Statics · Chapter 7

Internal Forces

When external loads act on beams, columns, and frames, forces and moments develop within each member. Knowing these internal loadings is essential to choosing a material and cross section that will not fail. This chapter introduces the method of sections, shear and moment diagrams, and the analysis of flexible cables.

Learning objectives

Use the method of sections to find the internal loadings at a chosen point in a member.
Express the internal shear and bending moment along a member and plot them as shear and moment diagrams.
Analyze the forces and the shape of cables under concentrated, distributed, and self-weight loadings.
Choose a cut segment and apply a stated internal-force sign convention.
Construct piecewise shear and moment functions and verify their diagram behavior.
Distinguish a parabolic deck-loaded cable from a self-weight catenary.

7.1Internal Loadings Developed in Structural Members

To verify that a member can carry its load, the designer needs the loading acting inside the member, not just the external reactions. The method of sections exposes these internal loadings: pass an imaginary cut perpendicular to the member's axis at the point of interest, separate the member into two segments, and the loadings hidden at the cut become external forces on each free-body diagram.

Method of Sections

A technique for finding internal loadings at a point. The member is cut at that point, one segment is isolated as a free-body diagram, and the equilibrium equations are applied to solve for the exposed internal force and moment resultants.

The Three (and Four) Internal Resultants

At any cut in a coplanar problem, three internal resultants can appear. The component perpendicular to the cross section is the normal force N; the component tangent to the section is the shear force V; and the couple resisting rotation is the bending moment M. In three dimensions a fourth resultant — a torsional (twisting) moment — also acts about the member's axis. These resultants generally act at the centroid of the cross-sectional area.

Normal Force, Shear Force, Bending Moment

The normal force prevents the two segments from translating along the axis; the shear force prevents translation across the axis; and the bending moment prevents relative rotation. By Newton's third law each acts in opposite senses on the two segments.

Fig. 7–1Sectioning a cantilever beam at B exposes the internal normal force N, shear V, and bending moment M as equal-and-opposite resultants on the two segments.

Sign Convention

A consistent two-dimensional sign convention keeps results comparable. The widely accepted choice: a normal force is positive when it puts the segment in tension; a shear force is positive when it tends to rotate the segment clockwise; and a bending moment is positive when it bends the segment concave upward. Loadings in the opposite sense are negative.

Positive Sign Convention (2-D)

Positive N → tension on the cut face. Positive V → clockwise rotation of the isolated segment. Positive M → concave-upward (sagging) bending. A negative result from the equilibrium equations simply means the true sense is opposite to the one assumed on the diagram.

Course Note

Draw the sign-convention sketch in the corner of every problem before you start. Most lost marks here come not from the algebra but from assuming a direction and forgetting that a negative answer just flips the arrow — it does not mean you made a mistake.

— Instructor

Procedure for Analysis

Support reactions. If the chosen segment touches a support, find the reactions first. Choosing the segment away from the supports often lets you skip this step.
Free-body diagram. Keep every distributed load, couple, and force in its true location, then cut the member. Isolate the segment with fewer loads and show N, V, and M in their positive senses.
Equilibrium. Sum moments at the cut to remove N and V and solve M directly; sum forces along and across the axis for N and V. A negative scalar reverses the assumed sense.

Key Point

A member can carry up to four internal resultants — normal force, shear force, bending moment, and torsional moment. They generally vary from point to point and are always recoverable with the method of sections.

7.2Shear and Moment Equations and Diagrams

Design requires knowing how V and M vary along the entire member, not just at one cut. To capture this, section the beam at an arbitrary distance \(x\) and write V and M as functions of \(x\). Plotting these functions gives the shear diagram and the bending-moment diagram.

Shear and Moment Functions

Expressions V(x) and M(x) obtained by sectioning the beam at a variable position \(x\) and applying equilibrium to the segment of length \(x\). Because loading can change abruptly, separate functions must be written for each region between concentrated forces, applied couples, or changes in the distributed load.

Beam Classification

A simply supported beam is pinned at one end and roller-supported at the other; a cantilever is fixed at one end and free at the other. Beams are typically long, straight, prismatic members loaded perpendicular to their axes.

Fig. 7–2The shear and moment diagrams record how V and M change region by region; discontinuities occur where concentrated loads or couples act.

