A tabular analysis of the transactions made by Roberta Mendez & Co., a certified public accounting firm, for the month of August is shown below

$\delta ={\int }^0\genfrac{}{}{0ex}{}{P(x)dx}{A(x)E(x)}$ .

Internal Force. The internal axial force varies along the member since it is dependent on the weight W ( y ) of a segment of the member below any section, Fig. 4–8 b . Hence, to calculate the displacement, we must use

$\delta ={\int }^0\genfrac{}{}{0ex}{}{P(x)dx}{A(x)E(x)}$ .

At the section located a distance y from its free end, the radius x of the cone as a function of y is determined by proportion; i.e.,

$\genfrac{}{}{0ex}{}{x}{y}=\genfrac{}{}{0ex}{}{r^0}{L};x=\genfrac{}{}{0ex}{}{r^0}{L}y$

The volume of a cone having a base of radius x and height y is

$V=\genfrac{}{}{0ex}{}{1}{3}\pi y{x}^2=\genfrac{}{}{0ex}{}{\pi r^0}{3L^2}{y}^3$

Since $W=\gamma V$, the internal force at the section becomes

$+\uparrow \sum F^y=0;P(y)=\genfrac{}{}{0ex}{}{\gamma \pi r^0}{3L^2}{y}^3$

Displacement. The area of the cross section is also a function of position y , Fig. 4–8 b . We have

$A(y)=\pi x^2=\genfrac{}{}{0ex}{}{\pi r^0}{L^2}{y}^2$

Applying $\delta ={\int }^0\genfrac{}{}{0ex}{}{P(x)dx}{A(x)E(x)}$ between the limits of $y=0$ and $y=L$ yields

\delta =\int_{0}^{L}{\frac{P(y)dy}{A(y)E} }=\int_{0}^{L}{\frac{[(\gamma \pi r^2_0/3L^2)y^3]dy}{[(\pi r^2_0/L^2)y^2]E} }δ=∫0L​A(y)EP(y)dy​=∫0L​[(πr0 need an explanation for this answer? contact us directly to get an explanation for this answer

1. The company issued shares of stock to stockholders for $25,000 cash.
2. The company purchased $7,000 of equipment on account.
3. The company received $8,000 of cash in exchange for services performed.
4. The company paid $850 for this month’s rent.