Strength of Materials

A. Stress types

Axial, Bending, Torsion, and Shear Stress

The four elementary stress formulas — P/A, Mc/I, Tr/J, VQ/Ib — plus Hooke's law and Poisson's ratio that every Strength-of-Materials question is built on.

Almost every Strength-of-Materials point on the FE Other Disciplines exam comes back to four formulas and the discipline of keeping them straight: axial

\sigma=P/A

, bending

\sigma=Mc/I

, torsion

\tau=Tr/J

, and transverse shear

\tau=VQ/Ib

. Each one says the same thing — internal force intensity equals load over a geometric property — but the geometric property changes with how the load is carried. Master what

A

,

I

,

J

, and

Q

mean and you can write down any of these from memory under exam pressure. The NCEES FE Reference Handbook collects them in the Mechanics of Materials chapter; this concept builds the fluency to deploy them without flipping pages.

Normal stress, shear stress, and strain

Stress is force intensity on an internal surface. A force perpendicular to the cut produces normal stress

\sigma

(tension positive, compression negative); a force tangent to the cut produces shear stress

\tau

. Strain is the geometric response: normal strain

\varepsilon=\delta/L

is fractional change in length, and shear strain

\gamma

is the change in a right angle, in radians. These pair up —

\sigma

with

\varepsilon

,

\tau

with

\gamma

— and the whole subject is the bookkeeping that links a load to a stress to a strain to a deformation.

\sigma=\frac{F}{A},\qquad \tau=\frac{V}{A},\qquad \varepsilon=\frac{\delta}{L},\qquad \gamma=\frac{\tau}{G}

Hooke's law and Poisson's ratio

Within the elastic (linear) region, stress is proportional to strain:

\sigma=E\varepsilon

for normal action and

\tau=G\gamma

for shear, where

E

is the modulus of elasticity (Young's modulus) and

G

the shear modulus. A bar stretched axially also contracts laterally; the ratio of lateral to longitudinal strain is Poisson's ratio

\nu

(about

0.30

for steel). The three elastic constants are not independent — they are tied by

G=E/[2(1+\nu)]

— so the handbook table only needs to list two of them.

\sigma=E\varepsilon,\qquad \tau=G\gamma,\qquad \nu=-\frac{\varepsilon_{lat}}{\varepsilon_{long}},\qquad G=\frac{E}{2(1+\nu)}

Axial stress: σ = P/A

When a load

P

passes through the centroid of a cross section, it spreads uniformly and the normal stress is simply

\sigma=P/A

. The matching axial strain is

\varepsilon=\sigma/E

, which is the seed of the elongation formula

\delta=PL/(AE)

developed in the deformations concept. Tension is positive, compression negative; the area

A

is the full cross-sectional area resisting the line of force.

\sigma=\frac{P}{A},\qquad \varepsilon=\frac{\sigma}{E}=\frac{P}{AE}

Bending stress: σ = Mc/I

A bending moment

M

produces normal stress that varies linearly across the depth — zero at the neutral (centroidal) axis, maximum at the extreme fibers. The handbook writes the fiber stress as

\sigma_x=-My/I

(the minus sign makes top fibers compressive for positive sagging

M

); for the magnitude at the outer fiber use

\sigma=Mc/I

, where

c

is the distance to the farthest fiber and

I

the centroidal moment of inertia. The grouping

S=I/c

is the elastic section modulus, so

\sigma_{max}=M/S

.

\sigma_x=-\frac{My}{I},\qquad \sigma_{max}=\frac{Mc}{I}=\frac{M}{S},\qquad S=\frac{I}{c}

Torsion: τ = Tr/J

A torque

T

on a circular shaft creates shear stress that grows linearly from zero at the axis to a maximum at the surface:

\tau=Tr/J

, where

J

is the polar moment of inertia and

r

the radial distance. For a solid circular shaft

J=\pi d^4/32

; for a hollow shaft subtract the bore,

J=\pi(d_o^4-d_i^4)/32

. This form is valid for solid or thick-walled circular sections only — non-circular bars warp and need different (handbook-supplied) treatment.

\tau=\frac{Tr}{J},\qquad J_{solid}=\frac{\pi d^4}{32},\qquad J_{hollow}=\frac{\pi\left(d_o^4-d_i^4\right)}{32}

Transverse shear: τ = VQ/Ib

The shear force

V

in a beam produces a horizontal/vertical shear stress that is zero at the top and bottom fibers and maximum at the neutral axis — the opposite distribution from bending stress. The general formula is

\tau=VQ/(Ib)

, where

Q=A'\bar{y}'

is the first moment about the neutral axis of the area beyond the level of interest, and

b

is the width there. For a solid rectangle this peaks at

\tau_{max}=1.5\,V/A

; for a solid circle,

\tau_{max}=4V/(3A)

.

