Mechanics of Materials

A. Shear and moment diagrams

Shear/Moment Diagrams, Beam Bending and Transverse Shear

Build shear and moment diagrams from the load-shear-moment relations, then size beams with the flexure formula σ = Mc/I and the transverse shear formula τ = VQ/Ib.

Beam bending is the centerpiece of the Mechanics of Materials section, and it always starts the same way: find where the bending moment is largest, because that is where the beam is most likely to fail. The shear and moment diagrams are the map that tells you, and the flexure formula

\sigma = Mc/I

turns that moment into a stress you can compare to an allowable. Candidates lose points by drawing diagrams that violate the load-shear-moment calculus, or by mixing up the centroidal

I

with

c

, or by forgetting that the same beam also carries a transverse shear stress that peaks at the neutral axis — exactly where the bending stress is zero. The FE Reference Handbook (Mechanics of Materials, "Beams" and "Stresses in Beams") gives the sign conventions and formulas; this concept makes them automatic.

The load-shear-moment relations

Distributed load

w(x)

, shear

V(x)

, and bending moment

M(x)

are linked by calculus: the slope of the shear diagram at any point equals minus the distributed load, and the slope of the moment diagram equals the shear. Integrating, the change in shear between two sections equals minus the area under the load curve, and the change in moment equals the area under the shear curve. These relations let you build either diagram by adding areas instead of writing equations for every segment — and they give the single most useful fact on the exam: the moment is a local maximum or minimum exactly where the shear crosses zero.

\dfrac{dV}{dx} = -w(x), \qquad \dfrac{dM}{dx} = V(x)

Building the diagrams

Start by finding the reactions from global equilibrium. Then sweep left to right: a downward point load steps the shear down by its magnitude, a uniform load ramps the shear linearly, and an applied couple steps the moment (not the shear). Between loads the diagrams follow the slope rules — constant load gives a sloped shear and a parabolic moment. Locate every point where

V = 0

; the moment there is a candidate for

M_{\max}

. The diagram closes back to zero at a free end, which is a built-in arithmetic check.

Bending sign convention

The handbook convention: a bending moment is positive when it bends the beam concave up (compression in the top fibers, tension in the bottom), and shear is positive when the right portion tends to slide down relative to the left. Getting this convention consistent matters because the flexure formula carries it: with

\sigma_x = -My/I

, a positive moment puts

y > 0

(above the neutral axis) into compression. For maximum-stress sizing you usually drop the sign and use magnitudes, but for combined-loading problems the sign decides whether bending adds to or subtracts from an axial stress.

The flexure formula

Bending stress varies linearly across the depth, zero at the neutral (centroidal) axis and maximum at the extreme fiber. The flexure formula gives the peak:

\sigma = Mc/I

, where

c

is the distance from the neutral axis to the outermost fiber and

I

is the centroidal second moment of area about the bending axis. The grouping

S = I/c

is the elastic section modulus, so

\sigma = M/S

— handy because tabulated steel shapes list

S

directly. For a rectangle,

I = bh^3/12

and

c = h/2

, giving

S = bh^2/6

.

\sigma_{\max} = \dfrac{Mc}{I} = \dfrac{M}{S}, \qquad S = \dfrac{I}{c}

Transverse shear stress

Bending is accompanied by a transverse shear stress that the flexure formula does not capture:

\tau = VQ/(Ib)

. Here

V

is the shear force at the section,

I

is the same centroidal moment of inertia,

b

is the width of the cross section at the level where you want

\tau

, and

Q = A'\bar{y}'

is the first moment of the area beyond that level about the neutral axis. Unlike bending stress,

\tau

is zero at the top and bottom fibers and maximum at the neutral axis. For a rectangular section the peak works out to a clean

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

; for a wide-flange beam the web carries almost all of it, so

\tau \approx V/A_{\text{web}}

.

\tau = \dfrac{VQ}{Ib}, \qquad \tau_{\max,\,\text{rect}} = \dfrac{3V}{2A}

Exam strategy

Always solve reactions first, then draw the shear diagram, then the moment diagram by accumulating shear areas — never guess

M_{\max}

from the load picture. The maximum moment lives where shear crosses zero; for a uniformly loaded simple span that is midspan with

M_{\max} = wL^2/8

. Plug

M_{\max}

into

\sigma = Mc/I

for bending strength, and check shear separately with

\tau = VQ/(Ib)

using

V_{\max}

(often at a support). Keep

I

and

c

about the same centroidal axis, and watch units: with

M

in N·m and

I

in m

^4

,

c

in m, the stress comes out in Pa. In USCS keep everything in inches and lbf so stress lands in psi.

Key equations

Load-shear relation

\dfrac{dV}{dx} = -w(x)

Slope of the shear diagram equals minus the distributed load (downward $w$ positive).

Shear-moment relation

\dfrac{dM}{dx} = V(x)

Slope of the moment diagram equals the shear; $M$ is extremum where $V=0$ .

Diagram areas

V_2 - V_1 = -\!\int w\,dx, \qquad M_2 - M_1 = \int V\,dx

Change in shear = minus area under load; change in moment = area under shear.

Flexure formula

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

Bending stress at fiber distance $y$ from the neutral axis; $I$ = centroidal moment of inertia.

Maximum bending stress

\sigma_{\max} = \dfrac{Mc}{I} = \dfrac{M}{S}

$c$ = distance to extreme fiber; $S = I/c$ = elastic section modulus.

