Statics and Dynamics

Chapter 7 — Internal Forces and Moments

Vector Mechanics for Engineers: Statics, 11th Edition

7.1 Internal Forces in Members

1. What Are Internal Forces?

Internal forces and moments are the forces and moments transmitted between neighboring portions of a body. They develop when external loads act on a member.

To determine internal forces, use the method of sections: imagine cutting through the member and isolating one side. Replace the material removed by the internal forces and moments acting at the cut.

For a planar member, the internal resultants are:

  • Normal force (N): Acts along the member's longitudinal axis. It produces tension or compression.

  • Shear force (V): Acts perpendicular to the member's longitudinal axis and tends to make the cut faces slide relative to each other.

  • Bending moment (M): Tends to bend the member.

For a three-dimensional member, the internal resultants may include three force components (N, Vy, Vz) and three moment components (T, My, Mz). T represents the twisting moment, or torque, about the member's longitudinal axis.

2. Method of Sections

Use the method of sections when internal forces are required at a specific location.

Procedure:

  1. Draw the free-body diagram of the entire member.

  2. Calculate any unknown external reactions needed.

  3. Pass an imaginary cut through the point of interest.

  4. Isolate whichever portion has fewer loads or unknowns.

  5. Replace the cut with the appropriate internal forces and moments.

  6. Apply the equilibrium equations.

  7. Interpret the signs of the results.

For a planar problem:

ΣFx = 0
ΣFy = 0
ΣM = 0

The internal resultants on opposite sides of the same cut are equal in magnitude and opposite in direction.

3. Sign Conventions

Use one consistent sign convention throughout a problem. A common convention for the left portion of a planar beam is:

  • Positive N: Acts away from the cut, representing tension.

  • Negative N: Represents compression.

  • Positive V: Acts upward on the left cut face.

  • Positive M: Acts counterclockwise on the left cut face.

The right cut face has equal-and-opposite internal resultants.

A negative result means the actual direction is opposite the assumed positive direction. For example, N = -8 kN means an 8 kN compressive force when tension is positive.

Sign conventions can differ between textbooks and instructors, so always follow the convention used in the problem.

4. Example: Finding Internal Forces and Moments

A horizontal member has a 4 kN force directed leftward at its left end and a 10 kN downward force located 2 m from the left end. The member is cut at x = 3 m. Isolate the left segment.

Normal force:
ΣFx = 0

Using a tension-positive convention, the internal normal force is N = -4 kN, meaning compression.

Shear force:
ΣFy = 0

V - 10 = 0

V = 10 kN

Bending moment:
Take moments about the cut, using counterclockwise as positive.

M + 10(1) = 0

M = -10 kN·m

The negative moment indicates a clockwise internal moment under the stated convention.

Always draw the cut-face forces explicitly before writing equilibrium equations. This prevents sign errors.

7.2 Beams

1. What Is a Beam?

A beam is a structural member designed primarily to support loads perpendicular to its longitudinal axis. Its internal shear force and bending moment generally vary along its length.

2. Common Supports and Reactions

Roller support:

  • Typically provides one reaction perpendicular to the supporting surface.

Pin support:

  • Provides horizontal and vertical reactions in a planar problem.

  • Does not provide a reaction moment.

Fixed support:

  • Provides horizontal and vertical reactions and a reaction moment in a planar problem.

Internal hinge:

  • Transmits force components.

  • Does not transmit a bending moment across the hinge.

Always draw the complete beam's free-body diagram and calculate the external reactions before finding internal forces or drawing shear and moment diagrams.

3. Concentrated and Distributed Loads

A concentrated force is modeled as acting at a single point.

A distributed load w(x) is force per unit length, commonly expressed in N/m or kN/m.

The resultant of a distributed load is:

FR = ∫w(x) dx

Its location is the centroid of the load distribution:

x̄ = ∫xw(x) dx / ∫w(x) dx

Common distributed-load shapes:

Uniform load:

  • Resultant: FR = wL

  • Location: Midpoint of the loaded length.

