Stress is defined as

Fluid Mechanics · FE Reference Handbook section

Fluid Mechanics

21 formulas

10 exam-style examples

~60 min

All Fluid Mechanics lectures

Learning objectives

What you must be able to do before leaving this section.

This chapter section covers Stress is defined as within Fluid Mechanics. Read it the way you would read a textbook chapter: the theory first so the relations mean something, then every equation with its use and its trap, then 10 fully worked examples with the arithmetic shown line by line, and finally a self-check you should be able to answer without notes.

Explain, in your own words, what stress is defined as describes physically and when it applies.
State every one of the 21 relations the handbook lists here and name each symbol with its unit.
Select the correct relation from the wording of an exam stem within 20 seconds.
Carry a complete solution from givens to a "most nearly" answer with the correct unit.
Recognise the distractors generated by the unit trap: γ = 62.4 lb/ft³ or 9.81 kN/m³; convert psi to feet of head early.

Lecture

Why this section exists. Stress is defined as is the part of Fluid Mechanics that lets you connect a pipeline, jet or submerged surface to a number you can defend. Before any equation is useful you must be able to picture the physical situation it describes; the schematic below is that picture.

How the theory is built. The handbook prints results, not derivations. Each relation in this section comes from one governing principle applied to the idealised system: state the principle, impose the stated assumptions, and the printed equation follows. Knowing which assumption each relation rests on is what lets you reject a wrong answer choice in seconds.

How it is examined. Items from this page are written as continuity plus energy, with one head-loss or force term. Roughly two thirds are direct substitution, one third require one intermediate quantity from a neighbouring relation, and a small number are conceptual — testing whether you know the assumption, not the arithmetic.

The habit that earns the points. Unit discipline. γ = 62.4 lb/ft³ or 9.81 kN/m³; convert psi to feet of head early. Every relation below is dimensionally consistent only when that rule is honoured, and the distractor set is deliberately built from candidates who ignored it. Write the unit next to every number you substitute, every time.

How to study this page. Read the theory, then cover the formula cards and try to reproduce each relation from its description. Then work the examples with the solution hidden, revealing one line at a time. Finish with the self-check questions; if you cannot answer one, return to the matching formula card.

Crane lowering a steel plate girder onto bridge bearings while ironworkers guide it. — Photo 1. Where this shows up in practice: stress is defined as.
Capstone Studio instructional photograph

Fluid Mechanics — Stress is defined as: reference schematic for orienting the symbols used in this section.

Theory, developed

Read this before the equations — it is what makes them memorable.

The physical situation

Every item from this section describes a pipeline, jet or submerged surface. Sketch it before you compute — a labelled sketch with the givens on it converts a wordy stem into a solvable problem and exposes the quantity the examiner left out on purpose.

The governing principle

The 21 relations on this page are consequences of one principle applied to that idealised system. Identify which quantity is conserved, balanced, or defined, and the correct equation follows without memorisation.

Assumptions and limits of validity

Each printed relation carries silent assumptions — linearity, steady state, uniformity, small deformation, or standard conditions, depending on the subject. Conceptual exam items are written by violating exactly one of these, so read the sentence above the equation as carefully as the equation itself.

Solution procedure you should automate

1) Read the last sentence of the stem to identify the requested quantity. 2) Locate the relation on this page whose left-hand side is that quantity. 3) Tabulate the givens with units and mark the missing symbol. 4) If a symbol is missing, find the one relation that produces it. 5) Rearrange symbolically, substitute once, evaluate, and round only at the end.

