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All sheets

Mechanics of materials / Sheet 04

Bending and deflection

Formula reference for mechanical engineering interviews. Bending stress in a determinate beam is independent of material; deflection is not. Check both, and say which one governed.

Scope: Straight prismatic beamsSections: 11Units: SI, N and mmRevision: A

42 formulas

The flexure equation

M / I = σ / y = E / R

Same three-part structure as the torsion equation.

Bending stress

σ = M y / I

y is measured from the neutral axis, not along the span

Max bending stress

σ_max = M c / I = M / S

c is the distance to the extreme fibre

Section modulus

S = I / c

The single number that ranks sections for strength

Curvature

1 / R = M / E I

EI is the flexural rigidity

Combined axial and bending

σ = P/A ± M c / I

Eccentric columns, C-frames, brackets

Neutral axis location

through the centroid

True for homogeneous linear elastic sections in pure bending

Section properties

Section	I	S = I/c	Notes
Rectangle b × h	b h³ / 12	b h² / 6	depth cubed, width linear
Solid round	π d⁴ / 64	π d³ / 32	J = 2I
Hollow round	π (d_o⁴ - d_i⁴) / 64	divide by d_o/2	best stiffness per mass
Hollow rectangle	(B H³ - b h³) / 12	divide by H/2	box sections
Triangle, base b	b h³ / 36	about the centroid	centroid at h/3

Parallel axis theorem

I = I_c + A d²

The whole basis of I-beam and box section efficiency

Depth sensitivity

I ∝ h³, S ∝ h²

Doubling depth: 8× stiffness, 4× strength

Width sensitivity

I ∝ b, S ∝ b

Doubling width: 2× both. Depth wins every time

Transverse shear

Shear stress

τ = V Q / (I t)

Q = A′ȳ, first moment of the area beyond the cut

Rectangle

τ_max = 1.5 V / A

At the neutral axis, zero at the surfaces

Solid circle

τ_max = 4 V / 3 A

At the neutral axis

I-beam

τ ≈ V / A_web

The web carries almost all the shear

Shear flow

q = V Q / I

Sizes fasteners and welds in built-up beams

Bending stress peaks at the outer fibre where shear is zero, and shear peaks at the neutral axis where bending stress is zero. They rarely combine. The exception is the web to flange junction of an I-beam, where both are significant and a von Mises check is warranted.

Load, shear and moment

Load to shear

dV / dx = -w

Slope of the shear diagram is the load intensity

Shear to moment

dM / dx = V

Slope of the moment diagram is the shear

Governing equation

E I d⁴y/dx⁴ = -w

Integrate down: V, then M, then θ, then y

Maximum moment

where V crosses zero

First thing to locate on any beam problem

Point of contraflexure

where M = 0

Curvature reverses; useful for splice locations

Standard deflection cases

Case	δ max	M max	Slope
Cantilever, point load at tip	P L³ / 3EI	P L, at wall	P L² / 2EI
Cantilever, UDL	w L⁴ / 8EI	w L² / 2, at wall	w L³ / 6EI
Cantilever, moment at tip	M L² / 2EI	M, constant	M L / EI
Simply supported, central load	P L³ / 48EI	P L / 4, centre	P L² / 16EI
Simply supported, UDL	5 w L⁴ / 384EI	w L² / 8, centre	w L³ / 24EI
Fixed both ends, central load	P L³ / 192EI	P L / 8, at ends	0 at ends
Fixed both ends, UDL	w L⁴ / 384EI	w L² / 12, at ends	0 at ends
Propped cantilever, UDL	w L⁴ / 185EI	w L² / 8, at fixed end	0 at fixed end

Cantilever tip stiffness

k = 3 E I / L³

Use for natural frequency estimates

Simply supported stiffness

k = 48 E I / L³

Central load, central deflection

Fixing both ends of a uniformly loaded beam cuts maximum deflection to one fifth and maximum moment to two thirds, and moves the peak moment from midspan to the supports.

Deflection methods

Double integration

E I y″ = M(x)

Integrate twice, apply boundary conditions

Macaulay's method

singularity terms ⟨x - a⟩ⁿ

Point loads and discontinuous loading

Moment-area, first theorem

Δθ = area of the M/EI diagram

Change in slope between two points

Moment-area, second theorem

δ = first moment of that area

Fast for cantilevers

Castigliano

δ = ∂U/∂P, U = ∫ M² dx / 2EI

Frames, curved members, indeterminate structures

Superposition

add standard cases

Linear elastic small deflections only

Transformed section

n = E₁ / E₂

Flitched and composite beams

Scaling and ratios

Deflection vs length

δ ∝ L³ (point), L⁴ (UDL)

Span dominates; check it before the section

Stress vs length

σ ∝ L (point), L² (UDL)

Much weaker dependence than deflection

Material swap

σ unchanged, δ ∝ 1/E

Determinate beams only; aluminium deflects about 3× steel

Shear vs bending

span/depth below about 10

Shear starts to govern; beam theory also weakens

Deflection limits

L/360 live, L/240 total

Structures. Machines and robotics run far tighter

Plastic bending

Plastic moment

M_p = Z S_y

Z is the plastic section modulus

Rectangle

Z = b h² / 4

Compare S = b h² / 6

Shape factor

Z / S

1.5 rectangle, 1.7 solid round, about 1.12 for an I-beam

The shape factor is the reserve between first yield and a fully plastic hinge. An I-beam has almost none, because its material already sits at the extreme fibre, which is exactly why it is efficient elastically.

Worked chain: sizing a beam

Step 1

reactions, then V and M diagrams

Locate M_max where V crosses zero

Step 2

S_required = M_max / σ_allow

Strength check; pick a trial section

Step 3

δ from the standard case

Compare with the deflection limit

Step 4

τ = VQ/It near the supports

Only governs for short deep beams

Step 5

state which check governed

Usually deflection, and saying so is the point

Numbers worth memorizing

Quantity	Steel	Aluminium	Ratio
E (GPa)	200	69	2.9
Density (kg/m³)	7850	2700	2.9
Specific stiffness E/ρ	25.5	25.6	1.0
Deflection, same geometry	1×	2.9×	-
Bending stress, same geometry	1×	1×	-

Specific stiffness is identical, so equal-mass equal-stiffness swaps are impossible in pure tension. In bending aluminium wins, because the thicker section it allows raises I faster than the lower E costs you.

Concepts that decide the interview

Depth is worth cubed, width only linear. Doubling depth gives 8 times the stiffness and 4 times the strength; doubling width gives 2 times both. Quote the exponents, not just the direction.
Length beats everything. Deflection scales with L³ or L⁴ while stress scales with only L or L². When something is too floppy, look at span before section.
Material at the neutral axis does nothing for bending but is exactly where shear peaks. That designs the I-beam: flanges carry moment through the Ad² term, the web carries shear.
You can put a hole through a beam near the neutral axis at midspan, but not near the supports, where shear is maximum.
Bending stress in a determinate beam is independent of E, because M comes from statics alone. Deflection is not. In indeterminate beams load redistributes toward the stiffer members in proportion to EI.
Deflection, not stress, usually sizes the part, and far more so in robotics where positional accuracy rather than failure is the constraint.
End fixity is a large, free win, exactly like bracing a column. Change the boundary conditions before you change the part.
Superposition requires linear elastic material and small deflections. Once a fibre yields or geometry changes materially, adding standard cases is no longer legitimate.

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Sheet 04 of 07 / Rev AStraight prismatic linear elastic beams, small deflections