Civil Engineering - Strength of Materials - Discussion

@All.

A) If we convert all the necessary data into "mm" & keeping load in "kgs",
We get,
(3000*5000)/(314.16*2.1e^6) = 0.0227 mm.

B) If we covert the data in "cm", we get
(3000*500)/(3.142*2.1e^6) = 0.2273 cm.

C) If we convert only length in "mm" & keeping dia in "cm" we get,
(3000*5000)/3.142*2.1e^6) = 2.274 unknown units and wrong procedure.

D) If we convert load in "N" & all dimensions in "cm", we get
(3000*9.81*500)/(3.142*2.1e^6) = 2.230 cm.

E) If "E" is given in Kgs/cm^2, we get
(3000*5000)/(314.16*2.1e^8) = 0.0002274 mm, which is not in the given answer.

But the given answer is in "mm", and the correct one is "C" which is in "mm"

We know that,

Modulus of Elasticity (E) = Stress/Strain.
= (P/A) / (dL/L).
Or, E = PxL/AxdL.
Or, dL = PxL/AxE

Given, P = 3000 Kg. A = (π/4) x 202 mm2 L = 5000 mm E = 2.1 x 106 Kg/mm2
So, dL = 3000x5000/{(π/4)x20x20x2.1x106 } mm,
= 3000x5000x7x4x10 / 22x20x20x21x106 mm [ Put π = 22/7]
1/44 mm,
= 0.022727 mm.

Given;

dl = (PxL)/(A.xE)
P = 3000kg, L=5m = 500cm.
A = 3.14x (2/2)>^2 = 3.14 cm^2.
E = 2.1x10^6 it/cm^2.

Putting the values in the formula.
dl = (3000kg x 500cm)/(3.14cm^2x2.1x10^6kg/cm^2).

Deduct the dimensions and zero we get;
dl= (15/3.14x2.1x10)=15/65.94=0.2275mm.
So, Answer :B

dia=2cm, A=3.1416cm^2,
P=3000kg, L=5m=500cm, (E=2.1*10^6 Kg/cm^2)

Now, (PL/AE)={(3000*500)/(3.1416*2.1*10^6)} {(kg*cm)/(cm^2*(kg/cm^2))}

PL/AE= {(15*10^5)/(3.1416*21*10^5)} (cm)
PL/AE=0.22736cm =2.274~ 2.275mm.
So, answer is 2.275mm.

I agree @ Subham,

Because they given 'e' in kg/cm^2, So only we have to convert length in cm, after calculation we get the answer in cm, then multiplied by "10" that gives the answer in mm.

So finally option "A" right answer.

Dia = 20 mm
Lenght = 5000 mm
Load (P) = 3000 kg.

Note;

{E = 2.1x10^6 N/mm^2} (given)
Convert E into kg/mm^2
So, E = 0.21x10^6 kg/mm^2 (Approx).

Delta = PL/AE = (3000 x 5000 x 4) / (pie x 20 x 20 x 0.21 x 10^6)
Delta = 0.2273 mm.

A = 314.16,
P=3000kg=3000*10=30000N,
L=5m=5000mm,
E=2.1*10^6.

Elongation= Pl/AE.
= (30000* 5000)/(314.16 * 2.1 * 10^6).
= 0.2275 mm.
So, B is the right answer.

Force = 3000 kg.
Converting it into Newton,
Force = 3000*9.18 N,

dL = PL/AE.
dL = (3000 * 9.81 * 5000)/((π/4)*20*2.1*10^6).
dL = 0.222mm.

P = 3000kg = 30000N,
L = 5000mm,
D = 20mm,
E = 2.1*10^6 N/mm^2.
So elongation = 30000*5000*4/3.14*400*2.1*10^6,
Ans = .2274... = 0.2275.

A = 314.16 mm^2.
Elongation= Pl/AE.
= (3000 * 5 * 10^3)/(314.16 * 2.1 * 10^6).
= 0.02275 mm.