the spring balance to measure the weights. One end of the spring is attached to the rigid support while the other end, which is free, carries the weights to be measured. The maximum weight attached to the spring balance is 1500 N and the length of the scale should be approximately 100 mm. The spring index can be taken as 6. The spring is made of oil-hardened and tempered steel wire with ultimate tensile strength of 1360 N/mm? and modulus of rigidity of 81 370 N(mm?. The permissible shear stress in the spring
the spring balance to measure the weights. One end of the spring is attached to the rigid support while the other end, which is free, carries the weights to be measured. The maximum weight attached to the spring balance is 1500 N and the length of the scale should be approximately 100 mm. The spring index can be taken as 6. The spring is made of oil-hardened and tempered steel wire with ultimate tensile strength of 1360 N/mm? and modulus of rigidity of 81 370 N(mm?. The permissible shear stress in the spring
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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file:///C:/Users/ecemb/Desktop/makel/Machine%20design%20-%20Bhandari%20-%20BY%20Civildatas.com.pdf
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Example 10.5 A helical tension spring is used in
the spring balance to measure the weights. One end
of the spring is attached to the rigid support while
the other end, which is free, carries the weights to
be measured. The maximum weight attached to the
spring balance is 1500 N and the length of the scale
should be approximately 100 mm. The spring index
can be taken as 6. The spring is made of oil-hardened
and tempered steel wire with ultimate tensile
strength of 1360 N/mm? and modulus of rigidity of 81
370 N/mm. The permissible shear stress in the spring
wire should be taken as 50% of the ultimate tensile
The pea
given by,
T= 0.5 3
4C-
K =
4C-4
From Eq
8PC
T=K
nd2
:-
strength. Design the spring and calculate
(i) wire diameter;
(ii) mean coil diameter;
(iii) number of active coils;
(iv) required spring rate; and
(v) actual spring rate.
Step II Mear
De
Step III Numa
From Eq. (10.
8 PD
8=
Solution
Given P= 1500 N C=6 S 1360 N/mm2
Gd
%3D
G= 81370 N/mm2
T0.5 S
o search
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