Internal forces in a body

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.

Figure 3.65: Internal forces in a body

All internal forces in the section are usually replaced by a force-couple system $\bf F_{k}^{}$,$\bf M_{k}^{}$ in the centroid C of the cut K. (see Fig.3.65). The force $\bf F_{k}^{}$ consists of the axial force $\bf N$ (its line of action is perpendicular to the plane K and shearing force $\bf V$ lying in the plane K. Accordingly, couple $\bf M_{k}^{}$ consists of two components the first of which is referred to as the torque $\bf T$ (its line of action is perpendicular to the plane K) and the second is called the bending moment $\bf M_{b}^{}$ 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

Figure 3.66: Internal forces in a beam

$${\frac{{\rm d}V}{{\rm d}x}} {\frac{{\rm d}M}{{\rm d}x}}$$ (3.39)

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.

Exercise 3.10.1   Sample problem The beam of the lenght l = 0.7 m is shown in Fig. 3.67. It is loaded by the force F = 400 N and by partly non-uniformly distributed load characterized by w₁ = 50 Nm^-1 and w₂ = 400  Nm^-1. The angle $\alpha$ = 45^o. Determine inner forces at section A-A.

Figure 3.67: Exercise 3.10.1. Internal forces in a beam

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

$${\textstyle\frac{1}{2}} {\textstyle\frac{2}{3}}$$ (3.40)

They act in the centroids of the respective areas.

The moment equilibrium equation with respect to B yields

$$\alpha {\frac{l}{2}} {\frac{2 l}{9}} {\frac{l}{3}}$$ (3.41)

and the result is S = 242.67 N.

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

W^x₁ = $${\textstyle\frac{1}{2}}$$ l w₁,        W^x₂ = $${\textstyle\frac{1}{2}}$$$${\textstyle\frac{1}{6}}$$ l $${\textstyle\frac{1}{4}}$$ w₂

and - for the section A-A -

N	=	S cos$$\alpha$$ = 171.56  N
V	=	S sin$$\alpha$$ - Wx1 - Wx2 = 148.23  N
Mo	=	$${\frac{l}{2}}$$ S sin$$\alpha$$ - $${\frac{l}{4}}$$ Wx1 - $${\textstyle\frac{1}{3}}$$$${\textstyle\frac{1}{6}}$$ l Wx2 = 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 = ${\frac{l}{3}}$ = 0.2 m from the left side.

Exercise 3.10.2   Internal forces in a beam The simply supported beam (see Fig. 3.68) has the length l = 0.6 m, a = 0.2 m. It is loaded by the force F = 200 N, by the torque M = 20 Nm, and by uniformly distributed load w = 100 Nm^-1.
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.

Figure 3.68: Exercise 3.10.2. Internal forces in a beam

Solution
x = 0.2 m, M_omax = 34.83 Nm.

Exercise 3.10.3   Internal forces in a beam The beam shown in Fig. 3.69 is loaded by the forces F₁ = 400 N, F₂ = 500 N, by the torque M = 90 Nm, and by the distributed load w = 5000 Nm^-1. Further we know a = 0.3 m, $\alpha$ = 30^o. Draw shear and moment diagrams. Indicate the section where the bending moment reaches its maximum value and compute it.

Figure 3.69: Exercise 3.10.3. Internal forces in a beam

Solution
x = 0.193 m, M_omax = 93.44 Nm^-1

Exercise 3.10.4   Internal forces in a beam The beam shown in Fig. 3.70 is loaded by the torque M = 60 Nm, by uniform distributed load w₁ = 400 Nm^-1, and by linearly distributed load w₂ = 1066 Nm^-1. The length l = 0.3 m. Draw shear and moment diagrams. Indicate the section where the bending moment reaches its maximum value and compute it.

Figure 3.70: Exercise 3.10.4. Internal forces in a beam

Solution
x = 0.3 m, M_omax = - 48 Nm

Exercise 3.10.5   Internal forces in a beam The beam shown in Fig. 3.71 is loaded by linearly distributed load w₀ = 1000 Nm^-1. The length a = 0.1 m. Draw shear and moment diagrams. Indicate the section where the bending moment reaches its maximum value and compute it.

Figure 3.71: Exercise 3.10.5. Internal forces in a beam

Solution
x = 0.43 m, M_omax = 23.59 Nm

Exercise 3.10.6   Internal forces in a beam The simply supported beam according to Fig. 3.72 is loaded by sinus-shape distributed load. It is known that l = 0.7 m, w₀ = 800 Nm^-1. Draw shear and moment diagrams. Indicate the section where the bending moment reaches its maximum value and compute it.

Figure 3.72: Exercise 3.10.6. Internal forces in a beam

Solution
x = 0.35 m, M_omax = 39.72 Nm