Procedure for Analysis

Support reactions. Find all reactions and resolve forces parallel and perpendicular to the beam axis.
Shear and moment functions. Use an origin at the left end and separate \(x\) coordinates for each region between load discontinuities. Section the beam, draw the segment with V and M positive, then sum forces for V and moments (about the cut) for M.
Diagrams. Plot V(x) and M(x); positive values go above the axis, negative below.

7.3Relations Between Distributed Load, Shear, and Moment

When many loads act on a beam, writing separate equations for every region becomes tedious. A faster approach uses the differential relations between the distributed load \(w\), the shear \(V\), and the moment \(M\). These let the diagrams be built by tracking slopes and areas instead of solving repeated free-body diagrams.

PRINCIPLE — Slope of the Shear Diagram

Distributed load equals the slope of V

At any point free of concentrated loads, the slope of the shear diagram equals the intensity of the distributed loading (taken positive upward).

Load–Shear Relation

\[ \frac{dV}{dx} = w(x) \]

Change in Shear = Area Under the Load Curve

\[ \Delta V = \int w(x)\,dx \]

PRINCIPLE — Slope of the Moment Diagram

Shear equals the slope of M

The slope of the moment diagram at any point equals the shear there. Consequently the moment reaches a maximum or minimum where the shear is zero.

Shear–Moment Relation

\[ \frac{dM}{dx} = V \]

Change in Moment = Area Under the Shear Diagram

\[ \Delta M = \int V\,dx \]

Discontinuities at Concentrated Loads and Couples

The differential relations fail exactly where point loads or couples act. A concentrated force makes the shear diagram jump by the force's magnitude (\(\Delta V = F\)); a couple makes the moment diagram jump by its magnitude (\(\Delta M = M_0\), upward for a clockwise couple).

Course Note

Think of building the diagrams as integration by eye: load → shear → moment, each one integrated from the one above it. If the load is a polynomial of degree n, the shear is degree n+1 and the moment is degree n+2. So a uniform load (constant) gives a sloped shear line and a parabolic moment — knowing the expected curve shape catches errors instantly.

— Instructor

7.4Cables

Flexible cables and chains carry loads efficiently in suspension bridges, transmission lines, and guy wires. In analysis they are assumed perfectly flexible (no bending resistance, so tension is always tangent to the cable) and inextensible (constant length). Three loading cases follow: concentrated loads, a load distributed along the horizontal, and the cable's own weight.

Cable Assumptions

Perfect flexibility means the internal force is pure tension directed along the cable. Inextensibility means a loaded cable holds a fixed geometry, so any segment can be treated as a rigid body for equilibrium.

Cable Under Concentrated Loads

A weightless cable carrying point loads forms straight segments, each in constant tension. Equilibrium at each support and load point supplies the equations; one extra geometric fact (a known total length or a known sag) is needed to close the system.

Cable Under a Distributed Load

For a load \(w = w(x)\) measured along the horizontal, equilibrium of a differential element yields three differential equations. Integrating them gives the cable's shape and shows that its horizontal tension component is constant.

Constant Horizontal Tension Component

\[ T\cos\theta = F_H = \text{constant} \]

Cable Shape Under a Horizontal-Distributed Load

\[ y = \frac{1}{F_H}\int\!\left(\int w(x)\,dx\right)dx \]

Cable Under Its Own Weight

When the cable's weight matters, the load is a function of arc length \(s\) rather than horizontal distance \(x\). Carrying through the integration produces the catenary curve.

RESULT — Catenary Curve

Shape of a cable hanging under its own weight

A uniform cable of self-weight \(w_0\) per unit length, with horizontal tension \(F_H\), hangs in a catenary:

Catenary Deflection Curve

\[ y = \frac{F_H}{w_0}\left[\cosh\!\left(\frac{w_0}{F_H}x\right) - 1\right] \]

Course Note

Remember the distinction that trips up most students: a load distributed along the horizontal (like a bridge deck) gives a parabola, while a load distributed along the cable's own length (its weight) gives a catenary. For shallow sags the two are nearly identical, which is why suspension-bridge cables are often analyzed as parabolas.

— Instructor

Physical meaning

Internal resultants are the force and couple that the material on one side of a cut must transmit to the material on the other. Shear is the local rate at which bending moment changes; the area under a shear diagram is therefore a change in bending moment, not merely a plotting convention.

