## Supply All Kinds of Springs

Kathysia industrial Co., Ltd was reorganized from Zhengzhou spring factory, which founded in 1957, having more than 60 years of professional experience in this field. Kathysia is a designated spring products supplier by the Chinese ministry of machinery industry, and a member of China spring association.

We Supply all kinds of Springs for all applications, If you can’t find a stock spring for your application, or would like to match a competitor’s spring not offered through this website, Our engineering staff is ready to answer your questions or assist in designing the right spring for your application.

### Extension spring info

is a higher probability that the spring will deform or the loop will fail. Determining Loads at any Length To determine loads at lengths other than the maximum extended length multiply the amount of extension beyond the free length by the spring rate and add the initial tension. If a spring is 1" long, has a rate of 2 lbs per inch and the initial tension is .6 lbs, what is the load at 1 3/4 inches?

Extension Spring Force Calculator. When designing for a particular spring, design for critical loads and rates to be within the central 60% deflection and load range. Standard and custom springs are available. Exceeding manufactures maximum deflection almost always reduces usable spring life cycles.

A horizontal spring-mass system oscillates on a frictionless table. If the ratio of the mass to the spring constant is 0.038 kg m/N, and the maximum speed of the mass was measured to be 17.60 m/s, find the maximum extension of the spring. 3.4 cm 3.4 m 0.67 m 67 cm

Mar 21, 2008 The rigidity appearing on the spring, F = mg = 0.60 x 9.8 N we've, F = kx therefore, extension,x = F/ok = 0.60 x 9.8/15 = 0.39 m ability ability, U =

Nov 04, 2010 A mass m is hung vertically from a spring of spring constant k. Show that the maximum extension of the spring A is given by A = 2mg / k Thanks for any help

### Extension Spring Force Formula and Calculator

Most extension springs are wound with an initial tension. This tension is the force that holds the spring coils wound together. The spring rate tends to be constant over the central 60 percent of the deflection range. When designing for a particular spring, design for critical loads and rates to be within the central 60

(a) the spring is relaxed (b) the body is at the equilibrium point (c) the spring is at its maximum extension (d) the body is somewhere between the equilibrium position and the maximum of the spring extension F=ma Æa=F/m, but F=-kx Æa=-kx/m, where x is measured from equilibrium ÆMaximum -a| when x is maximum, which is either (a) or (c).

With an extension spring, there is no such safety geometry since the spring is in tension. For this and other reasons, extension spring maximum working stresses are typically limited to three-fourths (3/4) of those for compression springs of similar geometry and material.

Below is a graph of F versus x for an elastic spring. Determine: A) the spring constant. B) the spring’s maximum amount of elastic potential energy. C) the change in elastic potential energy when the spring extends from 3 cm to 4 cm. 6. A spring that obeys Hooke’s law has the following F-versus-x graph.

### Extension Springs

Body Length - is the measure of the spring length, excluding the loops. Extended Length - is the length at full rated extension. Maximum Load is the load at full extension. Loop Length - measures the length of a loop from the end of a body to the inside diameter of the loop.

international standards for extension spring design do this - the stress is calculated directly from the applied load. To establish whether initial tension stress has any influence on either elastic limit or fatigue performance springs were made to the same nominal design with a) Maximum initial tension 28.7 N b) Minimum initial tension 5.0 N

If an extension spring is required to work dynamically, it must be remembered that extension springs have approximately 20% lower performance with regard to fatigue than compression springs. The lower the working stresses the greater the expected life of the spring.

The period of oscillation of a spring and mass system. The period of oscillation of a spring-and-mass system is 0.50 s and the amplitude is 5.0 cm. What is the magnitude of the acceleration at the point of maximum extension of the spring? A. 7.9 m/s 2 B. 6.2 m/s 2 C. 5.8 m/s 2 D. 0.5 m/s 2 E. 6.7 m/s 2 Section: 10.5 Simple Harmonic Motion 32.

2 Answers. This is needed so that at the end of the extension, the spring remains at rest. Once we understand this, we note that the force at extension x is F (x)=kx. Then, the work done along the path is ∫xmax0F (x)dx=1 2kx2max The latter of course is precisely the potential energy of the spring.

### A 1500 g block is attached to a vertical spring whose

Jan 27, 2014 (a) When the spring is extended to the maximum extent, the block is at the lowest position. Before this instant, the block was going down, and after this instant, it will start moving up. At this instant, its speed is zero, which is the same as the initial speed of zero. We use conservation of energy to find the extension of the spring.

Ln (Maximum Loaded length): Maximum acceptable extension length for a spring Fn (Maximum Force): Maximum acceptable force which can operate on a spring. R (Spring Rate/Stiffness): the change in load per unit deflection in pounds per inch (lb/in) or Newtons per millimeter (N/mm).

Spring (device) An extension or compression spring's rate is expressed in units of force divided by distance, for example or N/m or lbf/in. A torsion spring is a spring that works by twisting; when it is twisted about its axis by an angle, it produces a torque proportional to the angle. A torsion spring's rate is in units of torque divided by angle,

Outer Diameter. The outer diameter (coil width) of your extension spring is calculated by adding two wire diameters to the inner diameter. You must measure the space that will be surrounding the spring’s body to set this tolerance.

Dec 31, 2012 What is the maximum extension of the spring ? ans 2mg/k using WE theorem why is it not mg/k I get this by using force equation Ie. f=ma but a=0 so f=0 So, mg-kx=0 Therefore x=mg/k Can you explain what went wrong with my approach?also can you explain why the spring doesn't stop extending extending when mg=kx (X is the extension in the spring)

### Hooke’s Law – density, elastic limit

The extension of a spring or wire is directly proportional to the force applied provided the limit of proportionality is not exceeded. When a graph of force against extension is plotted for a material which obeys Hooke’s law it looks like the graph below. The gradient of this graph is the spring constant (k) which is measured in Nm-1.

In most cases the stretch of the extension spring should be half of the door height. This means for a 7 foot high door (84 inches) the spring stretch would be 42 inches. Most extension springs for 7 foot high doors are 25 inches long with a 42 inch stretch. For an 8 foot high door the spring would be 28 inches long with a 48 inch stretch.

Drawbar extension springs are a type of pitched extension springs with the loads attached to the further ends of the spring, connected through the center of the spring with a metal loop. When a static load is applied, the spring will compress accordingly and provide a set