a helical spring has N turns of coil of diameter D, and a second spring made of same material and of same wire diameter has N/2 turns of coil of diameter 2D. if stiffness of first spring is k, then stiffness of second spring is
Spring stiffness is a property that relates load to deflection. Let k = stiffness and P = load and x = deflection, then P = kx The stiffer the spring, k, the smaller the deflection under a constant load. k can be measured for springs and otherdevices, such as beams, with simple load deflection devices or machines that measure deflection as function of laod
It is related. Flexural modulus is the modulus of elasticity (E) in bending and the higher it is the higher the bending stiffness. Technically, bending stiffness is the product of the flexural modulus and the material bending moment of inertia, I, that is EI.
The flexural stiffness of a structural beam (E*I/L) is represented as the product of the modulus of elasticity (E) and the second moment of area (I) divided by the length (L) of the member.
All bearings and bushings have a basic stiffness. When you apply load on the shaft the shaft will move, and so the stiffnes can be calculated from the ratio of the applied force to the absolute value of movement.
If the length of a spring is halved, the stiffness remains the same. Stiffness of a spring is determined by its material and construction, not by its length. Cutting the length in half does not change the material properties that govern stiffness.
Increasing the spring stiffness will result in a higher natural frequency. This is because a stiffer spring will require more force to displace it, leading to faster oscillations and a higher frequency. Conversely, decreasing the spring stiffness will lower the natural frequency of the system.
a helical spring has N turns of coil of diameter D, and a second spring made of same material and of same wire diameter has N/2 turns of coil of diameter 2D. if stiffness of first spring is k, then stiffness of second spring is
The stiffness of a spring can be measured by calculating its spring constant, which is the force required to deform the spring by a certain distance. This can be done by applying a known force to the spring and measuring the resulting displacement, then using Hooke's Law (F = kx) to determine the spring constant. Another method is to measure the frequency of oscillation of the spring when subjected to a known mass, as the stiffness is inversely proportional to the period squared.
If the spring is cut in half, its stiffness will increase and it will stretch less for the same load. The new stretch will depend on the new stiffness of the spring. Without knowing the exact stiffness of the original spring and the new one, it is difficult to determine the exact stretch without calculations.
it the ratio of load applied on the displacement of spring.. stiffness=load/change in length.
The amount of force required to stretch a spring by 49 inches depends on the stiffness or spring constant of the spring. The formula to calculate this force is F = k * x, where F is the force, k is the spring constant, and x is the displacement of the spring (in this case, 49 inches). Without knowing the spring constant, the force required cannot be determined.
Spring stiffness is a property that relates load to deflection. Let k = stiffness and P = load and x = deflection, then P = kx The stiffer the spring, k, the smaller the deflection under a constant load. k can be measured for springs and otherdevices, such as beams, with simple load deflection devices or machines that measure deflection as function of laod
The spring constant remains the same regardless of the length of the spring. It is a physical property of the spring material and design, representing its stiffness. Cutting the length of the spring in half will not change its spring constant.
It may relax, or creep, meaning its stiffness is less
Yes, a spring will stretch when a weight is added to it due to the force of gravity acting on the weight, causing the spring to deform. The amount of stretch will depend on the weight added and the stiffness of the spring.
Elastic potential energy depends on the spring constant (stiffness of the spring) and the displacement from equilibrium (how far the spring is stretched or compressed).