Notation used in this section

x ^1 h	Quantity produced by "x ^1 h = limit DF/DA" — read its definition and unit from the handbook line directly above the equation.
x ]1g	Quantity produced by "x ]1g = surface stress vector at Point 1" — read its definition and unit from the handbook line directly above the equation.
∆F	Quantity produced by "∆F = force acting on infinitesimal area ∆A" — read its definition and unit from the handbook line directly above the equation.
∆A	Quantity produced by "∆A = infinitesimal area at Point 1" — read its definition and unit from the handbook line directly above the equation.
τn	Quantity produced by "τn = – P" — read its definition and unit from the handbook line directly above the equation.
τt	Quantity produced by "τt = µ(dv/dy) (one-dimensional; i.e., y)" — read its definition and unit from the handbook line directly above the equation.
τn and τt	Quantity produced by "τn and τt = normal and tangential stress components at Point 1, respectively" — read its definition and unit from the handbook line directly above the equation.
P	Quantity produced by "P = pressure at Point 1" — read its definition and unit from the handbook line directly above the equation.
µ	Quantity produced by "µ = absolute dynamic viscosity of the fluid" — read its definition and unit from the handbook line directly above the equation.
dv	Quantity produced by "dv = differential velocity" — read its definition and unit from the handbook line directly above the equation.
dy	Quantity produced by "dy = differential distance, normal to boundary" — read its definition and unit from the handbook line directly above the equation.
v	Quantity produced by "v = velocity at boundary condition" — read its definition and unit from the handbook line directly above the equation.

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

DA " 0
where
where
N•s/m2 [lbm/(ft-sec)]
For a thin Newtonian fluid film and a linear velocity profile,
where
For a power law (non-Newtonian) fluid
where
n < 1 ≡ pseudo plastic
n > 1 ≡ dilatant

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1

Shear stress and drag force in a thin oil film — Stress is defined as

A flat plate of contact area 0.80 m² slides at 3.20 m/s over a stationary surface separated by an oil film 4.50 mm thick. The oil has dynamic viscosity 0.0288 Pa·s and the velocity profile across the film is linear. Compute the shear stress on the plate and the force needed to keep it moving.

Given

μ = 0.0288 Pa·s
U = 3.20 m/s
h = 4.50 mm = 4.50e-3 m
A = 0.80 m²

Find

Shear stress τ and the drag force F

Start with the thinking

Fluid stress is defined as force per unit area acting tangentially; for a Newtonian fluid it is proportional to the velocity gradient.
A linear profile makes du/dy = U/h, so no calculus is needed.

Step-by-step solution

Formula
$τ = μ \frac{d u}{d y}$
Velocity gradient
$d u / d y = U / h = 3.20/4.50 e - 3 = 711.1 s^{- 1}$
Substituting
$τ = 0.0288 (711.1) = 20.5 Pa$
Formula
$F = τ A$
Substituting
$F = 20.5 (0.80) = 16.4 N$

Answer: τ = 20.5 Pa and F = 16.4 N

Why the other options are there

0.000415 Pa (multiplied by film thickness)
25.6 N (divided by area)

Reference: FE Reference Handbook — Fluid Mechanics → Stress is defined as

Example 2

Shear stress and drag force in a thin oil film — Stress is defined as (2)

A flat plate of contact area 0.90 m² slides at 3.60 m/s over a stationary surface separated by an oil film 2.50 mm thick. The oil has dynamic viscosity 0.0298 Pa·s and the velocity profile across the film is linear. Compute the shear stress on the plate and the force needed to keep it moving.

Given

μ = 0.0298 Pa·s
U = 3.60 m/s
h = 2.50 mm = 2.50e-3 m
A = 0.90 m²

Find

Shear stress τ and the drag force F

Start with the thinking

Fluid stress is defined as force per unit area acting tangentially; for a Newtonian fluid it is proportional to the velocity gradient.
A linear profile makes du/dy = U/h, so no calculus is needed.

Step-by-step solution

Formula
$τ = μ \frac{d u}{d y}$
Velocity gradient
$d u / d y = U / h = 3.60/2.50 e - 3 = 1, 440 s^{- 1}$
Substituting
$τ = 0.0298 (1, 440) = 42.9 Pa$
Formula
$F = τ A$
Substituting
$F = 42.9 (0.90) = 38.6 N$

Answer: τ = 42.9 Pa and F = 38.6 N

Why the other options are there

0.000268 Pa (multiplied by film thickness)
47.7 N (divided by area)

Reference: FE Reference Handbook — Fluid Mechanics → Stress is defined as

Example 3

Shear stress and drag force in a thin oil film — Stress is defined as (3)

A flat plate of contact area 1.15 m² slides at 3.50 m/s over a stationary surface separated by an oil film 1.50 mm thick. The oil has dynamic viscosity 0.0846 Pa·s and the velocity profile across the film is linear. Compute the shear stress on the plate and the force needed to keep it moving.