Problem-solving strategy

State the beam or cable model, axes, and sign convention before sectioning.
Find reactions when the chosen segment requires them, then choose the side with fewer loads.
At a beam cut use force balance for N and V and moments about the cut for M.
Break diagram equations at every point load, couple, or load-law change.
Check jumps, slopes, endpoints, units, and the physical direction of each result.

EXAMPLE 1Cut in a cantilever shelf rail

A horizontal cantilever rail projects 2.0 m from a wall and supports a 3.0 kN downward point load at its free end. Determine the internal normal force, shear force, and bending moment at a section 0.50 m from the wall by isolating the free-end segment.

Recognition strategy

A section result is requested at one location, so a free-body cut is faster than a full diagram. The free-end segment contains only the point load and has no axial loading.

Given information

Cantilever length is 2.0 m; a 3.0 kN downward point load acts at the free end.
The cut is 0.50 m from the wall, leaving 1.50 m between the cut and load.

Model and interpretation

Use the right segment. At the left cut assume positive N to the right, positive V upward, and positive M clockwise; this clockwise right-face action is the positive sagging-moment convention used here.

Step-by-step solution

1. Locate the lever arm. The load lies \(2.0-0.50=1.50\ \text{m}\) from the cut, so its moment must be included at that true distance.

2. Solve axial equilibrium. No horizontal external force acts on the free segment, hence \(\sum F_x=N=0\), and the rail carries no normal force at this section.

3. Solve vertical equilibrium. \(\sum F_y=V-3.0=0\), therefore \(V=+3.0\ \text{kN}\) for the assumed upward cut-face arrow.

4. Solve moment equilibrium. \(\sum M_{\rm cut}=M-3.0(1.50)=0\), giving \(M=+4.50\ \text{kN}\!\cdot\!\text{m}\) in the assumed clockwise, positive-sagging sense on the right segment.

Verification

Units: The shear equation is in kN and the moment is kN times m; \(3.0\ \text{kN}(1.50\ \text{m})=4.50\ \text{kN}\!\cdot\!\text{m}\).

Signs and direction: The positive moment arrow on the right free segment is clockwise, which is the stated positive-sagging convention; its action-reaction partner on the left segment is counterclockwise.

Plausibility: A point load produces constant 3.0 kN shear along the rail, while moment grows toward the wall; 4.50 kN·m is less than the wall moment of 6.0 kN·m.

Final result: \( \boxed{N=0,\quad V=+3.0\ \text{kN},\quad M=+4.50\ \text{kN}\!\cdot\!\text{m}} \)

Engineering interpretation

The rail has no axial demand in this ideal loading case, but bending is material-critical near the wall. Connection rotation, fastener pull-out, local wall capacity, and allowable deflection remain design checks.

EXAMPLE 2Uniformly loaded access beam

A simply supported 6.0 m access beam carries a uniform downward load of 4.0 kN/m over its complete span. Determine the reactions, write \(V(x)\) and \(M(x)\) for \(0\le x\le6.0\) m from the left support, and locate the maximum bending moment.

Recognition strategy

A uniform load requires a variable cut and the load–shear–moment relationship. Symmetric geometry and loading make reactions equal, then one left-side expression covers the entire span.

Given information

Span \(L=6.0\ \text{m}\), load \(w=4.0\ \text{kN/m}\), and simple supports at both ends.
Coordinate x starts at the left support; downward distributed loading is taken negative.

Model and interpretation

Replace the full load by 24.0 kN at midspan for reactions, then retain \(wx\) on a variable left segment. Positive sagging moment and upward shear are used.

Step-by-step solution

1. Find support reactions. The total load is \(4.0(6.0)=24.0\ \text{kN}\); symmetry gives \(A_y=B_y=12.0\ \text{kN}\) upward.

2. Write the shear function. On a left segment, \(\sum F_y=12.0-4.0x-V=0\), hence \(V(x)=12.0-4.0x\ \text{kN}\).

3. Write the moment function. Taking moments about the cut gives \(M(x)=12.0x-[4.0x](x/2)=12.0x-2.0x^2\ \text{kN}\!\cdot\!\text{m}\).

