Axially loaded members

Axially loaded members are everywhere in civil engineering: truss chords and diagonals, hanger rods, tie-backs for retaining walls, columns, piles and post-tensioning tendons. Lesson 1 showed how to find the stress in such a member. This lesson finds how much it stretches or shortens, which matters for serviceability and for fit-up during construction, and then uses deformation to solve problems that equilibrium alone cannot: statically indeterminate members and members restrained against thermal movement.

Elongation of a prismatic bar

Combine the three ingredients from lesson 1. Equilibrium gives the internal force \(P\) and the stress \(\sigma = P/A\). The material law gives the strain \(\varepsilon = \sigma/E\). Geometry gives the elongation \(\delta = \varepsilon L\). Putting them together:

The product \(AE\) is called the axial rigidity. Doubling the area or using a material twice as stiff halves the elongation. The ratio \(AE/L\) is the axial stiffness \(k\), the force needed to produce a unit elongation, so the bar behaves like a spring with \(P = k\,\delta\). Thinking of members as springs is a powerful idea, and it is the starting point of the stiffness method used in structural analysis software.

Use a consistent sign convention: tension and elongation positive, compression and shortening negative. Then a compressive force automatically gives a negative \(\delta\).

Bars with several segments

When the load, area or material changes along the bar, split it into segments in which all three are constant, find the internal force in each segment with the method of sections, and add the elongations:

Drawing an axial force diagram, a plot of internal force along the bar, makes the bookkeeping much easier. Each external load causes a jump in the diagram equal to the load.

When the area or the internal force varies continuously, the sum becomes an integral, \(\delta = \int_0^L P(x)\,dx / [A(x)\,E]\). A classic case is a tall hanging bar or cable under its own weight: the internal force grows linearly from zero at the bottom to the full weight at the top, and the elongation works out to half of what it would be if the whole weight hung from the bottom end. Tapered members, such as some light poles and piles, also need the integral form.

Statically indeterminate axial members

A member is statically indeterminate when equilibrium equations alone cannot give all the forces. A bar fixed between two rigid walls and loaded in the middle is the simplest example: there are two unknown reactions but only one useful equilibrium equation. A reinforced concrete column, in which steel bars and concrete share a load, is another.

The extra equations come from compatibility, a statement about how the deformations must fit together. The method has four steps that never change:

• Equilibrium. Write the equilibrium equations in terms of the unknown forces.
• Compatibility. Write a geometric condition the deformations must satisfy, such as the total elongation being zero between fixed walls, or two parts shortening by the same amount.
• Force–deformation. Replace each deformation with \(PL/AE\).
• Solve the equations together, then check that the answers make physical sense.

The composite post shows a general rule: when parts are forced to deform together, the stiffer part attracts more load. This is why reinforcing steel in a concrete column is so effective, and why a designer must think about stiffness, not just strength, when different materials work together. In real concrete members, creep and shrinkage gradually shift more of the load onto the steel over time, a refinement covered in concrete design courses.

Thermal stress

Most materials expand when heated. A free bar of length \(L\) whose temperature rises by \(\Delta T\) lengthens by

where \(\alpha\) is the coefficient of thermal expansion. For steel \(\alpha \approx 12 \times 10^{-6}\) per degree Celsius, and concrete is similar, which is one reason the two materials work well together. Aluminum expands about twice as much, roughly \(23 \times 10^{-6}\) per degree Celsius.

A free bar develops no stress when its temperature changes. Stress appears only when the movement is restrained. If a bar is fixed between rigid walls and heated, compatibility requires the thermal elongation to be canceled by a mechanical shortening, \(\alpha\,\Delta T\,L - PL/(AE) = 0\), so the restrained stress is

Note that the length and area cancel. A steel bar fully restrained and heated by 40 °C develops \(200\,000 \times 12 \times 10^{-6} \times 40 = 96\ \text{MPa}\) of compression, whatever its size. That is a large fraction of the yield strength, which is why bridges have expansion joints and bearings, why long concrete slabs have control joints, and why continuously welded rail must be installed at a carefully chosen neutral temperature to avoid buckling on hot days.

US bridges must accommodate a wide range: the AASHTO LRFD specifications give design temperature ranges that, in cold climates, exceed 70 °C (about 125 °F) for steel bridges. A 100 m steel span moving through 70 °C changes length by about \(12 \times 10^{-6} \times 70 \times 100\,000 \approx 84\ \text{mm}\), roughly 3.3 in., all of which the joints and bearings must accept.

Stress concentrations

The formula \(\sigma = P/A\) gives the average stress on the net section. Near a hole, notch or sharp change in width, the actual stress peaks well above the average. The ratio of the peak stress to the nominal stress is the stress concentration factor \(K\):

For a small circular hole in a wide plate under tension, \(K\) is about 3. Fillets with generous radii have much lower factors than sharp corners. Charts of \(K\) for common geometries are published in handbooks such as Peterson's Stress Concentration Factors.

For ductile materials under steady load, stress concentrations are usually not critical: once the peak reaches yield, the material there yields and the load spreads to the surrounding area. Design codes for static strength of steel members therefore use the net-section average. But under repeated loading, fatigue cracks start at stress concentrations, and for brittle materials the peak governs failure. Highway bridges, crane runways and machine parts must be detailed with this in mind.

Computing a multi-segment bar

The bookkeeping for a bar with many segments is ideal for a short program. The code below takes the external loads and segment properties, builds the axial force in each segment by summing the loads to its right, and adds the elongations. It reproduces the stepped bar example and then adds a third segment.

# Axial force and displacement along a bar fixed at its left end
# Segments: (length mm, area mm^2, E MPa); loads at the right end of each segment (N, + to the right)
def solve_bar(segments, loads):
    total = 0.0
    for i, (L, A, E) in enumerate(segments):
        P = sum(loads[i:])               # internal force = sum of loads to the right of the cut
        d = P * L / (A * E)
        total += d
        print(f"Segment {i+1}: P = {P/1000:7.1f} kN   stress = {P/A:7.1f} MPa   "
              f"delta = {d:+.3f} mm   tip so far = {total:+.3f} mm")
    return total

print("Stepped bar from the example:")
solve_bar([(600, 800, 200e3), (400, 400, 200e3)], [-100e3, 60e3])

print("\nWith an aluminum extension carrying a further 20 kN:")
solve_bar([(600, 800, 200e3), (400, 400, 200e3), (300, 500, 70e3)], [-100e3, 60e3, 20e3])

Stepped bar from the example:
Segment 1: P =   -40.0 kN   stress =   -50.0 MPa   delta = -0.150 mm   tip so far = -0.150 mm
Segment 2: P =    60.0 kN   stress =   150.0 MPa   delta = +0.300 mm   tip so far = +0.150 mm

With an aluminum extension carrying a further 20 kN:
Segment 1: P =   -20.0 kN   stress =   -25.0 MPa   delta = -0.075 mm   tip so far = -0.075 mm
Segment 2: P =    80.0 kN   stress =   200.0 MPa   delta = +0.400 mm   tip so far = +0.325 mm
Segment 3: P =    20.0 kN   stress =    40.0 MPa   delta = +0.171 mm   tip so far = +0.496 mm

Note that adding a load at the far end changes the force in every segment to its left. That is the essence of the method of sections, and the program makes it hard to forget.