The internal forces in a section of a body are those forces which hold together two parts of a given body separated by the section. Both parts of the body remain in equilibrium. It follows that internal forces which exist at a section are equivalent to all external forces acting on the particular part of the body.

All internal forces in the section are usually replaced by a force-couple
system
, in the centroid C of the cut K.
(see Fig.3.65). The
force consists of the *axial force* (its line of action
is perpendicular to the plane K and *shearing force* lying in
the plane K. Accordingly, couple consists of two components
the first of which is referred to as the *torque* (its line of
action is perpendicular to the plane K) and the second is called the
*bending moment* lying in the plane K.

Now, we will restrict our attention to the case in which a body is
loaded in just one plane. Moreover we will analyze the internal forces in
a very common engineering structure which is referred to as a *beam*. Beams
are usually long straight slender prismatic members designed to support transversal
loads. The loads may be either concentrated at specific points, or
distributed along the entire length or a portion of the beam. We will
limit our analysis to beams which are statically determinate supported.
The aim of an analysis is to obtain shear *V* and bending moment
*M* in all cuts K of the beam.

First we determine the reactions at the supports of the beam. Than we cut the beam at K and use the free-body diagram of one of the two parts of the beam. We adopt the sign convention according to Fig. 3.66. The result of our analysis should be a shear diagram and bending moment diagram representing the shear and the bending moment at any section of the beam. For doing so we use so called Schwedler theorem saying

where | w |
is the distributed load per unit length assumed positive if directed |

downwards | ||

V |
is the shear | |

M |
is the bending moment | |

x |
is the coordinate of the cut oriented from left to right. |

We note that the cuts of the beam where the bending moment is maximum or minimum are also the cuts where the shear is zero.

**Solution**

First we determine the force *S* in the rope. The free-body diagram is shown in
Fig. 3.67. The distributed loads are substituted by forces

W_{1} = l w_{1} = 35 N, W_{2} = l w_{2} = 93.33 N |
(3.40) |

They act in the centroids of the respective areas.

The moment equilibrium equation with respect to B yields

l S sin - W_{1} - W_{2} - F = 0 |
(3.41) |

and the result is

Second we use the free-body diagram of the right-hand part of the beam
(see Fig. 3.67) for determining the internal forces
*N*, *V*, *M*_{o}.

We have

N |
= | S cos = 171.56 N |

V |
= | S sin - W^{x}_{1} - W^{x}_{2} = 148.23 N |

M_{o} |
= | S sin - W^{x}_{1} - l W^{x}_{2} = 56.76 Nm |

Third we construct shear and moment diagrams according to definitions
(see Fig. 3.67).
The maximum value of the bending moment
*M*_{omax} = 34.83 *Nm*
occurs in the section where *V* = 0, namely where
*x* = = 0.2 m from the left side.

Determine the shear and moment equations for the beam. Draw shear and moment diagrams. Indicate the section where the bending moment reaches its maximum value.

**Solution**

*x* = 0.2 *m*, *M*_{omax} = 34.83 Nm.

**Solution**

*x* = 0.193 m,
*M*_{omax} = 93.44 *Nm*^{-1}

**Solution**

*x* = 0.3 m,
*M*_{omax} = - 48 *Nm*

**Solution**

*x* = 0.43 m,
*M*_{omax} = 23.59 *Nm*

**Solution**

*x* = 0.35 m,
*M*_{omax} = 39.72 *Nm*