Strength Of Materials Miscellaneous
- A helical compression spring made of a wire of circular cross-section is subjected to a compressive load. The maximum shear stress induced in the cross-section of the wire is 24 MPa. For the same compressive load, if both the wire diameter and the mean coil diameter are doubled, the maximum shear stress (in MPa) induced in the cross-section of the wire is ___.
-
View Hint View Answer Discuss in Forum
τmax = 8WD kw πd2 τmax ∝ 1 d2
[ ∴ W1 = W2 = W ; C1 = C2 ; ( Kw)1 = ( Kw)2 = ( Kw) ]τ2 = 
d1 
2 τ1 d2 τ2 = d 2 24 2d
⇒ τ2 = 6 MPaCorrect Option: A
τmax = 8WD kw πd2 τmax ∝ 1 d2
[ ∴ W1 = W2 = W ; C1 = C2 ; ( Kw)1 = ( Kw)2 = ( Kw) ]τ2 = d1 2 τ1 d2 τ2 = d 2 24 2d
⇒ τ2 = 6 MPa
- The spring constant of a helical compression spring DOES NOT depend on
-
View Hint View Answer Discuss in Forum
Stiffness of helical spring
k = Gd4 64R3 n
where G is Shear modulus
d is Wire diameter
R is Wire radius
n is No. of active turns.Correct Option: B
Stiffness of helical spring
k = Gd4 64R3 n
where G is Shear modulus
d is Wire diameter
R is Wire radius
n is No. of active turns.
- A weighing machine consists of a 2 kg pan resting on a spring. In this condition, with the pan resting on the spring, the length of the spring is 200 mm. When a mass of 20 kg is placed on the pan, the length of the spring becomes 100 mm. For the spring, the undeformed length L and the spring constant k (stiffness) are
-
View Hint View Answer Discuss in Forum
Let initial length is δ without load, and stiffness is k
∴ 2g = k(δ – 0.2) ...(i)
22g = k(δ – 0.1) ...(ii)
Solving equations (i) and (ii), we get
δ = 210 mm and k = 1960 N/mCorrect Option: B
Let initial length is δ without load, and stiffness is k
∴ 2g = k(δ – 0.2) ...(i)
22g = k(δ – 0.1) ...(ii)
Solving equations (i) and (ii), we get
δ = 210 mm and k = 1960 N/m
- The deflection of a spring with 20 active turns under a load of 1000 N is 10 mm. The spring is made into two pieces each of 10 active coils and placed in parallel under the same load. The deflection of this system is
-
View Hint View Answer Discuss in Forum
Where; n = 20 P = 1000 N
∆x = 10 mm
F = K∆xK = P = 100 N /mm ∆x
When a spring is cut into two pieces, no of coils get halved
∴ Stiffness of each half coiled get doubled
The stiffness when connected in parallel = 2k + 2k = 4k
P = K∆x∆x = P = 1000 = 2.5 mm k 4 × 100 Correct Option: D
Where; n = 20 P = 1000 N
∆x = 10 mm
F = K∆xK = P = 100 N /mm ∆x
When a spring is cut into two pieces, no of coils get halved
∴ Stiffness of each half coiled get doubled
The stiffness when connected in parallel = 2k + 2k = 4k
P = K∆x∆x = P = 1000 = 2.5 mm k 4 × 100
- The figure shows arrangements of springs. They have stiffness k1, and k2 as marked.
Which of the following arrangements offers a stiffness = 2k1k2 ? k1 + 2k2
-
View Hint View Answer Discuss in Forum
k = 2k1k2 k1 + 2k2 1 = k1 + 2k2 = 1 + 1 k (2k1)k2 2k2 k1 1 = 1 + 1 k k1 2k2 ∴ k = 2k1k2 k1 + 2k2 Correct Option: D
k = 2k1k2 k1 + 2k2 1 = k1 + 2k2 = 1 + 1 k (2k1)k2 2k2 k1 1 = 1 + 1 k k1 2k2 ∴ k = 2k1k2 k1 + 2k2
