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Mechanics of Materials

7% of exam

Shear and moment diagrams, axial, torsion, bending, shear, and thermal stresses and deformations, and combined and principal stresses with Mohr's circle.

3 concepts

A. Shear and moment diagrams

Shear and Moment Diagrams

Sign conventions, the load–shear–moment calculus, and a clean procedure for drawing V and M diagrams and locating the maximum bending moment.

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B. Stresses and strains

Axial, Torsion, Bending, and Shear Stress with Deformations

The four load cases — axial, torsion, bending, transverse shear — their stress formulas, Hooke's law, and the deformations (elongation, twist, thermal) each produces.

This is the load-bearing core of the Mechanics of Materials topic and a heavy scorer on the FE Civil exam: four canonical load cases — axial, torsion, bending, and transverse shear — each with a stress formula and a companion deformation. Master the pattern and a large fraction of structural and geotechnical follow-on problems become bookkeeping. Where candidates lose points is mixing up which

A

, which second moment (

I

vs.

J

), and which distance (

c

,

r

,

Q

) belongs to which case. Everything below follows the notation and formulas in the FE Reference Handbook's Mechanics of Materials chapter, so on exam day your scratch work will match the page you are allowed to open.

Stress, strain, and Hooke's law

Normal (axial) stress is force spread over the area it acts on,

\sigma = P/A

, with tension positive. Normal strain is the fractional length change,

\varepsilon = \delta/L

. In the linear-elastic range the two are tied by Young's modulus through Hooke's law,

\sigma = E\varepsilon

. Shear behaves the same way with its own modulus:

\tau = G\gamma

, where

\gamma

is the angle change. The three elastic constants are not independent — Poisson's ratio

\nu

(minus the ratio of lateral to longitudinal strain) links them by

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

, so steel's

E=29{,}000\,\text{ksi}

and

\nu=0.30

give

G\approx 11{,}200\,\text{ksi}

.

\sigma = E\varepsilon, \qquad \tau = G\gamma, \qquad G = \dfrac{E}{2(1+\nu)}

Axial loading and elongation

A member carrying load

P

along its axis develops uniform stress

\sigma = P/A

and stretches by

\delta = PL/(AE)

— a direct integration of

\varepsilon = \sigma/E

over the length. The group

AE

is the axial stiffness: a stiff member (large area or modulus) moves little. When the section, load, or material changes along the length, sum the pieces,

\delta = \sum P_iL_i/(A_iE_i)

. This is the formula behind truss-joint displacements and bolt/anchor stretch.

\sigma = \dfrac{P}{A}, \qquad \delta = \dfrac{PL}{AE}

Thermal deformation

A temperature change expands or contracts a member by

\delta_t = \alpha L\,\Delta T

, independent of stress, where

\alpha

is the coefficient of thermal expansion (steel

\approx 6.5\times10^{-6}/^\circ\text{F}

). The trap that wins exam questions: if the member is free to move, it sees deformation but NO stress; if it is restrained between rigid supports, it develops a thermal stress because the would-be expansion is forced back. Setting the restrained case equal to an equal-and-opposite mechanical strain gives

\sigma = E\alpha\,\Delta T

for full restraint.

\delta_t = \alpha L\,\Delta T, \qquad \sigma_{\text{restrained}} = E\,\alpha\,\Delta T

Torsion of circular shafts

For a solid or thick-walled circular shaft, torque produces a shear stress that grows linearly from zero at the center to a maximum at the surface,

\tau = Tr/J

, with the polar moment of inertia

J = \pi d^4/32

for a solid round bar (

J = J_{outer}-J_{inner}

for a tube). The shaft twists through an angle

\phi = TL/(GJ)

— note the parallel to axial

\delta=PL/(AE)

, with

GJ

as the torsional stiffness. Use the outer radius for maximum stress and the actual radius for stress at an interior fiber.

\tau = \dfrac{Tr}{J}, \qquad \phi = \dfrac{TL}{GJ}, \qquad J_{solid} = \dfrac{\pi d^4}{32}

Bending stress

Pure bending produces normal stress that varies linearly with distance

y

from the neutral axis:

\sigma = -My/I

, where

I

is the second moment of area of the cross section about its neutral (centroidal) axis. The extreme-fiber stress, the one you check against allowables, is

\sigma_{max} = Mc/I

with

c

the distance to the farthest fiber. Defining the elastic section modulus

S = I/c

collapses this to

\sigma_{max} = M/S

— the form structural tables are built around. Top fibers go into compression and bottom into tension for a positive (sagging) moment.

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

Transverse (beam) shear

The same shear force that lives on your shear diagram produces a horizontal/vertical shear stress through the depth of the beam,

\tau = VQ/(Ib)

. Here

Q = A'\bar{y}'

is the first moment about the neutral axis of the area beyond the layer where you want the stress,

b

is the section width there, and

I

is the full-section second moment. The stress is zero at the top and bottom fibers and maximum at the neutral axis. For a rectangle this collapses to a handy shortcut,

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

; for a solid circular section,

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

.