Triangular load increasing from zero to wmax:

  • Resultant: FR = (1/2)wmaxL

  • Location: L/3 from the larger-intensity end, or 2L/3 from the zero-intensity end.

Trapezoidal load:

  • Can be treated as a rectangle plus a triangle.

  • Find the resultant and centroid of each part, then combine them.

Equivalent resultants are useful for calculating reactions. When drawing shear and moment diagrams, account for how the original load varies along the beam.

4. Shear-Force and Bending-Moment Diagrams

A shear-force diagram represents V(x), while a bending-moment diagram represents M(x).

Using a common convention in which downward distributed load w(x) is positive:

dV/dx = -w(x)

dM/dx = V(x)

These equations determine the shapes of the diagrams.

No distributed load:

  • Shear is constant.

  • Moment varies linearly.

Uniform distributed load:

  • Shear varies linearly.

  • Moment varies quadratically.

Linearly varying distributed load:

  • Shear varies quadratically.

  • Moment varies cubically.

Concentrated force:

  • Causes a jump in the shear diagram.

  • Does not cause a moment jump unless a concentrated couple is also present.

Applied concentrated couple:

  • Does not cause a jump in the shear diagram.

  • Causes a jump in the bending-moment diagram.

Where V = 0, the bending moment may have a local maximum or minimum. Check the moment values on either side to determine whether the point is a maximum or minimum.

5. Example: Simply Supported Beam with a Center Load

A simply supported beam has a 6 m span and a 12 kN downward load at its midpoint.

By symmetry:

Ay = By = 6 kN

Shear:

  • From the left support to the center load: V = +6 kN.

  • At the center load, shear drops by 12 kN.

  • From the center load to the right support: V = -6 kN.

  • At the right support, the reaction returns the shear to zero.

Maximum bending moment:

Mmax = 6 kN × 3 m

Mmax = 18 kN·m

The maximum occurs at the center of the beam.

7.3 Relations Among Load, Shear, and Bending Moment

The relationships between distributed load, shear, and bending moment are:

dV/dx = -w(x)

dM/dx = V(x)

Integrating over a beam segment:

V(x2) - V(x1) = -∫ from x1 to x2 of w(x) dx

M(x2) - M(x1) = ∫ from x1 to x2 of V(x) dx

The change in shear equals the negative signed area under the distributed-load diagram.

The change in bending moment equals the signed area under the shear diagram.

Important interpretations:

  • Positive shear means the bending-moment diagram has a positive slope under this convention.

  • Negative shear means the bending-moment diagram has a negative slope.

  • Zero shear indicates a stationary value of the bending moment, although it may be a maximum, minimum, or flat region.

  • A point force causes a shear jump.

  • A point couple causes a moment jump.

  • With no distributed load, shear remains constant and bending moment changes linearly.

To draw diagrams:

  1. Calculate the support reactions.

  2. Divide the beam into segments at concentrated loads, couples, supports, and changes in distributed loading.

  3. Write equilibrium equations for each segment.

  4. Determine V(x) and M(x).

  5. Check the diagram's slopes, jumps, endpoints, and units.

  6. Confirm that the area under the shear diagram matches the change in bending moment.

7.4 Cables

1. Properties of Cables

Cables are flexible members that carry tension only. They cannot resist compression or bending moment.

The cable's shape depends on how the load is distributed.

For a cable segment with tension T at an angle θ measured from the horizontal:

Tx = T cos θ

Ty = T sin θ

At a joint where multiple cable segments meet, apply force equilibrium:

ΣFx = 0

ΣFy = 0

Cable tension acts away from the point being isolated, along the cable segment.

2. Cables Carrying Concentrated Loads

When a cable supports concentrated loads at discrete points, the cable generally consists of straight segments between those points.

At each load point:

  1. Isolate the joint or point.

  2. Draw the tensions along the adjoining cable segments.

  3. Resolve the tensions into horizontal and vertical components.

  4. Apply equilibrium.

  5. Use the cable geometry to relate the angles to the segment dimensions.

If there are no distributed horizontal loads, the horizontal component of tension remains constant along the cable. The total tension changes as the cable's slope changes.

