Mechanics of Materials · Chapter 1

Stress

Internal loadings, the concept of stress, and the design of simple axially and shear-loaded members.

Chapter Objectives

Review the statics principles used to find internal resultant loadings in a body.
Define normal and shear stress and the general state of stress at a point.
Apply the average normal stress and average shear stress equations to axially loaded and directly sheared members.
Size simple connections using allowable stress design (ASD) and limit state design (LRFD).

Physical meaning

Interpret the result in terms of transmitted load and deformation rather than formula symbols alone.

Problem-solving strategy

Draw the system.
State assumptions.
Select a relation.
Compute consistently.
Audit the result.

1.1Introduction

Mechanics of Materials

The branch of mechanics concerned with the internal effects produced in a solid body by external loads. Two quantities are central: stress, which relates to the strength of the material the body is made from, and strain, which measures how much the body deforms. Most design formulas in engineering codes rest on these two ideas.

The subject has deep historical roots: early load experiments on rods and beams date to the seventeenth century (Galileo), and the theory matured in the nineteenth century through the work of investigators such as Saint-Venant, Poisson, Lamé, and Navier. Modern extensions include the theories of elasticity and plasticity, supported by computational methods for complex geometries.

1.2Equilibrium of a Deformable Body

External Loads

Surface loads act through direct contact: a load applied over a very small area is idealized as a concentrated force, while loading spread over a larger area is a distributed load. For a coplanar distributed load, the resultant equals the area under the loading diagram and acts through the centroid of that area.

Body forces act without contact — gravity being the most common. The gravitational body force is represented by the weight W acting through the body's center of gravity.

Support Reactions

The rule for any support: if it prevents translation in a direction, it exerts a force in that direction; if it prevents rotation, it exerts a couple moment. Hence a roller produces one normal force, a pin produces two force components, and a fixed support produces two force components plus a moment.

Equations of Equilibrium (Vector Form)

\[ \sum \mathbf{F} = \mathbf{0} \qquad \sum \mathbf{M}_O = \mathbf{0} \]

Force balance prevents translation; moment balance about any point \(O\) prevents rotation.

Equations of Equilibrium (Scalar, 3-D and Coplanar)

\[ \Sigma F_x = 0,\;\; \Sigma F_y = 0,\;\; \Sigma F_z = 0 \qquad \Sigma M_x = 0,\;\; \Sigma M_y = 0,\;\; \Sigma M_z = 0 \] \[ \text{Coplanar } (x\text{–}y)\text{:}\quad \Sigma F_x = 0,\;\; \Sigma F_y = 0,\;\; \Sigma M_O = 0 \]

Always draw the free-body diagram first so every known and unknown load is accounted for.

Internal Resultant Loadings — the Method of Sections

To expose what is happening inside a body, pass an imaginary cut through the point of interest, separate the body, and draw the free-body diagram of one part. The distributed internal forces on the cut face are replaced by a resultant force and a resultant moment acting at the centroid of the sectioned area, and these resultants are found from equilibrium of the chosen segment.

The Four Internal Resultants

Normal force, \(N\) — acts perpendicular to the cut area; produced when external loads pull the two segments apart or push them together.

Shear force, \(V\) — lies in the plane of the area; produced when loads tend to slide one segment over the other.

Torsional moment (torque), \(T\) — twists one segment relative to the other about an axis perpendicular to the area.

Bending moment, \(M\) — bends the body about an axis lying in the plane of the area.

For coplanar loadings only \(N\), \(V\), and \(M\) exist at the section.

Sectioning a coplanar member exposes the internal resultants: normal force N, shear force V, and bending moment M.

FIG. 1-1 (original diagram)

Procedure for Analysis — Internal Loadings

1. Support reactions. If needed, find the reactions on the whole body first (or choose a segment that avoids them).

2. Section. Pass an imaginary cut through the point where the internal loads are required and draw the free-body diagram of one segment, showing the unknown resultants \(N\), \(V\), \(M\) (and \(T\) in 3-D) at the centroid of the cut.

3. Equilibrium. Sum moments at the cut point — this eliminates \(N\) and \(V\) and gives \(M\) directly — then sum forces for \(N\) and \(V\). A negative result means the load acts opposite to the assumed direction.

1.3Stress

Stress

Stress describes the intensity of internal force — force per unit area — acting on a specific plane passing through a point. It is obtained by shrinking a small area \(\Delta A\) on the sectioned surface toward zero and taking the limit of force over area. Two material assumptions underlie the definition: the material is continuous (no voids) and cohesive (no cracks or separations).

