Beams and Simple Spans

How Bridges Work

Chapter 2 · Beams and Simple Spans

The oldest bridge is a log laid across a stream, and the idea has not changed: a straight member rests on a support at each end and carries the load between them. How Bridges Work 1 found the reactions at the two ends. This chapter asks what happens inside the beam between them: how bending and shear build up along its length, why a deep beam beats a shallow one, why an I-beam has the shape it does, and why a plain beam runs out of strength long before it reaches the spans of the Forth bridges.

What was run

Every Python snippet in this chapter and in the three exercise solutions was executed, and the output shown in the comments is what Python printed. You only need Python 3 and its standard library.

Simply Supported

A simply supported beam rests on a support at each end. The supports push up but do not grip the beam, so the ends are free to tilt as the beam sags. The distance between the supports is the span. This is the simplest case to analyse, and the one that every bridge-building student meets first. Two quantities describe what the load does to it along its length:

Quantity	Symbol	What it is	Largest where
Shear force	V	the sliding force across a cut through the beam, equal to the net upward force on one side of the cut	at the supports
Bending moment	M	the turning effect at a cut, equal to the sum of force × distance for everything on one side	at midspan, for a symmetric load

For the two cases you will meet most often, the maximum values come from standard formulas, where w is a uniform load in kN per metre and P a single load at midspan in kN, on a span L in metres:

Load	Max shear (at the supports)	Max moment (at midspan)
Uniform load `w` over the whole span	wL / 2	wL² / 8
Point load `P` at midspan	P / 2	PL / 4

def udl_beam(w, span):                    # w in kN/m, span in m
    return w * span**2 / 8, w * span / 2  # (max moment kN m, max shear kN)

def point_beam(p, span):                  # one load p (kN) at midspan
    return p * span / 4, p / 2

print(udl_beam(20, 10))      # (250.0, 100.0)
print(point_beam(200, 10))   # (500.0, 100.0)
print(udl_beam(20, 20))      # (1000.0, 200.0)
    

Two lessons are in those three lines. First, how a load is spread matters: 20 kN/m over 10 m is a total of 200 kN, and it produces a moment of 250 kN·m, while the same 200 kN concentrated at midspan produces twice that (500 kN·m). Spreading a load out is kind to a beam. Second, span is expensive: with the same load per metre, doubling the span from 10 m to 20 m multiplies the moment by four (250 to 1000 kN·m), because the moment grows with the square of the span.

The shape of the two diagrams is worth picturing. For a uniform load, shear starts at its maximum at one support, falls in a straight line through zero at midspan, and reaches its maximum again, with the opposite sign, at the other support. The moment is zero at both ends (the beam is free to tilt there, so nothing resists turning) and builds in a smooth curve to its peak at midspan.

uniform load on a simply supported beam

  ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓
 ═══════════════════════════      beam
 ▲                         ▲
 A                         B

 shear V      +wL/2 ╲
                      ╲        zero at midspan
                       ╲
                        ╲ −wL/2

 moment M         ╭───────╮
                ╭─╯       ╰─╮    peak wL²/8 at midspan
              ╭─╯           ╰─╮
 zero at A  ──╯                 ╰──  zero at B
    

Bending Stress: Why Depth Matters

How Bridges Work 1 said that bending is tension and compression working together. Here is how. As a beam sags, the top face is squeezed and the bottom face is stretched. Somewhere between them is a layer that is neither: the neutral axis. The stress in the material is zero there and grows in proportion to the distance from it, reaching its largest at the top and bottom faces. The formula that links everything is:

The bending formula

stress = M × y / I, where M is the bending moment, y is the distance from the neutral axis to the face you are looking at, and I is the second moment of area of the cross-section, a number that measures how well the shape resists bending. For a rectangle of width b and depth h, I = b×h³/12.