Given

μ = 0.0846 Pa·s
U = 3.50 m/s
h = 1.50 mm = 1.50e-3 m
A = 1.15 m²

Find

Shear stress τ and the drag force F

Start with the thinking

Fluid stress is defined as force per unit area acting tangentially; for a Newtonian fluid it is proportional to the velocity gradient.
A linear profile makes du/dy = U/h, so no calculus is needed.

Step-by-step solution

Formula
$τ = μ \frac{d u}{d y}$
Velocity gradient
$d u / d y = U / h = 3.50/1.50 e - 3 = 2, 333 s^{- 1}$
Substituting
$τ = 0.0846 (2, 333) = 197.4 Pa$
Formula
$F = τ A$
Substituting
$F = 197.4 (1.15) = 227.0 N$

Answer: τ = 197.4 Pa and F = 227.0 N

Why the other options are there

0.000444 Pa (multiplied by film thickness)
171.7 N (divided by area)

Reference: FE Reference Handbook — Fluid Mechanics → Stress is defined as

Example 4

Shear stress and drag force in a thin oil film — Stress is defined as (4)

A flat plate of contact area 0.55 m² slides at 5.60 m/s over a stationary surface separated by an oil film 4.00 mm thick. The oil has dynamic viscosity 0.0066 Pa·s and the velocity profile across the film is linear. Compute the shear stress on the plate and the force needed to keep it moving.

Given

μ = 0.0066 Pa·s
U = 5.60 m/s
h = 4.00 mm = 4.00e-3 m
A = 0.55 m²

Find

Shear stress τ and the drag force F

Start with the thinking

Fluid stress is defined as force per unit area acting tangentially; for a Newtonian fluid it is proportional to the velocity gradient.
A linear profile makes du/dy = U/h, so no calculus is needed.

Step-by-step solution

Formula
$τ = μ \frac{d u}{d y}$
Velocity gradient
$d u / d y = U / h = 5.60/4.00 e - 3 = 1, 400 s^{- 1}$
Substituting
$τ = 0.0066 (1, 400) = 9.2 Pa$
Formula
$F = τ A$
Substituting
$F = 9.2 (0.55) = 5.1 N$

Answer: τ = 9.2 Pa and F = 5.1 N

Why the other options are there

0.000148 Pa (multiplied by film thickness)
16.8 N (divided by area)

Reference: FE Reference Handbook — Fluid Mechanics → Stress is defined as

Example 5

Shear stress and drag force in a thin oil film — Stress is defined as (5)

A flat plate of contact area 0.10 m² slides at 0.80 m/s over a stationary surface separated by an oil film 1.50 mm thick. The oil has dynamic viscosity 0.0868 Pa·s and the velocity profile across the film is linear. Compute the shear stress on the plate and the force needed to keep it moving.

Given

μ = 0.0868 Pa·s
U = 0.80 m/s
h = 1.50 mm = 1.50e-3 m
A = 0.10 m²

Find

Shear stress τ and the drag force F

Start with the thinking

Fluid stress is defined as force per unit area acting tangentially; for a Newtonian fluid it is proportional to the velocity gradient.
A linear profile makes du/dy = U/h, so no calculus is needed.

Step-by-step solution

Formula
$τ = μ \frac{d u}{d y}$
Velocity gradient
$d u / d y = U / h = 0.80/1.50 e - 3 = 533.3 s^{- 1}$
Substituting
$τ = 0.0868 (533.3) = 46.3 Pa$
Formula
$F = τ A$
Substituting
$F = 46.3 (0.10) = 4.6 N$

Answer: τ = 46.3 Pa and F = 4.6 N

Why the other options are there

0.000104 Pa (multiplied by film thickness)
462.9 N (divided by area)

Reference: FE Reference Handbook — Fluid Mechanics → Stress is defined as

Example 6

Shear stress and drag force in a thin oil film — Stress is defined as (6)

A flat plate of contact area 0.80 m² slides at 5.70 m/s over a stationary surface separated by an oil film 4.00 mm thick. The oil has dynamic viscosity 0.0620 Pa·s and the velocity profile across the film is linear. Compute the shear stress on the plate and the force needed to keep it moving.

