ENG1011 Chap.10 Axial Member Design and Euler Buckling
Axial Member Design and Euler Buckling
Axial Member Design and Euler Buckling is Week 7, where analysis becomes design. Once the method of joints gives a member force, the member must be sized for it. In tension, a member fails by yielding, so its capacity is the squash load, the yield stress times the cross-sectional area.
In compression, a short member also yields, but a slender member fails first by buckling sideways at the Euler load, pi squared times the elastic modulus times the second moment of area, divided by the square of the effective length. The effective length factor depends on the end conditions: 1.0 when both ends may rotate and 0.5 when rotation is restrained at both ends.
The second moment of area measures how far the material sits from the bending axis, so a member buckles about its weaker axis unless that direction is braced, and hollow sections resist buckling far better than solid ones of the same area.
The capacity of any compression member is the smaller of its squash and buckling loads, and the design check compares that with the force it must carry.
Week 7 questions combine geometry and capacity: compute the area and second moment of area of a section, then the squash load and the buckling loads about each axis, and finally decide the largest safe load.
The repeater practice test follows exactly that sequence for a hollow circular member with different end restraint about its two axes. The same reasoning sizes the compression members of each team's bridge in the Week 7 practical.
What this chapter covers
- 01
Squash load for tension and short compression members
- 02
Euler buckling of slender members
- 03
Effective length factors for end restraint
- 04
Second moment of area for rectangles and tubes
- 05
Buckling about the weaker axis
- 06
The capacity as the smaller of two limits
- 07
Sizing a member for a given force
Worked example · free
Capacity of a square aluminium strut
- 1Area: 25 times 25 = 625 mm squared, so the squash load is 200 times 625 = 125,000 N.
- 1Second moment: 25 to the fourth over 12 = 32,552 mm to the fourth, the same about both axes.
- 1Euler load with k = 1: pi squared times 70,000 times 32,552 divided by 1500 squared = 9,995 N.
- 1Capacity is the smaller value, about 10.0 kN, so buckling governs by a wide margin.
Key terms
- Squash load
- The axial load at which a member's whole cross-section reaches the yield stress.
- Euler buckling load
- The compressive load at which a slender pin-ended member bows sideways, set by its stiffness and length.
- Effective length
- The actual length multiplied by a factor that reflects how the ends are restrained.
- Second moment of area
- A section property measuring how far material lies from a bending axis, in units of length to the fourth power.
Axial Member Design and Euler Buckling FAQ
How do I know whether yielding or buckling will govern?
Calculate both loads and compare them. The smaller one is the capacity and identifies the failure mode. Short, stocky members usually yield first, while long, slender members almost always buckle at loads well below their squash load.
Which axis does a rectangular member buckle about?
The axis about which its second moment of area is smallest, which is the axis parallel to its wider face. A thin strip therefore bends across its thin dimension, unless bracing or end conditions prevent movement in that direction.
Why do end conditions matter so much?
Restraining both ends against rotation halves the effective length, and because the Euler load depends on the square of that length, the buckling load rises fourfold. Questions sometimes give different restraint about each axis, so each needs its own effective length factor.
Why are tubes used for compression members?
A tube places its material far from the axis, which raises the second moment of area much more than it raises the area. For the same mass, a tube therefore resists buckling far better than a solid rod.
Exam move
For every compression member, write the squash load and the Euler load about each axis side by side, then circle the smallest. Practise the second moment of area formulas until they are instant, and always check whether the end restraint differs between the two axes before choosing the effective length factor.
Keep a short table of second moment of area formulas and effective length factors, and practise filling in all four capacity numbers for one member in under five minutes.
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