Normal Stress at a Point

\[ \sigma_z = \lim_{\Delta A \to 0} \frac{\Delta F_z}{\Delta A} \]

\(\sigma\) (sigma) = intensity of force acting perpendicular to \(\Delta A\).
A "pull" on the area is tensile stress; a "push" is compressive stress.

Shear Stress at a Point

\[ \tau_{zx} = \lim_{\Delta A \to 0} \frac{\Delta F_x}{\Delta A} \qquad \tau_{zy} = \lim_{\Delta A \to 0} \frac{\Delta F_y}{\Delta A} \]

\(\tau\) (tau) = intensity of force acting tangent to \(\Delta A\).
First subscript = orientation of the area (its outward normal); second subscript = direction of the stress.

General State of Stress

Sectioning the body with three mutually perpendicular planes isolates a cubic volume element around the point. Its state of stress consists of three components on each face — one normal stress and two shear stresses — characterizing how the material is loaded at that point.

Units of Stress

SI: newtons per square metre, the pascal (\(1\ \text{Pa} = 1\ \text{N/m}^2\)). Because the pascal is small, engineering values use kPa, MPa, or GPa. Note \(1\ \text{N/mm}^2 = 1\ \text{MPa}\).

US customary: pounds per square inch (psi) or kilopounds per square inch (ksi), where 1 kip = 1000 lb.

1.4Average Normal Stress in an Axially Loaded Bar

For a straight prismatic bar loaded through the centroid of its cross section, and made of homogeneous, isotropic material, the bar stretches uniformly and the normal stress is uniformly distributed over any cross section away from the ends. Equilibrium of the sectioned bar then yields the workhorse equation of this chapter.

Average Normal Stress

\[ \sigma = \frac{N}{A} \]

\(\sigma\) = average normal stress at any point on the cross section
\(N\) = internal resultant normal force, acting through the centroid of the section (found by the method of sections; for a two-force bar \(N = P\))
\(A\) = cross-sectional area where \(\sigma\) is evaluated

An axial load applied through the centroid produces a uniform normal stress distribution σ = N/A on the cross section.

FIG. 1-2 (original diagram)

Maximum Average Normal Stress

When a bar carries several axial loads, or its cross-sectional area changes, the internal force \(N\) differs from region to region. To find the largest stress, determine \(N\) in each region (an axial or normal force diagram — a plot of \(N\) versus position — helps), then evaluate \(\sigma = N/A\) wherever the ratio is largest. Tension is taken positive, compression negative.

Procedure for Analysis — Average Normal Stress

1. Internal loading. Section the member perpendicular to its axis at the point of interest and use equilibrium to find the internal axial force \(N\).

2. Average stress. Compute \(\sigma = N/A\). The stress acts perpendicular to the section: tensile \(N\) pulls on the section, compressive \(N\) pushes on it.

1.5Average Shear Stress

Shear stress is the stress component acting in the plane of a sectioned area. It dominates in simple connections — pins, bolts, glued joints, welds — where the applied load tends to slide one part of a member over another along a failure plane.

Average Shear Stress

\[ \tau_{avg} = \frac{V}{A} \]

\(\tau_{avg}\) = average shear stress, assumed uniform over the section
\(V\) = internal resultant shear force on the section, from equilibrium
\(A\) = area of the sheared section

Single Shear vs. Double Shear

In a single-shear (lap) joint, the connector resists the load across one shear plane, so \(V = F\). In a double-shear joint, two planes resist the load, so each plane carries \(V = F/2\). The equation \(\tau_{avg}=V/A\) is approximate — real shear distributions are non-uniform — but engineering codes accept it for sizing fasteners and bonded joints.

A bolt in a lap joint carries the full load on one shear plane; in a double lap joint each of the two planes carries half.

FIG. 1-3 (original diagram)

Principle · Complementary Property of Shear

Force and moment equilibrium of a volume element require that a shear stress acting on one face of the element be accompanied by equal-magnitude shear stresses on all four faces bordering that direction:

\[ \tau_{zy} = \tau'_{zy} = \tau_{yz} = \tau'_{yz} = \tau \]

The four stresses point either toward or away from each other at opposite edges of the element. Thus shear stress never exists in isolation on a single plane.

Procedure for Analysis — Average Shear Stress

1. Internal shear. Section the member along the plane where shear is to be determined and find the internal shear force \(V\) on that plane from equilibrium.