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

Reading the pattern

Notice the symmetry across all four cases. Each stress is (a generalized load) times (a distance) over (a section property):

\sigma=P/A

,

\tau=Tr/J

,

\sigma=Mc/I

,

\tau=VQ/(Ib)

. Each axial-type deformation is (load

\times

length) over (a stiffness):

\delta=PL/(AE)

for stretch,

\phi=TL/(GJ)

for twist. Keeping that template in your head means you never have to ask whether torsion uses

I

or

J

— it uses

J

, the polar one — or whether bending uses

c

or

r

. Match the section property to the load type and the rest follows.

Exam strategy

First identify the load case (axial, torsion, bending, or transverse shear), then pull the matching pair of formulas. Watch three things: the area or second moment must belong to the actual cross section (use

J=\pi d^4/32

,

I=\pi d^4/64

,

I=bh^3/12

from memory); the distance is

c

to the extreme fiber for

\sigma_{max}

, the radius for

\tau

, and

Q

for shear; and units must be consistent (kip and in give ksi). For deformation problems,

AE

and

GJ

are the stiffnesses — large means small movement. If a member is restrained against thermal growth, expect a stress

E\alpha\,\Delta T

; if free, expect movement and zero stress.

Key equations

Axial normal stress

\sigma = \dfrac{P}{A}

Uniform stress on a cross section from axial load $P$ over area $A$ ; tension positive. Units: kip/in² = ksi.

Normal strain

\varepsilon = \dfrac{\delta}{L}

Fractional change in length; dimensionless (in/in).

Hooke's law (axial and shear)

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

$E$ = modulus of elasticity, $G$ = shear modulus; valid in the linear-elastic range only.

Elastic-constant relation

G = \dfrac{E}{2(1+\nu)}

Links $E$ , $G$

Axial deformation

\delta = \dfrac{PL}{AE}

Elongation of a prismatic member; $AE$ is the axial stiffness. Sum segments for stepped members.

Thermal deformation

\delta_t = \alpha L\,(T - T_o)

Free-expansion length change; $\alpha$

Fully restrained thermal stress

\sigma = E\,\alpha\,\Delta T

Stress when expansion is completely prevented between rigid supports; compressive for heating.

Torsional shear stress

\tau = \dfrac{Tr}{J}

Linear from center; max at surface ( $r$ = outer radius). $J$ = polar moment of inertia.

Polar moment of inertia (solid round)

J = \dfrac{\pi d^4}{32} = \dfrac{\pi r^4}{2}

Angle of twist

\phi = \dfrac{TL}{GJ}

Total twist in radians; $GJ$ is the torsional stiffness. Mirrors $\delta=PL/(AE)$

Bending (flexure) stress

\sigma_x = -\dfrac{My}{I}, \quad \sigma_{max} = \dfrac{Mc}{I} = \dfrac{M}{S}

Transverse shear stress

\tau = \dfrac{VQ}{Ib}

$Q=A'\bar{y}'$

Rectangular shear shortcut

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

Worked examples

Axial stress and elongation of a steel rod

Problem. A solid steel rod, diameter $d = 1.0\,\text{in}$ and length $L = 6\,\text{ft}$ , carries an axial tension $P = 15\,\text{kip}$ . Take $E = 29{,}000\,\text{ksi}$ . Find the axial stress and the elongation.

Solution. Area:

A = \dfrac{\pi}{4}d^2 = \dfrac{\pi}{4}(1.0)^2 = 0.785\,\text{in}^2

\delta = \dfrac{PL}{AE} = \dfrac{15(72)}{0.785(29{,}000)} = 0.0474\ \text{in}

Torsion of a solid shaft

Problem. A solid circular steel shaft of diameter $d = 2.0\,\text{in}$ and length $L = 4\,\text{ft}$ transmits a torque $T = 30\,\text{kip\text{-}in}$ . Take $G = E/[2(1+\nu)] = 11{,}200\,\text{ksi}$

Bending and transverse shear in a rectangular beam

Problem. A rectangular timber beam, width $b = 4\,\text{in}$ and depth $h = 8\,\text{in}$ , carries, at one section, a bending moment $M = 200\,\text{kip\text{-}in}$ and a shear

Common pitfalls

•Using $I$ where torsion needs $J$ . Bending uses the centroidal second moment $I=\pi d^4/64$ ; torsion uses the polar $J=\pi d^4/32 = 2I$ .
•Plugging diameter into $\tau = Tr/J$ instead of radius. The distance in $Tr/J$ is the radius ( $r=d/2$ ).
•Forgetting that an unrestrained member sees thermal deformation but ZERO stress; a fully restrained member sees stress $E\alpha\Delta T$ but (ideally) zero net movement.
•Computing $Q$ over the whole section. $Q=A'\bar y'$ uses only the area beyond (above or below) the layer where you want the shear stress.
•Mixing kip-ft with inches in $\sigma=Mc/I$ . Convert the moment to kip-in so the result comes out in ksi.
•Treating bending shear stress as maximum at the extreme fiber. Transverse shear is zero at top/bottom and peaks at the neutral axis — the opposite of bending normal stress.
•Assuming $G$ is a free input. It is fixed by $G=E/[2(1+\nu)]$ (steel $\approx 11{,}150\,\text{ksi}$ ); do not pick an unrelated value. The FE Handbook material table lists a slightly different $G_{steel}\approx 11{,}500\,\text{ksi}$