Look at the cube in that formula. Depth appears as h³, width only as b. The same amount of material arranged deeper is far more resistant to bending:

def rect_i(b, h):
    return b * h**3 / 12                  # mm^4

def bend_stress(m_knm, i_mm4, y_mm):
    return m_knm * 1e6 * y_mm / i_mm4     # N/mm2 = MPa

M = 10                                    # kN m
for b, h in [(100, 100), (50, 200), (200, 50)]:
    i = rect_i(b, h)
    print(b, h, b * h, round(i), round(bend_stress(M, i, h / 2), 1))
# 100 100 10000 8333333 60.0
# 50 200 10000 33333333 30.0
# 200 50 10000 2083333 120.0
    

All three bars have the same cross-sectional area (10,000 mm²), so they use the same amount of steel and weigh the same. Yet under the same 10 kN·m moment the stress is 60 MPa in the square bar, 30 MPa when the bar is turned on edge (50 wide, 200 deep), and 120 MPa when it is laid flat. Same material, four times the stress between best and worst orientation. This is why floor joists stand on edge and why a plank across a ditch bends less when you turn it on its side.

The I-Beam: Material Where It Works

Look again at the stress picture. The material near the neutral axis is hardly stressed: all the heavy work is at the top and bottom faces. A solid rectangle therefore wastes steel in the middle. The I-beam (and its relatives, the plate girder and the box girder) solves this by concentrating the material in two flanges far from the neutral axis, joined by a thin web that keeps them apart and carries the shear.

def i_section(bf, tf, tw, depth):
    inner = depth - 2 * tf
    return (bf * depth**3 - (bf - tw) * inner**3) / 12

def area_i(bf, tf, tw, depth):
    return 2 * bf * tf + tw * (depth - 2 * tf)

a = area_i(150, 12, 8, 300)               # flanges 150 x 12, web 8, depth 300
i = i_section(150, 12, 8, 300)
side = a ** 0.5                           # a square bar of the same area
i_sq = side**4 / 12
print(a, round(i))                        # 5808 88709184
print(round(side, 1), round(i_sq), round(i / i_sq, 1))
# 76.2 2811072 31.6
    

This I-section has a cross-section of 5,808 mm². A solid square bar with the same area is 76.2 mm on a side. The I-section's I is 31.6 times larger, which means it resists bending over thirty times better for exactly the same weight of steel. That is the whole reason for the shape.

The idea is old. When Robert Stephenson built the Britannia Bridge over the Menai Strait in the 1840s, the trains ran through enormous rectangular tubes of wrought-iron plate: box girders, which are hollow beams with the same logic as the I-section. The same thought appears in the box-shaped decks of many modern bridges. (This history is from general knowledge, not from a source checked for this chapter.)

What the formula leaves out

A tall thin web can buckle sideways, and a deep beam can twist and buckle away from its loaded plane unless it is braced. Real designs check these too. The formula here gives you the first and most important answer, the bending stress, and nothing more.

Deflection: How Much It Sags

A beam can be strong enough and still be unusable if it sags too much: a deck that dips visibly is uncomfortable to cross and cracks its surfacing. The middle of a simply supported beam under a uniform load sags by

Deflection of a uniformly loaded simple beam

δ = 5 w L⁴ / (384 E I), where E is the material's stiffness (Young's modulus; about 200,000 MPa for steel) and I is the same second moment of area as before. Deflection grows with the fourth power of the span.

E = 200e3                                 # MPa, steel

def deflection_mm(w, span_m, i_mm4):      # w in kN/m (the same as N/mm)
    L = span_m * 1000
    return 5 * w * L**4 / (384 * E * i_mm4)

for span in (4, 8, 16):
    d = deflection_mm(10, span, 8.0e7)
    print(span, round(d, 1), round(span * 1000 / d))
# 4 2.1 1920
# 8 33.3 240
# 16 533.3 30
    

The three columns are the span in metres, the sag in millimetres, and the span divided by the sag. A 4 m span sags 2.1 mm (about span/1920): invisible. At 8 m it sags 33.3 mm (span/240), and at 16 m it sags 533 mm (span/30), more than half a metre. Each doubling of span multiplies the sag by sixteen. A common design target for live-load sag is around span/360, which the 8 m case already misses. Stiffness, not strength, often decides the size of a beam.