Given

μ = 0.0620 Pa·s
U = 5.70 m/s
h = 4.00 mm = 4.00e-3 m
A = 0.80 m²

Find

Shear stress τ and the drag force F

Start with the thinking

Fluid stress is defined as force per unit area acting tangentially; for a Newtonian fluid it is proportional to the velocity gradient.
A linear profile makes du/dy = U/h, so no calculus is needed.

Step-by-step solution

Formula
$τ = μ \frac{d u}{d y}$
Velocity gradient
$d u / d y = U / h = 5.70/4.00 e - 3 = 1, 425 s^{- 1}$
Substituting
$τ = 0.0620 (1, 425) = 88.4 Pa$
Formula
$F = τ A$
Substituting
$F = 88.4 (0.80) = 70.7 N$

Answer: τ = 88.4 Pa and F = 70.7 N

Why the other options are there

0.001414 Pa (multiplied by film thickness)
110.4 N (divided by area)

Reference: FE Reference Handbook — Fluid Mechanics → Stress is defined as

Example 7

Shear stress and drag force in a thin oil film — Stress is defined as (7)

A flat plate of contact area 1.15 m² slides at 6.00 m/s over a stationary surface separated by an oil film 5.50 mm thick. The oil has dynamic viscosity 0.0318 Pa·s and the velocity profile across the film is linear. Compute the shear stress on the plate and the force needed to keep it moving.

Given

μ = 0.0318 Pa·s
U = 6.00 m/s
h = 5.50 mm = 5.50e-3 m
A = 1.15 m²

Find

Shear stress τ and the drag force F

Start with the thinking

Fluid stress is defined as force per unit area acting tangentially; for a Newtonian fluid it is proportional to the velocity gradient.
A linear profile makes du/dy = U/h, so no calculus is needed.

Step-by-step solution

Formula
$τ = μ \frac{d u}{d y}$
Velocity gradient
$d u / d y = U / h = 6.00/5.50 e - 3 = 1, 091 s^{- 1}$
Substituting
$τ = 0.0318 (1, 091) = 34.7 Pa$
Formula
$F = τ A$
Substituting
$F = 34.7 (1.15) = 39.9 N$

Answer: τ = 34.7 Pa and F = 39.9 N

Why the other options are there

0.001049 Pa (multiplied by film thickness)
30.2 N (divided by area)

Reference: FE Reference Handbook — Fluid Mechanics → Stress is defined as

Example 8

Shear stress and drag force in a thin oil film — Stress is defined as (8)

A flat plate of contact area 0.15 m² slides at 1.20 m/s over a stationary surface separated by an oil film 3.00 mm thick. The oil has dynamic viscosity 0.0724 Pa·s and the velocity profile across the film is linear. Compute the shear stress on the plate and the force needed to keep it moving.

Given

μ = 0.0724 Pa·s
U = 1.20 m/s
h = 3.00 mm = 3.00e-3 m
A = 0.15 m²

Find

Shear stress τ and the drag force F

Start with the thinking

Fluid stress is defined as force per unit area acting tangentially; for a Newtonian fluid it is proportional to the velocity gradient.
A linear profile makes du/dy = U/h, so no calculus is needed.

