Question

In an experiment, bras and steel wires of length lm each with areas of cross section 1mm2 are used. The wires are connected in series and one end of the combined wire is connected to a rigid support and the other end is subjected to elongation. The stress required to produce a net elongation of 0.2mm is :
(Given, the Young's Modulus for steel and brass are respectively, $120 \times {10^9}N/{m^2}$ and $60 \times {10^9}N/{m^2}$ )

Answer 1

The relation between stress experienced by the elements and their physical properties i.e., length, Young’s modulus and area is used. Further, the final stress is calculated and at last the required stress i.e., forces per unit area is calculated.

Formula used:
\eqalign{ & k = \dfrac{{yA}}{\ell } \cr & {k_{eq}} = \dfrac{{{k_1}{k_2}}}{{{k_1} + {k_2}}} \cr & F = {k_{eq}} \cr}

Complete step by step answer:
As we know stress is defined as force per unit area within the material that arises from externally applied forces.
Here, we use the relation between the stress, Young’s modulus, cross section area and length of the respective wires, which is given by:
\eqalign{ & {k_{1}} = \dfrac{{{y_1}{A_1}}}{{{\ell _1}}} = \dfrac{{120 \times {{10}^9} \times A}}{1} \cr }
\eqalign{ & {k_2} = \dfrac{{{y_2}{A_2}}}{{{\ell _2}}} = \dfrac{{60 \times {{10}^9} \times A}}{1} \cr }
\eqalign{ & {k_{eq}} = \dfrac{{{k_1}{k_2}}}{{{k_1} + {k_2}}} = \dfrac{{120 \times 60 \times {{10}^9} \times A}}{{180}} \cr }
\eqalign{ & \Rightarrow {k_{eq}} = 40 \times {10^9} \times A \cr }
\eqalign{ & F = {k_{eq}} \cr }
\eqalign{ & \Rightarrow F = (40 \times {10^9})A.(0.2 \times {10^{ - 3}}) \cr }
\eqalign{ & \therefore \dfrac{F}{A} = 8 \times {10^6}N/{m^2} \cr}

Therefore, the stress experienced by the two, bras and steel wires of length lm each with areas of cross section and are connected in series and one end of the combined wire is connected to a rigid support and other end is subjected to elongation is given by above equation.

Additional Information: The maximum stress that a material can experience before it breaks is called the breaking stress or ultimate tensile stress. The word tensile means the material is under Tension.
An alloy can be defined as a combination of metals or metals combined with one or more other elements. If combining the metallic elements gold and copper produces red gold, gold and silver becomes white gold, and silver combined with copper produces sterling silver.
We already know that brass is an alloy of copper (Cu) and zinc (Zn). Copper and zinc is present in proportions which can be varied to achieve varying mechanical and electrical properties of the element. It is a substitution alloy which means that the atoms of the two constituents may replace each other within the same crystal structure.
Further, steel is composed of iron and carbon. It accounts for 90 percent of steel production. There are Low alloy steel which is alloyed with other elements, usually with molybdenum, manganese, chromium, or nickel, in amounts of up to 10 percent by weight to improve the hardenability of thick sections.

Note: There are different types of stress but the basic concept is the same i.e., stress is defined as force per unit area. Stress experienced by an element or an object depends on the area, length and also young modulus of the element or object.