The Span Limit of a Plain Beam

There is a harder limit still. A beam must carry its own weight, and that weight grows with the length. For the same section, the self-weight moment wL²/8 grows with the square of the span, so there is a span at which the beam uses up all its strength carrying nothing but itself. The code below works it out for a 5,000 mm² steel section with a moment of area of 8.0×10⁷ mm⁴ and a working stress limit of 165 MPa (roughly the 355 MPa steel of How Bridges Work 1 with a factor of safety a little over 2):

import math

RHO, G = 7850, 9.81                       # steel density, gravity
area, i, y = 5000, 8.0e7, 150             # mm2, mm4, mm
allow = 165                               # MPa, working stress

w = RHO * G * area * 1e-6 / 1000          # own weight in kN/m
print(round(w, 3))                        # 0.385
for span in (10, 20, 40, 80):
    print(span, round(w * span**2 / 8, 1))
# 10 4.8
# 20 19.3
# 40 77.0
# 80 308.0

m_cap = allow * (i / y) / 1e6             # moment the section can carry, kN m
print(round(m_cap, 1), round(math.sqrt(8 * m_cap / w), 1))
# 88.0 42.8
    

The beam weighs 0.385 kN per metre. Its own-weight moment is only 4.8 kN·m at 10 m, but 77 kN·m at 40 m and 308 kN·m at 80 m: each doubling quadruples it. The section can carry 88 kN·m, so it breaks under its own weight at 42.8 m, before a single vehicle drives on. In practice the limit is far shorter, because the live load takes much of the capacity (Exercise 3 shows this). Making the beam deeper helps, but not as much as you would like, because a deeper beam is also a heavier one.

That is why beam bridges are used for short and medium spans (the approach spans, the viaducts, the footbridges), and why every longer crossing uses another idea. An arch turns bending into compression (How Bridges Work 3). A truss builds a deep beam out of light struts and ties (How Bridges Work 4). A cable carries load in pure tension, with no bending at all (How Bridges Work 7 and 8). The Forth bridges are three different answers to a span that a plain beam could never reach.

Common mistakes

Mixed-up units, above all: I in mm⁴ needs the moment in N·mm (kN·m × 10⁶) and y in mm; putting b and h the wrong way round in b×h³/12 (the depth is the cubed one); using the moment formula for a point load on a uniform-load problem; and forgetting that the beam's own weight is part of the load.

Where This Leads

A beam carries load by bending, and bending is the least efficient way to use material: only the outer faces are working hard. The next chapter takes the oldest answer to that problem, the stone arch, where the whole shape is squeezed in compression, and looks at the Roman arches, the clapper bridges and the packhorse bridges of Scotland.

Exercise 1

A timber joist spans 4 m and carries a uniform load of 6 kN/m including its own weight. Its section is 75 mm × 225 mm. Write a program that works out the maximum moment and shear, then the maximum bending stress when the joist is on edge (225 mm deep) and when it is laid flat (75 mm deep). By what factor does the stress change, and why is it 3 when the moment of area changes by a factor of 9?

📄 View solution

Exercise 2

A steel beam spans 6 m and carries one 80 kN load at midspan (ignore its own weight). The stress limit is 165 MPa and the deflection limit is span/360; E = 200,000 MPa. For a point load at midspan, the sag is P×L³/(48×E×I). Test these I-sections (flange width, flange thickness, web thickness, depth in mm): (100, 8, 5, 200), (125, 10, 6, 250), (150, 12, 8, 300), (150, 14, 8, 350). For each, print the stress, the sag and whether it passes both limits. Predict first which is the smallest that passes.

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Exercise 3

Take I-sections with 150 × 12 mm flanges and an 8 mm web, at depths of 300, 450 and 600 mm. For each, find the moment capacity at 165 MPa, the beam's own weight per metre, and the longest simply supported span (a) when it carries only its own weight and (b) when it carries a further 10 kN/m. By how much does doubling the depth increase each span, and what does that tell you about the limits of a beam?

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Quick Reference

Uniform load: max shear wL/2 at the supports; max moment wL²/8 at midspan
Midspan point load: max shear P/2; max moment PL/4
Bending stress: M × y / I, largest at the outer faces; zero at the neutral axis
Rectangle: I = b×h³/12; depth counts cubed, width once
I-beam: flanges far from the neutral axis resist bending; the web resists shear
Deflection (uniform load): 5wL⁴/(384EI); steel E ≈ 200,000 MPa
Span: moment grows with L² and sag with L⁴, so a plain beam only suits short spans

Coming next

How Bridges Work 3 turns to arches and early bridges: the Roman arch, thrust and abutments, stone and brick construction, and the clapper and packhorse bridges of Scotland.