\tau=\frac{VQ}{Ib},\qquad Q=A'\bar{y}',\qquad \tau_{max,rect}=\frac{3V}{2A}

Exam strategy

First identify how the load is carried — pull/push (axial), couple about a transverse axis (bending), couple about the long axis (torsion), or transverse cut (shear) — then the formula picks itself. Compute the section property in consistent units before substituting: a

\text{mm}

section gives

I

in

\text{mm}^4

, so pair it with

N

and

\text{mm}

to land in

\text{MPa}

(

1\,\text{MPa}=1\,\text{N/mm}^2

). Watch the

32

vs

64

trap:

J=\pi d^4/32

for torsion but

I=\pi d^4/64

for bending of a circular section. When several actions act at once, each gives its own stress and you superpose them — the subject of the next concept.

Key equations

Axial (normal) stress

\sigma=\frac{P}{A}

Uniform normal stress from a centroidal axial load. $P$ in N (or lbf), $A$ in $\text{mm}^2$ (or $\text{in}^2$ ); tension positive.

Engineering strain

\varepsilon=\frac{\delta}{L}

Fractional change in length (dimensionless). $\delta$ = elongation, $L$ = original gauge length.

Hooke's law (1D)

\sigma=E\varepsilon,\qquad \tau=G\gamma

Linear-elastic relation. $E$ = modulus of elasticity, $G$ = shear modulus (both in $\text{GPa}$ or $\text{Mpsi}$ ); $\gamma$ in radians.

Poisson's ratio and modulus link

\nu=-\frac{\varepsilon_{lat}}{\varepsilon_{long}},\qquad G=\frac{E}{2(1+\nu)}

$\nu\approx0.30$ for steel; ties the three elastic constants together so only two are independent.

Bending stress (flexure formula)

\sigma=\frac{Mc}{I}=\frac{M}{S}

Outer-fiber bending stress. $M$ = moment, $c$ = distance to extreme fiber, $I$ = centroidal moment of inertia, $S=I/c$ = section modulus.

Fiber bending stress with sign

\sigma_x=-\frac{My}{I}

Linear variation across depth; $y$ measured from the neutral axis. The minus sign puts top fibers in compression for positive (sagging) $M$ .

Rectangular moment of inertia

I=\frac{bh^3}{12}

Centroidal $I$ of a rectangle, base $b$ , depth $h$ (bending about the horizontal centroidal axis).

Torsional shear stress

\tau=\frac{Tr}{J}

Circular shafts only. $T$ = torque, $r$ = radius to point ( $r$ at surface for max), $J$ = polar moment of inertia.

Polar moment of inertia (circular)

J=\frac{\pi d^4}{32}\;(\text{solid}),\qquad J=\frac{\pi\left(d_o^4-d_i^4\right)}{32}\;(\text{hollow})

Note the 32 (torsion) versus 64 for the bending $I=\pi d^4/64$ of the same circle.

Transverse shear stress

\tau=\frac{VQ}{Ib}

$V$ = shear force, $Q=A'\bar{y}'$ = first moment of area beyond the level, $I$ = full-section inertia, $b$ = width at that level.

Max transverse shear (common sections)

\tau_{max}=\frac{3V}{2A}\;(\text{rect}),\qquad \tau_{max}=\frac{4V}{3A}\;(\text{circle})

Peak shear at the neutral axis; handy shortcuts that avoid computing $Q$ for solid sections.

Shear flow

q=\frac{VQ}{I}

Force per unit length along a joint (e.g., bolted/glued built-up beams); $\tau=q/b$ .

Worked examples

Axial stress and Poisson contraction

Problem. A solid steel rod of diameter $25\,\text{mm}$ carries a tensile load $P=50\,\text{kN}$ . With $E=200\,\text{GPa}$ and $\nu=0.30$ , find the axial stress, the axial strain, and the change in diameter.

Solution. Area:

A=\tfrac{\pi}{4}(25)^2=490.9\,\text{mm}^2

. Stress:

\sigma=P/A=50{,}000/490.9=101.9\,\text{MPa}

(tension). Axial strain:

\varepsilon=\sigma/E=101.9/200{,}000=5.09\times10^{-4}

. Lateral strain:

\varepsilon_{lat}=-\nu\varepsilon=-0.30(5.09\times10^{-4})=-1.53\times10^{-4}

, so

\Delta d=\varepsilon_{lat}\,d=-1.53\times10^{-4}(25)=-3.82\times10^{-3}\,\text{mm}=-3.82\,\mu\text{m}

. Sanity:

\text{N}/\text{mm}^2=\text{MPa}

checks; diameter shrinks under tension, as Poisson requires. Answer:

\sigma=102\,\text{MPa}

,

\varepsilon=5.09\times10^{-4}

,

\Delta d=-3.82\,\mu\text{m}

.

