For the close coil helical spring of the maximum deflection is

A $$\frac{{{\text{W}}{{\text{D}}^3}{\text{n}}}}{{{{\text{d}}^4}{\text{N}}}}$$
B $$\frac{{2{\text{W}}{{\text{D}}^3}{\text{n}}}}{{{{\text{d}}^4}{\text{N}}}}$$
C $$\frac{{4{{\text{W}}^2}{{\text{D}}^3}{\text{n}}}}{{{{\text{d}}^4}{\text{n}}}}$$
D $$\frac{{8{\text{W}}{{\text{D}}^3}{\text{n}}}}{{{{\text{d}}^4}{\text{n}}}}$$

Correct Answer: $$\frac{{8{\text{W}}{{\text{D}}^3}{\text{n}}}}{{{{\text{d}}^4}{\text{n}}}}$$

Bissoy MCQ
#Civil Engineering #Theory Of Structures

Related Questions

When a closely-coiled helical spring of mean diameter (D) is subjected to an axial load (W), the deflection of the spring ($$\delta $$) is given by (where d = Diameter of spring wire, n = No. of turns of the spring and C = Modulus of rigidity for the spring material)
A $$\frac{{{\text{W}}{{\text{D}}^3}{\text{n}}}}{{{\text{C}}{{\text{d}}^4}}}$$
B $$\frac{{2{\text{W}}{{\text{D}}^3}{\text{n}}}}{{{\text{C}}{{\text{d}}^4}}}$$
C $$\frac{{4{\text{W}}{{\text{D}}^3}{\text{n}}}}{{{\text{C}}{{\text{d}}^4}}}$$
D $$\frac{{8{\text{W}}{{\text{D}}^3}{\text{n}}}}{{{\text{C}}{{\text{d}}^4}}}$$
A closely coiled helical spring is acted upon by an axial force. The maximum shear stress developed in the spring is $$\tau $$. The half of the length of the spring if cut off and the remaining spring is acted upon by the same axial force. The maximum shear stress in the spring in new condition will be
A $$\frac{\tau }{2}$$
B $$\tau $$
C $$2\tau $$
D $$4\tau $$
When an open coiled helical compression spring is subjected to an axial compressive load, the maximum shear stress induced in the wire is (where D = Mean diameter of the spring coil, d = Diameter of the spring wire, K = Wahl's stress factor and W = Axial compressive load on the spring)
A $$\frac{{{\text{WD}}}}{{\pi {{\text{d}}^3}}} \times {\text{K}}$$
B $$\frac{{2{\text{WD}}}}{{\pi {{\text{d}}^3}}} \times {\text{K}}$$
C $$\frac{{4{\text{WD}}}}{{\pi {{\text{d}}^3}}} \times {\text{K}}$$
D $$\frac{{8{\text{WD}}}}{{\pi {{\text{d}}^3}}} \times {\text{K}}$$
A close coil helical spring of mean diameter D consists of n coils of diameter d. If it carries an axial load W, the energy stored in the spring, is
A $$\frac{{4{\text{W}}{{\text{D}}^2}{\text{n}}}}{{{{\text{d}}^4}{\text{N}}}}$$
B $$\frac{{4{{\text{W}}^2}{\text{Dn}}}}{{{{\text{d}}^4}{\text{N}}}}$$
C $$\frac{{4{{\text{W}}^2}{{\text{D}}^3}{\text{n}}}}{{{{\text{d}}^4}{\text{N}}}}$$
D $$\frac{{4{{\text{W}}^2}{{\text{D}}^3}{{\text{n}}^2}}}{{{{\text{d}}^4}{\text{N}}}}$$
একটি স্টেপ আপ ট্রান্সফরমারে, গৌণ কুণ্ডলীর পাকসংখ্যা মুখ্য কুন্ডলীর পাকসংখ্যার দ্বিগুণ। যদি মূখ্য কুন্ডলীকে কোন পরিবর্তী বিদ্যুৎ লাইনের সাথে সংযুক্ত করা হয়, তাহলে মুখ্য কুণ্ডলীর তুলনায় গৌণ কুণ্ডলীতে- (In a step up transformer, the number of turns in the secondary coil is twice to that of primary coil. If you apply an alternating current in the primary coil, then in the secondary coil compared to the primary coil-)
A তড়িৎ প্রবাহের মান দ্বিগুণ হবে (current will be doubled)
B তড়িৎ প্রবাহের মান অর্ধেক হবে (current will be halved)
C তড়িৎ বিভবের মান অর্ধেক হবে (voltage will be halved)
D তড়িৎ প্রবাহের মান সমান থাকবে (Current will remain the same)
Two closely coiled helical springs 'A' and 'B' are equal in all respects but the number of turns of spring 'A' is half that of spring 'B'. The ratio of deflections in spring 'A' to spring 'B' is
A $$\frac{1}{8}$$
B $$\frac{1}{4}$$
C $$\frac{1}{2}$$
D 2
Two closely-coiled helical springs 'A' and 'B' are equal in all respects but the number of turns of spring 'A' is double that of spring 'B'. The stiffness of spring 'A' will be __________ that of spring 'B'.
A One-sixteenth
B One-eighth
C One-fourth
D One-half
Two closely coiled helical springs 'A' and 'B' are equal in all respects but the diameter of wire of spring 'A' is double that of spring 'B' The stiffness of spring 'B' will be __________ that of spring 'A'
A One-sixteenth
B One-eighth
C One-fourth
D One-half
Two closely-coiled helical springs 'A' and 'B' of the same material, same number of turns and made from same wire are subjected to an axial load W. The mean diameter of spring 'A' is double the mean diameter of spring 'B'. The ratio of deflections in spring 'B' to spring 'A' will be
A $$\frac{1}{8}$$
B $$\frac{1}{4}$$
C 2
D 4
One end of a helical spring is fixed while the other end carries the load W which moves with simple harmonic motion. The frequency of motion is given by (where $$\delta $$ = Deflection of the spring.)
A $$2\pi \sqrt {\frac{g}{\delta }} $$
B $$\frac{1}{{2\pi }}\sqrt {\frac{g}{\delta }} $$
C $$2\pi \sqrt {\frac{\delta }{g}} $$
D $$\frac{1}{{2\pi }}\sqrt {\frac{\delta }{g}} $$

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