Rectangle section properties

I = \dfrac{bh^3}{12}, \quad c = \dfrac{h}{2}, \quad S = \dfrac{bh^2}{6}

For a solid rectangle of width $b$ , depth $h$ , about the horizontal centroidal axis.

Transverse shear stress

\tau = \dfrac{VQ}{Ib}

$Q = A'\bar{y}'$ = first moment of area beyond the level; $b$ = width there. Max at the neutral axis.

Rectangle peak shear

\tau_{\max} = \dfrac{3V}{2A}

Maximum transverse shear in a rectangular section, $1.5\times$ the average $V/A$ .

Simple-span UDL maxima

V_{\max} = \dfrac{wL}{2}, \qquad M_{\max} = \dfrac{wL^2}{8}

Simply supported beam, full-span uniform load $w$ ; shear max at supports, moment max at midspan.

Worked examples

Maximum bending stress in a uniformly loaded beam

Problem. A simply supported timber beam spans $L = 6\,\text{m}$ and carries a uniform load $w = 4\,\text{kN/m}$ over the full span. The rectangular section is $b = 100\,\text{mm}$ wide by $h = 200\,\text{mm}$ deep. Find the maximum bending moment and the maximum bending stress.

Solution. Reactions: each support carries

R = wL/2 = 4(6)/2 = 12\,\text{kN}

. Shear is zero at midspan, so the maximum moment is

M_{\max} = wL^2/8 = 4(6)^2/8 = 18\,\text{kN·m}

. Section properties:

I = bh^3/12 = (0.100)(0.200)^3/12 = 6.667\times10^{-5}\,\text{m}^4

, and

c = h/2 = 0.100\,\text{m}

. Bending stress:

\sigma_{\max} = M_{\max}\,c/I = (18{,}000)(0.100)/(6.667\times10^{-5}) = 2.70\times10^{7}\,\text{Pa} = 27.0\,\text{MPa}

. Answer:

M_{\max} = 18.0\,\text{kN·m}

,

\sigma_{\max} = 27.0\,\text{MPa}

. Sanity check via section modulus:

S = bh^2/6 = (0.100)(0.200)^2/6 = 6.667\times10^{-4}\,\text{m}^3

, and

M/S = 18{,}000/6.667\times10^{-4} = 27.0\,\text{MPa}

— agrees.

\sigma_{\max} = \dfrac{M_{\max}c}{I} = \dfrac{(18{,}000)(0.100)}{6.667\times10^{-5}}

Maximum transverse shear stress

Problem. For the same $100\,\text{mm}\times200\,\text{mm}$ beam carrying $V_{\max} = 12\,\text{kN}$ at the support, find the maximum transverse shear stress two ways: with the rectangle shortcut and with $\tau = VQ/(Ib)$ .

Solution. Shortcut:

\tau_{\max} = \tfrac{3V}{2A}

, with

A = bh = (0.100)(0.200) = 0.020\,\text{m}^2

.

\tau_{\max} = 1.5(12{,}000)/0.020 = 9.00\times10^{5}\,\text{Pa} = 0.900\,\text{MPa}

. General formula at the neutral axis: the area above the NA is

A' = b(h/2) = (0.100)(0.100) = 0.0100\,\text{m}^2

, with centroid

\bar{y}' = h/4 = 0.050\,\text{m}

, so

Q = A'\bar{y}' = 0.0100(0.050) = 5.00\times10^{-4}\,\text{m}^3

. With

I = 6.667\times10^{-5}\,\text{m}^4

and

b = 0.100\,\text{m}

:

\tau = VQ/(Ib) = (12{,}000)(5.00\times10^{-4})/[(6.667\times10^{-5})(0.100)] = 0.900\,\text{MPa}

. Answer:

\tau_{\max} = 0.900\,\text{MPa}

(both methods agree). Sanity check: shear stress (

0.9\,\text{MPa}

) is far below the bending stress (

27\,\text{MPa}

), as expected for a slender beam where bending governs.

\tau_{\max} = \dfrac{3V}{2A} = \dfrac{3(12{,}000)}{2(0.020)}

Common pitfalls

•Locating $M_{\max}$ at the largest load instead of where $V = 0$ . The maximum moment occurs where shear crosses zero; find that point first.
•Treating an applied couple as a step in shear. A concentrated moment steps the moment diagram, not the shear diagram.
•Using a non-centroidal $I$ in $\sigma = Mc/I$ . $I$ and $c$ must both be referenced to the centroidal (neutral) axis of the bending cross section.
•Swapping bending and shear stress locations. Bending stress is max at the extreme fiber and zero at the NA; transverse shear is max at the NA and zero at the fibers.
•Using $b$ as the full width when computing $\tau$ in a flanged section. In $\tau = VQ/(Ib)$ , $b$ is the thickness at the level of interest (the web, not the flange).
•Confusing $S$ (section modulus, $I/c$ ) with $I$ . The flexure formula in the $M/S$ form needs $S$ , which already divides by $c$ .
•Unit drift: mixing mm and m so $I$ in mm $^4$ meets $c$ in m. Keep one length unit throughout (all m, or all mm) before computing stress.

References

NCEES FE Reference Handbook — Mechanics of Materials
Hibbeler, Mechanics of Materials — shear/moment diagrams, flexure and transverse shear
Gere & Goodno, Mechanics of Materials — beam bending theory

B. Stress transformations and Mohr’s circle

Stress Transformation, Mohr's Circle and Combined Loading

Transform plane stress to any orientation, find principal stresses and the maximum in-plane shear with Mohr's circle, and combine axial, bending, and torsional stresses on a critical element.