How to Calculate beams loads and support forces

For a beam in balance loaded with weights (or other load forces) the reaction forces “R”  at the supports equals the load forces “F”. The force balance can be expressed as:

F₁ + F₂ + ........ + F_n = R₁ + R₂                                                                  (1)

where

F = force from load (N or lbf)

R = forces from support (N or lbf)

In addition, for a beam in balance the algebraic sum of moments equals zero. The moment balance can be expressed as

F1 l₁ + F2 l₂ + .... + F_n l_n = R x₁ + R x₂                                                        (2)

where

l = the distance from the force to a common reference - usually the distance to one of the supports (m or ft)

x = the distance from the support force to a common reference - usually the distance to one of the supports (m or ft)

Example:
A beam with two symmetrical loads

A 9 m long beam with two supports is loaded with two equal and symmetrical loads F₁ and F₂, each 300 kg. The support forces F₃ and F₄ can be calculated:

(300 kg) (9.81 m/s²) + (300 kg) (9.81 m/s²) = R₁ + R₂

Therefore:

R₁ + R₂ = 5886 (N)

Note….!

Load due to the weight of a mass “m”  is (F = m.g) Newton's - where g = 9.81 m/s².

With symmetrical and equal loads the support forces also will be symmetrical and equal. Using

R₁ = R₂

The equation above can be simplified to

R₁ = R₂ = (5886 N) / 2

   = 2943 N

Example:

A beam with two not symmetrical loads

A 9 m long beam with two supports is loaded with two loads, 3 kg is located 3 m from the end (R₁), and the other load of 750kg is located 6 m from the same end. The balance of forces can be expressed as:

(300 kg) (9.81 m/s²) + (750 kg) (9.81 m/s²) = R₁ + R₂

Therefore:

R₁ + R₂ = 10300.5 (N)

The algebraic sum of the moments (2) can be expressed as:

(300 kg) (9.81 m/s²) (3 m) + (750 kg) (9.81 m/s²) (6 m) = R₁ (0 m) + R₂ (9 m)

So

R₂  = 5886 (N)

R₁ can be calculated as:

R₁  = (10300.5 N) - (5886 N)

    = 4414.5 N

References:

Hooke's Law

Spring

Springs are fundamental mechanical components which form the basis of many mechanical systems. A spring can be defined to be an elastic member which exerts a resisting force when its shape is changed. Most springs are assumed linear and obey the Hooke's Law,

where F is the resisting force, Δ is the displacement, and the k is the spring constant.

Basic Spring Types

Springs are of several types, the most plentiful of which are shown as follows,

Compression Spring

Figures : Spring Constant Dependencies [1]

We can expand the spring constant k as a function of the material properties of the spring. Doing so and solving for the spring displacement gives,

where G is the material shear modulus, n_a is the number of active coils, and D and d are defined in the drawing. The number of active coils is equal to the total number of coils n_t minus the number of end coils n* that do not help carry the load,

The value for n* depends on the ends of the spring. See the following illustration for different n* values:

Figure : An illustration of compression spring with different n* values [1]

Geometrical Factors

The spring index, C, can be used to express the deflection,

The useful range for C is about 4 to 12, with an optimum value of approximately 9. The wire diameter, d, should conform to a standard size if at all possible.

Shear Stress in the Spring

The maximum shear stress τ_max in a helical spring occurs on the inner face of the spring coils and is equal to,

where W is the Wahl Correction Factor which accounts for shear stress resulting from spring curvature,

To be continued

References :

1. http://www.efunda.com/DesignStandards/springs/spring_design.cfm
2.