Diagrams in Beams
From this text, you will learn what an axial (normal) force, a shear force, and a bending moment are, as well as the sign convention for these internal forces.
We will describe the relationships between a distributed load, the shear force, and the bending moment, and what changes to expect in the internal force diagrams for a given load on the beam.
Here is the usual sign convention for each of the internal forces:
1. Normal force (N)
2. Shear force (Q)
3. Bending moment (M)
Relationship between:
It may be constant or vary depending on the position x along the beam.
\( \frac{dQ(x)}{dx} = -q(x) \)
In other words, the change in shear force along the beam is caused by the distributed load.
\( \frac{dM(x)}{dx} = Q(x) \)
Thus, the change in the bending moment along the beam is due to the shear force.
1) The basic method for drawing internal force diagrams involves first writing out the internal force functions. In our courses, including free materials, you will find examples of solving a beam from start to finish using this method.
2) A more advanced method involves drawing diagrams without writing down the internal force functions, and for this method it is useful to know certain rules for drawing internal force diagrams:
3) An interesting relationship between Q(x) and M(x) — namely, that the bending moment diagram’s change corresponds to the area under the shear force diagram — can also be used for drawing diagrams. Calculations using this approach can be found in our YouTube materials, although it’s a method rarely shown by instructors.
Theory is one thing, but a better understanding will certainly come from working through examples:
A video course showing how to determine jumps on the diagrams
We will describe the relationships between a distributed load, the shear force, and the bending moment, and what changes to expect in the internal force diagrams for a given load on the beam.
Calculating beams
Types of internal forces
We have three basic types of internal forces that can occur in beams:
1. Normal (axial) forces – these result from forces acting along the beam’s axis. If the force is directed along the beam’s axis, causing it to stretch (tension) or compress, it is a normal force. Symbolically denoted as N.
2. Shear forces – these act perpendicular to the beam’s axis and cause internal shear stresses. Denoted as Q, they are responsible for the vertical displacement of different cross-sections of the beam relative to each other.
3. Bending moment – this is the moment of a force that causes the beam to bend. It affects the curvature of the beam and varies along its length. The bending moment is denoted as M.
2. Shear forces – these act perpendicular to the beam’s axis and cause internal shear stresses. Denoted as Q, they are responsible for the vertical displacement of different cross-sections of the beam relative to each other.
3. Bending moment – this is the moment of a force that causes the beam to bend. It affects the curvature of the beam and varies along its length. The bending moment is denoted as M.
Sign convention for internal forces
The sign convention for internal forces in beam mechanics is crucial for correctly interpreting their directions and magnitudes, and for consistently recording results in structural analysis.Here is the usual sign convention for each of the internal forces:
Fig. 1. Sign convention for internal forces
1. Normal force (N)
- Tension: A force that stretches the beam (acting outward from the cross-section, “away” from the center of the beam) is usually considered positive. In short: if the beam is in tension, the normal force N is positive.
- Compression: A force that compresses the beam (acting axially inward, causing compression) is considered negative.
- Compression: A force that compresses the beam (acting axially inward, causing compression) is considered negative.
2. Shear force (Q)
- Positive shear force: When the force acts upward on the left side of the cross-section or downward on the right side of the cross-section, the shear force is considered positive.
- Negative shear force: The force acting downward on the left side of the cross-section or upward on the right side of the cross-section is considered negative.
- Negative shear force: The force acting downward on the left side of the cross-section or upward on the right side of the cross-section is considered negative.
3. Bending moment (M)
- Positive bending moment (moments causing a “smile” shape): If the bending moment causes the beam to bend in a shape resembling a “smile” (i.e., the top fibers of the beam are in compression and the bottom fibers are in tension), the moment is positive.
- Negative bending moment (moments causing a “frown” shape): When the beam bends in the opposite direction (the top fibers in tension and the bottom fibers in compression — a “frown”), the bending moment is negative.
- Negative bending moment (moments causing a “frown” shape): When the beam bends in the opposite direction (the top fibers in tension and the bottom fibers in compression — a “frown”), the bending moment is negative.
Relationship between:
distributed load <-> shear force <-> bending moment
1. Distributed load q(x)
The distributed load q(x) is a force spread along the length of the beam, acting per unit length.It may be constant or vary depending on the position x along the beam.
2. Shear force Q(x)
The relationship between the distributed load and the shear force is:\( \frac{dQ(x)}{dx} = -q(x) \)
In other words, the change in shear force along the beam is caused by the distributed load.
- When the distributed load q(x) is positive (e.g., directed downward), the shear force Q(x) decreases along the length of the beam.
- When q(x) = 0 (no load), the shear force is constant.
- If the shear force is zero, then the bending moment is constant.
- When q(x) = 0 (no load), the shear force is constant.
- If the shear force is zero, then the bending moment is constant.
3. Bending moment M(x)
The relationship between the shear force and the bending moment is:\( \frac{dM(x)}{dx} = Q(x) \)
Thus, the change in the bending moment along the beam is due to the shear force.
- When the shear force Q(x) is positive, the bending moment M(x) increases.
- Where the shear force diagram Q(x) = 0, the bending moment reaches an extremum.
- Where the shear force diagram Q(x) = 0, the bending moment reaches an extremum.
Fig. 2. Relationship: distributed load – shear force – bending moment
Rules for drawing internal force diagrams
Let’s say there are three methods for drawing internal force diagrams:1) The basic method for drawing internal force diagrams involves first writing out the internal force functions. In our courses, including free materials, you will find examples of solving a beam from start to finish using this method.
2) A more advanced method involves drawing diagrams without writing down the internal force functions, and for this method it is useful to know certain rules for drawing internal force diagrams:
- If a transverse force is applied at any point on the beam, on the shear force diagram Q there is a jump equal to the magnitude of that force,
the jump is in the direction of the force when reading the diagram from left to right,
- If there is a distributed load on a span of the beam, then on the shear force diagram Q on that span there is a change by the resultant of the distributed load,
with the direction consistent with the resultant (when reading from left to right),
- If there is a concentrated moment applied at any point on the beam, then on the bending moment diagram M there is a jump equal to the magnitude of that moment,
the jump is positive or negative depending on whether it is clockwise or counterclockwise — see the sign convention.
the jump is in the direction of the force when reading the diagram from left to right,
- If there is a distributed load on a span of the beam, then on the shear force diagram Q on that span there is a change by the resultant of the distributed load,
with the direction consistent with the resultant (when reading from left to right),
- If there is a concentrated moment applied at any point on the beam, then on the bending moment diagram M there is a jump equal to the magnitude of that moment,
the jump is positive or negative depending on whether it is clockwise or counterclockwise — see the sign convention.
Fig. 3. Beam loading and jumps on Q and M diagrams
3) An interesting relationship between Q(x) and M(x) — namely, that the bending moment diagram’s change corresponds to the area under the shear force diagram — can also be used for drawing diagrams. Calculations using this approach can be found in our YouTube materials, although it’s a method rarely shown by instructors.
Theory is one thing, but a better understanding will certainly come from working through examples:
A video course showing how to determine jumps on the diagrams
and describing method 2) and 3) of drawing diagrams through 4 examples.
Set of 19 beams with short solutions
Below are short solutions + diagrams for each beam
Solutions
Below you will also find links to numerous solution examples for both straight and inclined beams, as well as hinged beams.