Lecture 12 som 10.03.2021

STRENGTH OF MATERIALS FOR MECHANICAL ENGINEERS

1. _ ELONGATION OF BAR
| DUE TO ITS SELF
WEIGHT

BIBIN CHIDAMBARANATHAN ES
OR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

2. ELONGATION OF BAR DUE TO ITS SELF WEIGHT

+ Consider a bar AB hanging freely under its own \\\\\\\ a
weight

A
Let Bar >
“ L = Length of the bar dx = WM i
+ A = Cross sectional Area of the bar
« E = Young’s modulus of the bar material :
<% wm = Specific weight of the bar material B

BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

3. Consider a small section d, of the bar at a distance x from B.
weight of the bar of length x, (P) = w.A.x

Elongation of the small section of the bar due to weight of the bar for a small section
of length x.

Ole=e2dx WAAAY
o
bly =—.d À
“E * Bar >
P ==
öl, ==>. dx dx
* AE L
wAx
ölz,= AE” bd x
& x
BIBINC 10550 deep Midi ENGINEERING /RIK cou GE Or Na IN AVE oL0G

4. Total elongation of bar may be found out by integrating the above equation
between the limits 0 and L

L

Wx

Total elongation (6l) = [Eu
0

A
L Bar —
a= Ef x.dx A
EJ*
o dx A
u L
aa
6l= —|—
El2], =
a= 2 Zo B
El2
w 1?

BIBIN.C / ASSOCIATE PRODESER / TIAL ENGINEERING / RK COLLECEO ENGINEERING AND TECHNOLOGY

5. _ w 1?
| AL 2E
Total weight (W) = wAL

Ww
Specific weight (w) = AL

WL
Elongation of bar due to weight of the bar (öl) = DAE

BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

6. Problem 01

A copper alloy wire of 1.5 mm diameter and 30 m long is hanging freely from a tower.
What will be its elongation due to self-weight? Take specific weight of copper and its
modulus of elasticity as 89.4 kN /m3and 90 GPa respectively.

Given data:
d=15mm L=30m=30x10%'mm — w = 89.2 kN/m? = 89.2 x 10% N/mm?

Modulus of elasticity (E) = 90 GPa = 90 x 10° N/mm?

To find:

Elongation due to self weight (öl)=?

BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

7. Formula :

. . w L?
Elongation due to self weight (öl) = =>

2E
Solution:
ol? d=15mm
Elongation due to self weight (öl) = DE Le 36m = 30510? mm
#1 = 89.2 x 107% x 30 x 10% @ = 89.2 x 106 N/mm?
u 2 x 90 x 103
E = 90 x 103 N/mm?
öl = 0.45 mm

Elongation due to self weight (öl) = 0.45 mm

BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

8. Problem 01
An alloy wire of 2 mm? cross sectional area and 12 N weight hangs freely under its own

weight. Find the maximum length of the wire, if its extension is not to exceed 0.6 mm.

Take E for wire material as 150 GPa.

Given data:
A=2mm? W=12N öl = 0.6mm

Modulus of elasticity (E) = 150 GPa = 150 x 10? N/mm?
To find:
Maximum length of the wire (L)=?

BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

9. Formula :

. . WL
Elongation due to self weight (öl) = DAE
Solution:
WL A=2mm?
Elongation due to self weight (Sl) = DAE
W=12N
12xL
0.6 = ——————
2x2x150x 103 öl = 0.6mm
E = 150 x 103 N/mm?
L = 30000 mm

Maximum length of the wire (L) = 30000 mm = 30m

BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

10. Problem 01

A steel wire ABC 16 m long having cross sectional area of «mm? weighs 20 N as shown
in fig. If the modulus of elasticity for the wire material is 200GPa, find the deflection at C
and B.

A
Given data: 8
L=16m=16x10?mm A=4mm? w=20N 2

B

Modulus of elasticity (E) = 200 GPa = 200 x 10° N/mm?
ES
To find:
c

Deflection at C (61, ) and B (öl; )=?

BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

11. Formula :

Deflection of wire at C due to self weight of wire AC (él, ) = PAE A
E
o
Deflection of wire at B (Slg) = Deflection of wire AB due to self weight +
Deflection due to weight of wire BC B
8
Deflection of wire at B (Slg) = öl, + Sl, u
€
5 . . . . _ Was X Lap
eflection of wire at B due to self weight of wire AB (Sl, ) = DE
. . A . _ Pec X Lap
Deflection of wire at B due to weight of wire BC (Sl, ) = A

BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

12. Solution:

Deflection of wire at C due to self weight of wire AC (Sl) = DAE

L=16x 10? mm

51, 20X16x 10°
© 2x4x 200 x 103 A=4mm?
W=20N
öl, =0.2mm E = 200 x 10? N/mm?

Deflection of wire at C due to self weight of wire AC (öl.) = 0.2mm

BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

13. ; ; ; ; _ Was X Las
Deflection of wire at B due to self weight of wire AB (él, ) = ==

2AE
W
Was == 10N W=20N
A
Lag = 8 X 10? mm 8
3 o
ól, = 10x8x10 À = dE
2 x 4 x 200 x 103 B
E = 200 x 103 N/mm? 2
o
öl, = 0.05mm
Cc

Deflection of wire at B due to self weight of wire AB (6l, ) = 0.05 mm

BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

14. 5 z A o PX Lap
Deflection of wire at B due to weight of wire BC (Sl, ) =

AE
pas 108
=5=
Lag = 8X 10% mm
x 10 x 8 x 10% A=4mm?
= AE
4x 200 x 103 W=20N
E=200x 103 N/mm?
öl, = 0.1mm

Deflection of wire at B due to weight of wire BC (dl, ) = 0.1mm

BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

15. Deflection of wire at B (6lg) = öl, + öl,
A

g
öl, = 0.05mm co
B
öl, = 0.1mm
öl; = 0.05+ 0.1 E
E

Deflection of wire at B (6lg) = 0.15 mm

BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

16. Thank You

BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY