Learning topic
Internal Forces and Stresses in Cross-Sections
Learn how to use the method of sections to find internal forces N, Q, M and T, construct force diagrams, and relate axial force to normal stress.
External loads produce internal forces within a bar that resist deformation. These internal actions are determined using the method of sections: imagine cutting the member at the required location and analyze the equilibrium of one of the resulting parts.
Internal force resultants
In the general case, a cross-section may carry an axial force $N$, shear forces $Q$, bending moments $M$, and a torque $T$. Under centric tension or compression, only the axial force $N$ acts.
- Make an imaginary cross-section through the bar at the location where the internal force is required.
- Remove one part of the bar and consider the equilibrium of the remaining part.
- Replace the action of the removed part by the internal axial force $N$ acting at the cut.
- Choose the positive direction of $N$ to correspond to tension and write the axial equilibrium equation: $\sum F_x=0$.
- Solve for $N$. A positive result corresponds to the assumed tensile direction; a negative result means the actual direction is opposite and the segment is in compression.
Sign convention for axial force
The axial force $N$ is taken as positive in tension and negative in compression. Its value is determined separately for each segment from equilibrium equations.
Normal stress
The internal force represents the resultant action of the material over the section, whereas stress describes the intensity of that action. For centric tension or compression of a straight member with a uniform cross-section, normal stress is uniformly distributed sufficiently far from local disturbances.
For centric tension or compression of a straight member with a uniform cross-section, the normal stress is:
$$\sigma=\frac{N}{A}$$
- $\sigma$ — normal stress, MPa;
- $N$ — axial force, N;
- $A$ — cross-sectional area, mm².
Since $1\ \text{N/mm}^2=1\ \text{MPa}$, using $N$ in newtons and $A$ in mm² gives the result directly in megapascals.
Example
Consider a bar with cross-sectional area $A=200\ \text{mm}^2$ carrying a tensile axial force $N=20\ \text{kN}$. Converting the force gives $N=20000\ \text{N}$. Therefore:
$$\sigma=\frac{20000}{200}=100\ \text{MPa}.$$
The positive stress corresponds to tension.
Learning outcome
After studying this topic, you should be able to cut a bar conceptually, determine the axial force from equilibrium, construct an $N$ diagram, and calculate normal stress in a cross-section.
About this topic
Internal force components in any cross-section are determined using the method of sections. This section details sign conventions and algorithms for constructing axial force (N), shear force (Q), bending moment (M), and torque (T) diagrams. It also explores the physical nature of stress as internal force intensity, divided into normal and shear stresses. Worked examples illustrate the connection between external loads and internal material response.