**ntroduction**

Normally a beam is analysed to obtain the maximum stress and this is compared to the material strength to determine the design safety margin. It is also normally required to calculate the deflection on the beam under the maximum expected load. The determination of the maximum stress results from producing the shear and bending moment diagrams. To facilitate this work the first stage is normally to determine all of the external loads.

**Nomenclature**

e = strain

σ = stress (N/m

E = Young's Modulus = σ /e (N/m

y = distance of surface from neutral surface (m).

R = Radius of neutral axis (m).

I = Moment of Inertia (m

Z = section modulus = I/y

M = Moment (Nm)

w = Distrubuted load on beam (kg/m) or (N/m as force units)

W = total load on beam (kg ) or (N as force units)

F= Concentrated force on beam (N)

S= Shear Force on Section (N)

L = length of beam (m)

x = distance along beam (m)

σ = stress (N/m

^{2})E = Young's Modulus = σ /e (N/m

^{2})y = distance of surface from neutral surface (m).

R = Radius of neutral axis (m).

I = Moment of Inertia (m

^{4}- more normally cm^{4})Z = section modulus = I/y

_{max}(m^{3}- more normally cm^{3})M = Moment (Nm)

w = Distrubuted load on beam (kg/m) or (N/m as force units)

W = total load on beam (kg ) or (N as force units)

F= Concentrated force on beam (N)

S= Shear Force on Section (N)

L = length of beam (m)

x = distance along beam (m)

**Calculation of external forces**

To allow determination of all of the external loads a free-body diagram is construction with all of the loads and supports replaced by their equivalent forces. A typical free-body diagram is shown below.

The unknown forces (generally the support reactions) are then determined using the equations for plane static equilibrium.

For example considering the simple beam above the reaction R

_{2}is determined by Summing the moments about R_{1}to zero R

R_{2}. L - W.a = 0 Therefore R_{2}= W.a / L_{1}is determined by summing the vertical forces to 0

W - R

_{1}- R_{2}= 0 Therefore R_{1}= W - R_{2}**Shear and Bending Moment Diagram**

The shear force diagram indicates the shear force withstood by the beam section along the length of the beam.

The bending moment diagram indicates the bending moment withstood by the beam section along the length of the beam.

It is normal practice to produce a free body diagram with the shear diagram and the bending moment diagram position below

For simply supported beams the reactions are generally simple forces. When the beam is built-in the free body diagram will show the relevant support point as a reaction force and a reaction moment....

The bending moment diagram indicates the bending moment withstood by the beam section along the length of the beam.

It is normal practice to produce a free body diagram with the shear diagram and the bending moment diagram position below

For simply supported beams the reactions are generally simple forces. When the beam is built-in the free body diagram will show the relevant support point as a reaction force and a reaction moment....

**Sign Convention**

The sign convention used for shear force diagrams and bending moments is only important in that it should be used consistently throughout a project. The sign convention used on this page is as below.

**Typical Diagrams**

A shear force diagram is simply constructed by moving a section along the beam from (say)the left origin and summing the forces to the left of the section. The equilibrium condition states that the forces on either side of a section balance and therefore the resisting shear force of the section is obtained by this simple operation

The bending moment diagram is obtained in the same way except that the moment is the sum of the product of each force and its distance(x) from the section. Distributed loads are calculated buy summing the product of the total force (to the left of the section) and the distance(x) of the centroid of the distributed load.

The sketches below show simply supported beams with on concentrated force.

The bending moment diagram is obtained in the same way except that the moment is the sum of the product of each force and its distance(x) from the section. Distributed loads are calculated buy summing the product of the total force (to the left of the section) and the distance(x) of the centroid of the distributed load.

The sketches below show simply supported beams with on concentrated force.

The sketches below show Cantilever beams with three different load combinations.

Note: The force shown if based on loads (weights) would need to be converted to force units i.e. 50kg = 50x9,81(g) = 490 N.

**Shear Force Moment Relationship**

Consider a short length of a beam under a distributed load separated by a distance δx.

The bending moment at section AD is M and the shear force is S. The bending moment at BC = M + δM and the shear force is S + δS.

The equations for equilibrium in 2 dimensions results in the equations.. Forces.

The equations for equilibrium in 2 dimensions results in the equations.. Forces.

S - w.δx = S + δS

Therefore making δx infinitely small then.. dS /dx = - w

Therefore making δx infinitely small then.. dS /dx = - w

Moments.. Taking moments about C

M + Sδx - M - δM - w(δx)

Therefore making δx infinitely small then.. dM /dx = S

^{2}/2 = 0Therefore making δx infinitely small then.. dM /dx = S

Therefore putting the relationships into integral form.

The integral (Area) of the shear diagram between any limits results in the change of the shearing force between these limits and the integral of the Shear Force diagram between limits results in the change in bending moment...

## 1 comments:

what does the +ive and -ive sign shows in the shear force diagram... answer at 222milesaway@gmail.com

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