Tension and compression in a truss bridge: what pulls, what pushes

Tension and compression with a rope and a stick

Hang a bucket from a rope and the rope is in tension. Press on the end of a long thin stick and it snaps sideways before much weight is on it. The first member is being pulled and the second pushed, and the second is far easier to ruin.

That small difference is the whole subject of tension and compression in a truss bridge. A member in tension is pulled along its length and tries to get longer; a member in compression is pushed along its length and tries to get shorter. Because a pin-jointed truss carries only axial force, every member falls into one of those two groups, and truss analysis is the job of deciding which, and by how much. A pulled member needs enough steel area. A pushed member needs that, and a good deal more besides, because what it fears is not the load but the sideways bend.

Engineers write tension as positive and compression as negative. Assume every unknown is a pull and solve the equations, for instance with the method of joints, and a negative answer shows the member is actually being pushed.

How to tell which a truss member carries

The pattern of tension and compression can mostly be guessed without solving the truss. Think of the truss as a beam. A beam resting on two supports with a downward load sags: the top fibres are squeezed and the bottom fibres stretched. A truss is a beam made of sticks, so the top chord is in compression and the bottom chord in tension.

In the Warren truss drawn above, 20 kN sits on the centre bottom joint. The top chord members F-G, G-H and H-I are cool, the bottom chords A-B to D-E are warm. The solver puts the centre top chord G-H at 20.0 kN in compression and the centre bottom chords B-C and C-D at 15.0 kN in tension.

The diagonals alternate. Following the load path helps: the 10 kN reaction at the left support climbs the end diagonal A-F, which is pushed, then crosses to the bottom chord through F-B, which is pulled, and so on across the truss. In a Warren the push-pull pattern simply repeats along the span.

Other layouts choose differently. A Pratt truss, patented in 1844 by Thomas and Caleb Pratt, points its diagonals so that they are in tension and its verticals in compression. The Howe truss, patented in 1840 by William Howe, reverses that: timber diagonals in compression, and slim wrought-iron vertical rods in tension. The two patents are mirror images, and which one suits a truss depends on what its members are made of and how long they are.

Count the signs: seven members are cool (the three top chords and the four diagonals A-F, B-G, H-D and I-E) and eight are warm (the four bottom chords and the other four diagonals). The largest force of all, 20.0 kN in G-H, is a push. Every tension member carries 15.0 kN or less, and in the designer every bar is the same 60 x 3 mm tube, so the push members are the ones with the least margin.

Why compression members buckle and tension members do not

Pull on a straight bar and any slight bend straightens out. Push on it and the same bend grows, because the load now acts at an offset from the centreline and makes the bend worse. At a certain load the bar gives up and bows out sideways. That load, for a pinned column, is Euler’s critical load:

P_cr = π² × E × I / L²

E is the stiffness of the material, I the second moment of area of the cross-section (how far the material sits from the centre of the bar), and L the length. Length is squared and sits in the denominator, so doubling a bar’s length cuts its buckling load to a quarter.

Take a 60 x 3 mm steel tube, 60 mm outer diameter and 3 mm wall, 3 m long, with E = 200 GPa. The second moment of area of a hollow circular tube is I = π/64 × (D⁴ − d⁴), with D the outer and d the inner diameter, here 60 mm and 54 mm:

I = π/64 × (60⁴ − 54⁴) = π/64 × 4 456 944 = 218 780 mm⁴

P_cr = π² × 200 000 × 218 780 / 3000² = 48.0 kN

The yield load of the same tube, 250 MPa times its steel area of 537 mm², is 134.3 kN. Buckling arrives first, at about a third of the yield load, and that is why the designer lets the lesser of the two decide. The centre top chord G-H carries 20.0 kN, and 20.0 / 48.0 is 41.7 percent of its capacity.

Compression members are therefore made stockier. At a fixed wall thickness a larger diameter raises I roughly with the cube of the diameter, and a shorter length helps even more. A tension member has no such problem. It fails when the stress reaches the yield strength of the steel, and a tube of 537 mm² does that at 134.3 kN regardless of its length.

Where tension members do fail is at their ends. A rod can be strong and the bolt, weld or glue that holds it can still let go first, which is exactly what happens in a popsicle-stick or toothpick model. The factor of safety page puts the buckling load of every member against its force.

Feeling tension and compression with a strip of card

A strip cut from a manila folder makes the point with nothing but scissors and a bag of coins. Clip one end of a narrow strip to a table edge and hang a small bag of coins from the other. It holds a surprising weight, because card in tension is strong along its fibres. Now stand the same strip on its end, flat side vertical, and press down gently from above with a finger: it bows and folds at a tiny load.

Roll the strip into a tube about as wide as a pencil, tape the seam, and press again. The tube carries many times the load that crumpled the flat strip, with exactly the same amount of card. Nothing about the card changed except where the material sits relative to the centre, which is the I in Euler’s formula. The file folder bridge page uses the same idea to build trusses whose compression members are tubes and whose tension members are flat strips.