2. Average stress. Compute \(\tau_{avg} = V/A\) over the sheared area, acting in the same direction as \(V\) on the section.

1.6Allowable Stress Design

Why Designs Need a Margin

The working load on a member must be kept below what it can actually support because of unavoidable uncertainties: fabrication and assembly tolerances, unanticipated vibration, impact or accidental loads, deterioration from corrosion or weathering, and material variability (notably in wood, concrete, and composites).

Factor of Safety

\[ \text{F.S.} = \frac{F_{fail}}{F_{allow}} \]

\(F_{fail}\) = load at which the member fails, established by experiment
\(F_{allow}\) = allowable (working) load; F.S. > 1

Factor of Safety in Terms of Stress

\[ \text{F.S.} = \frac{\sigma_{fail}}{\sigma_{allow}} \qquad\text{or}\qquad \text{F.S.} = \frac{\tau_{fail}}{\tau_{allow}} \]

Valid when load and stress are linearly related, as for \(\sigma = N/A\) and \(\tau_{avg} = V/A\) (the area cancels).
Typical values range from near 1 (aircraft, to save weight) up to 3 or more (e.g. nuclear components); codes and handbooks specify them.

Allowable Stress Design (ASD)

Design in which member sizes are chosen so that the stress under working loads does not exceed the allowable stress. ASD balances safety against economy and is the basis for sizing simple connections: base plates in bearing, embedded anchors in bond shear, pins and bolts in direct shear. The compressive stress between two bearing surfaces is called bearing stress.

Required Area — Normal Load and Shear Load

\[ A = \frac{N}{\sigma_{allow}} \qquad\qquad A = \frac{V}{\tau_{allow}} \]

Gives the minimum section area for a member carrying normal force \(N\) or shear force \(V\) at the allowable stress.

1.7Limit State Design (LRFD)

Limit State Design / LRFD

Whereas ASD lumps all uncertainty into a single factor of safety, limit state design — called load and resistance factor design (LRFD) in the United States — treats load uncertainty and material uncertainty separately, using statistics and probability.

Load Factors γ

Each load type is scaled by a load factor \(\gamma\) (gamma) reflecting how predictable it is. The dead load \(D\) (fixed self-weight) is well known, so \(\gamma_D = 1.2\) is typical; the live load \(L\) (people, vehicles, wind, snow, earthquake) is more variable, so e.g. \(\gamma_L = 1.6\). Codes prescribe load combinations; for example, a dead + live + snow combination might be \(R = 1.2D + 1.6L + 0.5S\) — the small snow factor reflecting the low probability that all maxima occur simultaneously.

Resistance Factors φ

The resistance factor \(\phi\) (phi) ≤ 1 reduces a member's nominal strength according to the consistency and quality of its material — e.g. \(\phi = 0.9\) for a steel tension member, with smaller values for concrete, whose behavior is less predictable.

LRFD Design Criterion

\[ \phi P_n \;\ge\; \sum \gamma_i R_i \]

\(P_n\) = nominal strength of the member (load causing failure or loss of serviceability)
\(\phi\) = resistance factor; \(\gamma_i, R_i\) = load factor and load of each type acting
The factored (reduced) strength must equal or exceed the factored (increased) load.

EXAMPLE 1Tie-rod average stress

A galvanized 20 mm diameter tie rod supports a centered 31.4 kN tensile service load. Determine the average normal stress in its uniform shank and identify the sense of the internal normal force.

Recognition strategy

This is centered axial loading of a prismatic member, so a cut exposes only normal force and the gross circular area supports an average normal-stress calculation.

Given information

Diameter is 20 mm, service load is 31.4 kN, and the shank is evaluated away from the threaded end and load introduction.

Model and interpretation

Take tensile normal force as positive. A cut normal to the rod axis gives internal force from equilibrium; assume the centered load is transmitted through the centroid.

Step-by-step solution

1. Plan. Isolate either half of the rod at a normal cut. Force equilibrium requires the exposed internal resultant to be 31.4 kN in tension.

2. Formulate. Write the gross section area as \(A=\pi d^2/4\) so diameter is not accidentally used as an area.

3. Calculate. Evaluate the area as \(A=\pi(20\ \text{mm})^2/4=314.16\ \text{mm}^2\).

4. Resolve. Divide the cut force by area: \(\sigma_{avg}=31400/314.16=99.95\ \text{MPa}\), rounded to 100 MPa tensile.

Verification

Units: Since \(\text{N}/\text{mm}^2=\text{MPa}\), the calculated 99.95 MPa has the correct dimensions for stress.

