What is The purpose of the body paragraph in an aggressive essay?

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<span>The body of the essayThe function of the essay's body is to fully develop the argument outlined in the introduction. Each paragraph within the body of the essay elaborates on one major point in the development of the overall argument (although some points may consist of a number of sub-points, each of which will need a paragraph). The main point in each paragraph needs to be clearly stated in the form of a topic sentence, which is then supported with evidence. Let's look at how the student's essay on why Italy in 1860 was seen as an unlikely nation developed (remember the argument is that Italy was seen as an unlikely nation because of the many obstacles in the way of unification). The example below shows only the first (or topic) sentence of each paragraph, and not the remaining sentences of each paragraph. This is to highlight the way the main points in the argument can be presented at the beginning of each paragraph to show the logical steps in the argument.</span>

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Stainless steel rod has a gap of 0.0371 mm between its end and a rigid wall. The modulus of elasticity E = 210 GPa. The coeffici

alexgriva [62]

Answer:

The value of stress in the rod is 220.9 MPa

Explanation:

As the data is not complete, a similar question is found online and the missing data is used from that question. Similar question is attached herewith.

From the given data

The allowed total strain is given as $\delta_a=0.0371 mm$

The length of the rod is missing in this data which in the additional data is l=290 mm

The Modulus of Elasticity is given as E=210 GPa

The coefficient of thermal expansion is given as α= 10 ×10⁻⁶ mm / mm⋅°C.

The initial temperature is room temperature which is given as T1=22C

The final temperature is given as T2=140 C

The strain due to stress is given as

$\delta_s=\dfrac{\sigma l}{E}$

$\delta_s=\dfrac{\sigma (0.290)}{210\times 10^9}\\\delta_s=1.38095\sigma \times 10^{-12}$

Now the thermal strain is given as

$\delta_c=l\alpha \Delta T\\\delta_c=0.290\times 10 \times 10^{-6}(T_2-T_1)\\\delta_c=0.290\times 10 \times 10^{-6}(140-22)\\\delta_c=0.0003422$

So now the equation of the allowed strain is given as

$\delta_a=\delta_s+\delta_c\\0.0371\times 10^{-3}=1.38095\sigma \times 10^{-12}+0.0003422\\\dfrac{1.38095}{10^{12}}\sigma+0.0003422=\frac{0.0371}{1000}\\\sigma=-\dfrac{10^{12}\times \:0.0003051}{1.38095}\\\sigma=-220934863.68 Pa\\\sigma=-220.9 MPa$