Do not assume that the total tension is the same in every segment.

3. Cables Carrying Distributed Loads

The cable shape depends on how the load is distributed.

Uniform load per unit horizontal distance:

  • Produces a parabolic cable shape.

Uniform load per unit cable length:

  • Produces a catenary cable shape.

For a cable carrying a uniform vertical load w per unit horizontal distance, with the lowest point at x = 0 and y = 0:

y = wx²/(2H)

Here, H is the constant horizontal component of tension.

This parabolic model assumes that the load is uniform along the horizontal projection, not along the actual curved cable length.

7.5 Catenary Cables

1. Definition

A catenary is the exact shape of an ideal flexible cable carrying a uniform load per unit of its own length, such as a cable carrying its own weight.

With the lowest point as the origin, the catenary equation is:

y = a[cosh(x/a) - 1]

where:

a = H/w

H is the horizontal tension component, and w is the load per unit cable length.

The hyperbolic cosine is:

cosh(u) = (e^u + e^(-u))/2

2. Symmetric Catenary

For a symmetric cable with supports at x = ±L/2, the sag f is:

f = a[cosh(L/(2a)) - 1]

The vertical tension component at position x is:

V = wx

The tension magnitude is:

T = √(H² + V²)

At the lowest point, x = 0, so V = 0 and T = H. Tension increases toward the supports.

3. Parabolic Cable Versus Catenary

Parabolic cable:

  • Uniform load per unit horizontal distance.

  • Shape: Parabola.

  • Equation: y = wx²/(2H), when the lowest point is the origin.

Catenary cable:

  • Uniform load per unit cable length.

  • Shape: Catenary.

  • Equation: y = a[cosh(x/a) - 1].

A parabola may approximate a shallow catenary, but the two models are based on different load assumptions.

Chapter 7: Key Equations to Memorize

Internal equilibrium:
ΣFx = 0
ΣFy = 0
ΣM = 0

Distributed-load resultant:
FR = ∫w(x) dx

Resultant location:
x̄ = ∫xw(x) dx / ∫w(x) dx

Load-shear relationship:
dV/dx = -w(x)

Shear-moment relationship:
dM/dx = V(x)

Uniform distributed load:
FR = wL

Triangular distributed load:
FR = (1/2)wmaxL

Parabolic cable under a uniform horizontal load:
y = wx²/(2H)

Catenary cable:
y = a[cosh(x/a) - 1]

Catenary parameter:
a = H/w

Tension components:
Tx = T cos θ
Ty = T sin θ

Chapter 7: Common Exam Mistakes

  1. Forgetting to include external reactions before cutting a member.

  2. Mixing up normal force, shear force, and bending moment.

  3. Changing sign conventions partway through a problem.

  4. Assuming a negative internal force means the calculations are wrong.

  5. Placing a triangular load's resultant at its midpoint instead of one-third of the length from the larger-intensity end.

  6. Forgetting that a concentrated force causes a shear jump.

  7. Forgetting that an applied couple causes a moment jump.

  8. Confusing the slope of the shear diagram with the slope of the moment diagram.

  9. Assuming cable tension is the same magnitude everywhere.

  10. Confusing a parabolic cable with a catenary.

  11. Using degrees rather than radians if a formula specifically requires radians.

  12. Mixing units, such as kN with meters and N with millimeters.

Chapter 7: Final Problem-Solving Checklist

Before submitting a solution, check:

  • Did I draw a complete free-body diagram?

  • Did I calculate the necessary external reactions?

  • Did I isolate the correct segment?

  • Did I include all internal force and moment components at the cut?

  • Did I apply equilibrium correctly?

  • Did I use a consistent sign convention?

  • Did I place distributed-load resultants at the correct locations?

  • Do the shear and moment diagrams have the correct jumps and slopes?

  • Did I distinguish between the parabolic and catenary cable models?

  • Are my final answers labeled with the correct units and physical meaning?