\sigma=\frac{50{,}000}{490.9}=102\ \text{MPa},\qquad \Delta d=-\nu\frac{\sigma}{E}\,d=-3.82\ \mu\text{m}

Bending stress in a rectangular beam

Problem. A beam with a $50\,\text{mm}\times150\,\text{mm}$ rectangular cross section (depth $150\,\text{mm}$ ) carries a bending moment $M=12\,\text{kN·m}$ . Find the maximum bending stress.

Solution. Inertia:

I=bh^3/12=50(150)^3/12=1.406\times10^{7}\,\text{mm}^4

. Extreme fiber:

c=h/2=75\,\text{mm}

.

\sigma=Mc/I=(12\times10^{6}\,\text{N·mm})(75)/(1.406\times10^{7})=64.0\,\text{MPa}

. Check with the section modulus:

S=I/c=1.875\times10^{5}\,\text{mm}^3

,

\sigma=M/S=12\times10^{6}/1.875\times10^{5}=64.0\,\text{MPa}

. Top fiber compresses, bottom fiber tensions. Answer:

\sigma_{max}=64.0\,\text{MPa}

.

\sigma=\frac{Mc}{I}=\frac{12\times10^{6}\,(75)}{1.406\times10^{7}}=64.0\ \text{MPa}

Torsion and transverse shear together

Problem. (a) A solid circular shaft of diameter $40\,\text{mm}$ transmits a torque $T=600\,\text{N·m}$ . Find the maximum shear stress. (b) Separately, a solid rectangular beam $50\,\text{mm}\times150\,\text{mm}$ ( $A=7500\,\text{mm}^2$ ) carries a shear force $V=20\,\text{kN}$ . Find the maximum transverse shear stress.

Solution. (a)

J=\pi d^4/32=\pi(40)^4/32=2.513\times10^{5}\,\text{mm}^4

,

r=20\,\text{mm}

.

\tau=Tr/J=(600\times10^{3}\,\text{N·mm})(20)/(2.513\times10^{5})=47.7\,\text{MPa}

. (b) Rectangular peak:

\tau_{max}=\tfrac{3V}{2A}=\tfrac{3(20{,}000)}{2(7500)}=4.00\,\text{MPa}

(at the neutral axis). Sanity: both land in MPa with N–mm units; torsion peaks at the surface, transverse shear peaks at mid-depth. Answers:

\tau_{torsion}=47.7\,\text{MPa}

,

\tau_{shear}=4.00\,\text{MPa}

.

\tau=\frac{Tr}{J}=\frac{600{,}000\,(20)}{2.513\times10^{5}}=47.7\ \text{MPa}

Common pitfalls

•Confusing $J=\pi d^4/32$ (torsion) with $I=\pi d^4/64$ (bending) for a circular section — they differ by a factor of two. Torsion uses the polar inertia.
•Using the full area in the transverse-shear formula. $Q=A'\bar{y}'$ is the first moment of only the area beyond the point of interest, and $\tau$ peaks at the neutral axis, not at the surface.
•Putting $c$ (extreme-fiber distance) where $y$ (any fiber) belongs, or vice versa: $\sigma=Mc/I$ gives the maximum; $\sigma=My/I$ gives the stress at a chosen depth.
•Mixing unit systems: keep $M$ in $\text{N·mm}$ when $I$ is in $\text{mm}^4$ . A common error is leaving $M$ in $\text{kN·m}$ — that is $10^6$ too small.
•Forgetting Poisson contraction signs: tension gives lateral shrinkage ( $\varepsilon_{lat}<0$ ). Plug the negative sign so diameter changes come out the right way.
•Treating $\tau=Tr/J$ as valid for non-circular shafts. Square and rectangular bars warp; their torsion needs the handbook's special factors, not this formula.
•Assuming $E$ and $G$ are independent inputs. If only $E$ and $\nu$ are given, get $G=E/[2(1+\nu)]$ before any shear-strain calculation.

References

NCEES FE Reference Handbook — Mechanics of Materials
NCEES FE Reference Handbook — Materials Science/Structure of Matter — engineering vs. true stress, Hooke's law, modulus definitions
Hibbeler, Mechanics of Materials — flexure and transverse-shear derivations

B. Combined loading and superposition

Combined Loading and Superposition

Add axial, bending, and torsional stresses at a point by superposition, handle eccentric loads, and find hoop and longitudinal stress in thin-walled pressure vessels.