References

NCEES FE Reference Handbook — Mechanics of Materials
NCEES FE Reference Handbook — Statics — Centroids and moments of inertia of areas
Hibbeler, Mechanics of Materials — Axial, torsion, flexure, and transverse-shear derivations, background only

D. Combined stresses, principal stresses, and Mohr's circle

Combined and Principal Stresses with Mohr's Circle

Plane-stress transformation, principal stresses and maximum shear, the Mohr's-circle construction, combined axial-bending-torsion states, and thin-walled pressure vessels.

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, and Poisson's ratio

\nu = -\varepsilon_{lat}/\varepsilon_{long}

. Steel:

\nu\approx 0.30

.

= coefficient of thermal expansion. Stress-free if unrestrained.

Solid circular shaft; for a tube subtract the inner from the outer value.

.

$I$ = second moment about the neutral axis, $c$ = distance to extreme fiber, $S=I/c$ = section modulus.

= first moment of the area beyond the layer;

b

= width there. Max at neutral axis.

Peak transverse shear at the neutral axis for common solid sections.

. Stress:

\sigma = P/A = 15/0.785 = 19.1\,\text{ksi}

(well below steel's yield

\approx 36\,\text{ksi}

, so elastic). Elongation (convert

L

to inches,

6\,\text{ft}=72\,\text{in}

):

\delta = \dfrac{PL}{AE} = \dfrac{15(72)}{0.785(29{,}000)} = \dfrac{1080}{22{,}777} = 0.0474\,\text{in}

. Sanity:

\delta = \varepsilon L = (\sigma/E)L = (19.1/29{,}000)(72) = 0.0474\,\text{in}

— consistent. Answer:

\sigma = 19.1\,\text{ksi}

,

\delta = 0.0474\,\text{in}

.

(steel,

\nu=0.30

). Find the maximum shear stress and the angle of twist.

Solution. Polar moment:

J = \dfrac{\pi d^4}{32} = \dfrac{\pi(2)^4}{32} = 1.571\,\text{in}^4

. Outer radius

r = 1.0\,\text{in}

. Max shear stress:

\tau = \dfrac{Tr}{J} = \dfrac{30(1.0)}{1.571} = 19.1\,\text{ksi}

. Angle of twist (

L = 48\,\text{in}

):

\phi = \dfrac{TL}{GJ} = \dfrac{30(48)}{11{,}200(1.571)} = \dfrac{1440}{17{,}593} = 0.0819\,\text{rad} = 4.69^\circ

. Sanity: shear stress at mid-radius would be half of 19.1, consistent with the linear

\tau\propto r

distribution. Note on

G

: the formula

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

gives

\approx 11{,}150\,\text{ksi}

(used here, rounded to

11{,}200

); the FE Handbook material table instead tabulates

G_{steel}\approx 11{,}500\,\text{ksi}

, which would give

\phi=0.0797\,\text{rad}=4.57^\circ

— about 3% lower. Either is acceptable; just state your source. Answer:

\tau_{max} = 19.1\,\text{ksi}

,

\phi = 0.0819\,\text{rad} = 4.69^\circ

.

\phi = \dfrac{TL}{GJ} = \dfrac{30(48)}{11{,}200(1.571)} = 0.0819\ \text{rad}

V = 12\,\text{kip}

. Find the maximum bending stress and the maximum transverse shear stress.

Solution. Second moment:

I = \dfrac{bh^3}{12} = \dfrac{4(8)^3}{12} = 170.7\,\text{in}^4

; extreme fiber

c = h/2 = 4\,\text{in}

. Bending stress:

\sigma_{max} = \dfrac{Mc}{I} = \dfrac{200(4)}{170.7} = 4.69\,\text{ksi}

(tension on bottom, compression on top). Maximum transverse shear at the neutral axis:

Q = A'\bar{y}' = (b\cdot \tfrac{h}{2})(\tfrac{h}{4}) = (4\cdot 4)(2) = 32\,\text{in}^3

, so

\tau = \dfrac{VQ}{Ib} = \dfrac{12(32)}{170.7(4)} = 0.563\,\text{ksi}

. Sanity: the rectangular shortcut gives

\tau_{max} = \dfrac{3V}{2A} = \dfrac{3(12)}{2(32)} = 0.563\,\text{ksi}

— a match. Answer:

\sigma_{max} = 4.69\,\text{ksi}

,

\tau_{max} = 0.563\,\text{ksi}

.

\tau_{max} = \dfrac{3V}{2A} = \dfrac{3(12)}{2(32)} = 0.563\ \text{ksi}

, so reconcile a ~3% answer difference by noting which source you used.