Step-by-step solution

Formula
$τ = μ \frac{d u}{d y}$
Velocity gradient
$d u / d y = U / h = 1.20/3.00 e - 3 = 400.0 s^{- 1}$
Substituting
$τ = 0.0724 (400.0) = 29.0 Pa$
Formula
$F = τ A$
Substituting
$F = 29.0 (0.15) = 4.3 N$

Answer: τ = 29.0 Pa and F = 4.3 N

Why the other options are there

0.000261 Pa (multiplied by film thickness)
193.1 N (divided by area)

Reference: FE Reference Handbook — Fluid Mechanics → Stress is defined as

Example 9

Shear stress and drag force in a thin oil film — Stress is defined as (9)

A flat plate of contact area 0.25 m² slides at 5.30 m/s over a stationary surface separated by an oil film 3.00 mm thick. The oil has dynamic viscosity 0.0468 Pa·s and the velocity profile across the film is linear. Compute the shear stress on the plate and the force needed to keep it moving.

Given

μ = 0.0468 Pa·s
U = 5.30 m/s
h = 3.00 mm = 3.00e-3 m
A = 0.25 m²

Find

Shear stress τ and the drag force F

Start with the thinking

Fluid stress is defined as force per unit area acting tangentially; for a Newtonian fluid it is proportional to the velocity gradient.
A linear profile makes du/dy = U/h, so no calculus is needed.

Step-by-step solution

Formula
$τ = μ \frac{d u}{d y}$
Velocity gradient
$d u / d y = U / h = 5.30/3.00 e - 3 = 1, 767 s^{- 1}$
Substituting
$τ = 0.0468 (1, 767) = 82.7 Pa$
Formula
$F = τ A$
Substituting
$F = 82.7 (0.25) = 20.7 N$

Answer: τ = 82.7 Pa and F = 20.7 N

Why the other options are there

0.000744 Pa (multiplied by film thickness)
330.7 N (divided by area)

Reference: FE Reference Handbook — Fluid Mechanics → Stress is defined as

Example 10

Shear stress and drag force in a thin oil film — Stress is defined as (10)

A flat plate of contact area 1.10 m² slides at 5.50 m/s over a stationary surface separated by an oil film 1.50 mm thick. The oil has dynamic viscosity 0.0652 Pa·s and the velocity profile across the film is linear. Compute the shear stress on the plate and the force needed to keep it moving.

Given

μ = 0.0652 Pa·s
U = 5.50 m/s
h = 1.50 mm = 1.50e-3 m
A = 1.10 m²

Find

Shear stress τ and the drag force F

Start with the thinking

Fluid stress is defined as force per unit area acting tangentially; for a Newtonian fluid it is proportional to the velocity gradient.
A linear profile makes du/dy = U/h, so no calculus is needed.

Step-by-step solution

Formula
$τ = μ \frac{d u}{d y}$
Velocity gradient
$d u / d y = U / h = 5.50/1.50 e - 3 = 3, 667 s^{- 1}$
Substituting
$τ = 0.0652 (3, 667) = 239.1 Pa$
Formula
$F = τ A$
Substituting
$F = 239.1 (1.10) = 263.0 N$

Answer: τ = 239.1 Pa and F = 263.0 N

Why the other options are there

0.000538 Pa (multiplied by film thickness)
217.3 N (divided by area)

Reference: FE Reference Handbook — Fluid Mechanics → Stress is defined as

Self-check

Answer these without notes before moving on.

Without looking, state the relation on this page whose left-hand side is the quantity most often requested, and name every symbol in it.
Which assumption, if violated, makes the main relation of this section invalid?
Given a pipeline, jet or submerged surface, what is the first quantity you would compute, and why that one first?
Which unit conversion in this subject most often produces a wrong answer choice, and what is its numerical factor?
Rework Example 1 above from the givens alone, without reading the solution lines.

Chapter summary

Stress is defined as contains 21 relations; you must be able to find this page in under 15 seconds.
Exam style: continuity plus energy, with one head-loss or force term.
Unit rule: γ = 62.4 lb/ft³ or 9.81 kN/m³; convert psi to feet of head early.
Work the 10 examples until the solution path, not the answer, is automatic.

Common traps in this section

γ = 62.4 lb/ft³ or 9.81 kN/m³; convert psi to feet of head early
Answering the intermediate quantity instead of the quantity requested.
Rounding intermediate values before the final step.
Using a relation from an adjacent handbook section that shares a symbol.
Skipping the sketch — most lost points on this page start with a misread geometry.