_1− 200 − 100 + R_2= 0      (4)
Since the bending moment should also be zero in the same region, we have, from Eq. (3),
R1(20) − 200(20 − 4) − 100(20 − 10) = 0        (5)
Equations (4) and (5) yield the reactions R_1= 210 \ lbf \ and \ R2= 90 \ lbf.
The reader should verify that substitution of the values of R_1 \ and \ R_2 \   into Eqs. (2) and (3) yield Figs. 3–5b and c.