Signs and direction: The loads pull outward from the cut, matching the declared positive tensile normal force and positive stress.

Plausibility: A force near 30 kN over approximately 300 square millimetres should be near 100 MPa, so the scale is credible.

Final result: \(\boxed{\sigma_{avg}=+100\ \text{MPa}\ \text{(tension)}}\)

Engineering interpretation

The result is a gross-shank average only. Thread-root area, corrosion loss, end eccentricity, and the material yield and fatigue limits need separate review before a connection is released.

EXAMPLE 2Double-shear clevis pin

A clevis joint transmits 42 kN through a 18 mm diameter pin. A central link is captured between two identical outer lugs, so two shear planes share the load. Determine average shear stress in the pin.

Recognition strategy

The central link tends to slide relative to the pair of outer lugs. This is direct double shear, requiring force sharing across two identified circular shear planes.

Given information

Transmitted load is 42 kN, pin diameter is 18 mm, and the symmetric lugs are assumed to share load equally.

Model and interpretation

Isolate the central link and use vertical equilibrium. Each plane carries half of the load; average shear assumes uniform traction on the circular plane.

Step-by-step solution

1. Plan. Count the two link-to-lug interfaces first; they are the two potential shear planes, not two pin cross sections in series.

2. Formulate. Equilibrium of the center link gives \(V_{plane}=P/2=42/2=21\ \text{kN}\).

3. Calculate. For one plane calculate \(A_s=\pi(18\ \text{mm})^2/4=254.47\ \text{mm}^2\).

4. Resolve. Compute \(\tau_{avg}=21000/254.47=82.53\ \text{MPa}\) on each plane.

Verification

Units: Newtons over square millimetres gives MPa, so 82.53 MPa is dimensionally a shear stress.

Signs and direction: Opposite faces exert opposite shearing tractions; the reported magnitude is positive because no signed pin-axis convention was requested.

Plausibility: Single shear would give about 165 MPa, exactly twice the double-shear result, which checks the load sharing.

Final result: \(\boxed{\tau_{avg}=82.5\ \text{MPa}\ \text{on each plane}}\)

Engineering interpretation

Pin shear does not alone govern a clevis. Bearing crush in lugs and link, pin bending due to clearance, net-section tearing, wear, and applicable connection rules must also be checked.

EXAMPLE 3ASD net-section hanger sizing

A flat steel hanger carries a 48 kN service tensile load. It has a 22 mm bolt hole and thickness 10 mm. With allowable net-section tensile stress of 140 MPa, determine the minimum plate width and select a whole-millimetre width.

Recognition strategy

The hole reduces the force-resisting net area, and the stated service load with allowable stress identifies an ASD net-section sizing check.

Given information

Service tension is 48 kN, thickness is 10 mm, hole diameter is 22 mm, and allowable net stress is 140 MPa.

Model and interpretation

Cut through the bolt-hole section and take tension positive. Use net area \(A_n=t(b-d_h)\); this simplified check neglects stagger, edge tear-out, and bolt bearing.

Step-by-step solution

1. Plan. Write the ASD condition \(P/A_n\leq\sigma_{allow}\) and solve symbolically before selecting a width.

2. Formulate. Required net area is \(A_{n,req}=48000/140=342.86\ \text{mm}^2\).

3. Calculate. Set \(10(b-22)=342.86\) and obtain the theoretical width \(b=56.29\ \text{mm}\).

4. Resolve. Select 57 mm upward; its net area is 350 square millimetres and its stress is \(48000/350=137.1\ \text{MPa}\).

Verification

Units: The net-area calculation uses millimetres squared, so the resulting force-per-area stress is in MPa.

Signs and direction: The applied load pulls across the hole section; positive tensile stress is compared to the tensile allowable value.

Plausibility: At 56 mm width the net area is 340 square millimetres and stress exceeds 140 MPa, whereas 57 mm passes.

Final result: \(\boxed{b_{min}=56.29\ \text{mm};\ \text{select }b=57\ \text{mm}}\)

Engineering interpretation

This net-section screen is not a complete hanger design. Bolt bearing, tear-out, edge distance, block shear, hole tolerance, corrosion, fatigue, and code-prescribed load combinations can control.

Chapter recap

Choose the model from the physical response and assumptions.
Keep force distinct from stress and displacement distinct from strain.
Before moving on, check units, signs, applicability, and competing failure modes.