Compute the tensile strength and ductility (%elongation, el) of a cylindrical copper rod if it is cold worked such that the diameter is reduced from 15.2 mm to 12.2 mm. note: materials become harder or stronger as they are plastically deformed. this is called as strain hardening. it is also called as work hardening and cold working. %cw

Therefore, as cold work is increased, yield strength increases, Tensile strength increases and Ductility (%EL) decrease

If we trace the intersection point between the X and Y axis,

We then derive %EL at 7%

Considering the given data, The %CW IS 35.5 and approximately 35.6

Explanation:

Recalling the formula for cold working ( %CW),

We have,  dr1² - dr2² ÷ dr1² × 100

Where dr1 = Initial Diameter of the copper rod 15.2

dr2 = Reduced diameter of the copper rod 12.2

Now, substituting the values into the equation,

We have, 15.2² - 12.2² ÷ 15.2² × 100

Now if we apply the principle of BODMAS, where

B= bracket, O = off,  D= division, M = Multiplication, A= Addition, S= Subtraction.

Considering this principle in solving the substituted vales,

We have,

%CW = 231.04 - 148.84 ÷ 231.04 × 100

=82.2 ÷231.04 ×100

82.2 ×100 = 8220

Therefore, 8220/231.04

%CW = 35.6

Now, for the Percentage elongation from the graph,

Let us plot a graph of Ductility (% EL) against % Cold work

Where ductility is on the y axis and %cold work is on the x axis

On the y axis, we have 0, 20, 40, in accending order while on the x axis, we have 0, 20, 40, 60,

A downward slope toward positive x axis

If we trace the intersection point between the X and Y axis,

We then derive %EL at 7%

Therefore, as cold work is increased, yield strength increases, Tensile strength increases and Ductility (%EL) decrease

connect the batteries in series and hope

490 n

f=125n a=2.5m/s2