Question

A mild steel rod of 12 mm diameter was tested for tensile strength with the gauge length of 60 mm. Following observations were recorded: Final length = 80 mm; Final diameter = 7 mm; Yield load = 3.4kN and Ultimate load = 6.1kN. Calculate: a) Yield stress. ​​​​​​​​​ ​ b) Ultimate tensile stress. ​​​​​​ c) Percentage reduction in area. ​​​​​ d) Percentage elongation.

Answer 1

Answer:

Explanation:

(i)Yields stress

=Wu/A

=3400/113

=30.1

(ii)Ultimate tensile stress

=Wu/A

=6100/113

=54N/mm sq (MPa)

(iii)Percentage reduction in area

=A-a / A

=113-38.5/113

=0.66 or 66%

(iv)Percentage elongation

=Max-min/min

=L-l ÷l ×100

=80-60÷60 ×100

=33.33%

Answer 2

Given :

initial diameter (D)  = 12 mm ; gauge length ( l)  = 60 mm ;  final length (L) = 80 mm;  initial diameter (d)   = 7 mm ; Yield load ( $W_{y}$  ) = 3.4 kN = 3400 N; ultimate load ($W_{u}$) = 6.1 kN = 6100 kN

To Find :

a) Yield stress. ​​​​​​​​​ ​ b) Ultimate tensile stress. ​​​​​​ c) Percentage reduction in area. ​​​​​ d) Percentage elongation in length.

Solution :

We know that original area of the rod (A) =  $\frac{\pi }{4}$ x $D^{2}$

                                                                    = $\frac{\pi }{4}$ x ($12^{2}$)

                                                                    = $\frac{\pi }{4}$ x 144 = 113 $mm^{2}$

Final Area (a) =   $\frac{\pi }{4}$ x $d^{2}$

                      =   $\frac{\pi }{4}$ x ($7^{2}$) = 38.5 $mm^{2}$

(a) We know, Yield Stress =  $\frac{W_{y}}{A}$

                                           = $\frac{3400}{113}$

                                           = 30.1 N/$mm^{2}$ = 30.1 MPa

(b) We know, the ultimate tensile stress = $\frac{W_{u} }{A}$

                                                                  = $\frac{6400}{113}$

                                                                  = 54 N/$mm^{2}$ = 54 MPa

(c) We know that percentage reduction in area = $\frac{A - a}{a}$ x 100%

                                                                              = $\frac{113 - 38.5}{113}$ x 100%

                                                                              = 66%

(d) We know that percentage elongation in length = $\frac{L - l}{L}$ x 100%

                                                                                    = $\frac{80 - 60}{80}$ x 100%

                                                                                    = 25%