4. Locate the extremum. Set \(V=dM/dx=0\): \(12.0-4.0x=0\), so \(x=3.00\ \text{m}\) and \(M_{\max}=12(3)-2(3)^2=18.0\ \text{kN}\!\cdot\!\text{m}\).

Verification

Units: In \(V=12.0-4.0x\), kN/m times m is kN; integrating that shear produces kN·m in the moment equation.

Signs and direction: Shear starts positive at the left reaction, crosses zero at midspan, and ends at \(-12.0\) kN before the right reaction returns it to zero.

Plausibility: Symmetric loading requires the peak moment at midspan and zero moment at both simple supports; \(M(0)=M(6.0)=0\) confirms both conditions.

Final result: \( \boxed{A_y=B_y=12.0\ \text{kN},\ V=12.0-4.0x\ \text{kN},\ M=12.0x-2.0x^2\ \text{kN}\!\cdot\!\text{m},\ M_{\max}=18.0\ \text{kN}\!\cdot\!\text{m}} \)

Engineering interpretation

The zero-shear location identifies the bending-demand peak for this continuous load. A selected beam still needs strength, lateral stability, deflection, vibration, connection, and code-prescribed load-combination checks.

EXAMPLE 3Deck-loaded cable tension

A level-span pedestrian-bridge cable carries a uniform deck load of 2.0 kN/m measured per horizontal metre over a 40 m span. Its specified midspan sag is 4.0 m. Using the parabolic cable model, determine the horizontal tension component, each vertical support component, and the maximum cable tension.

Recognition strategy

The load is uniform per horizontal distance, not per cable length, so the ideal cable is parabolic rather than catenary. Sag provides the geometry needed to determine horizontal pull.

Given information

\(w=2.0\ \text{kN/m}\), span \(L=40\ \text{m}\), equal support elevations, and sag \(h=4.0\ \text{m}\).
Cable is flexible, weightless relative to deck load, and inextensible; tension follows the tangent.

Model and interpretation

Use the lowest point as origin. For a symmetric parabolic cable, \(H=wL^2/(8h)\), each vertical reaction is \(wL/2\), and support tension is the vector magnitude of H and that vertical component.

Step-by-step solution

1. Determine horizontal pull. \(H=2.0(40^2)/[8(4.0)]=100\ \text{kN}\), a constant horizontal component throughout the ideal cable.

2. Determine vertical components. Total deck load is \(2.0(40)=80.0\ \text{kN}\), so symmetry gives \(V_A=V_B=40.0\ \text{kN}\) upward.

3. Combine components at a support. The steepest cable segment is at a support, where \(T_{\max}=\sqrt{H^2+V_A^2}=\sqrt{100^2+40^2}\ \text{kN}\).

4. Evaluate the maximum tension. \(T_{\max}=107.7\ \text{kN}\); the support tangent angle is \(\tan^{-1}(40/100)=21.8^\circ\), consistent with moderate sag.

Verification

Units: \(wL^2/h\) has units (kN/m)m²/m = kN, and the vector magnitude calculation combines two kN components only.

Signs and direction: Support reactions are upward on the cable while the cable pulls downward and inward on supports; the positive H is a component magnitude, not a separate support load direction.

Plausibility: The maximum tension must exceed H=100 kN and is only moderately larger because the 21.8° support slope is shallow; 107.7 kN meets both tests.

Final result: \( \boxed{H=100\ \text{kN},\quad V_A=V_B=40.0\ \text{kN},\quad T_{\max}=107.7\ \text{kN}} \)

Engineering interpretation

Reducing sag increases horizontal tension sharply, so tower and anchorage demand can dominate cable design. A real bridge requires cable self-weight, wind, temperature, erection stages, fatigue, dynamics, and code review beyond this static idealization.

Chapter recap

A cut exposes N, V, M, and in 3-D possibly torsion; signs describe the chosen face and convention.
Use equilibrium locally and use load–shear–moment slopes and areas between discontinuities.
Before moving on: calculate reactions, write piecewise functions, identify a maximum moment from zero shear, and distinguish cable load models.

Content adapted and condensed from a standard engineering statics treatment of internal forces. Figures are original schematics drawn for MechForge. Section exercises, fundamental problems, and review questions are intentionally omitted from this reference page.