Over the last seven steps, we've taken the stresses in a steel member apart one piece at a time. In this final step, we'll put everything together on a portal frame under code load combinations. Rather than introduce anything new, we'll establish a routine for checking stresses in any frame, and then run it from start to finish, so that you leave with a process you can apply to the frames you model yourself.
A portal frame to check. Orbit, read the results and hover the members; nothing you do here changes the model.
Members can copy this model into their own projects and work the step in it. Join or sign in.
1.0 The model
Our model is a portal frame with pinned bases, spanning 8 m. The two columns (col-left and col-right) are UC 203 x 203 x 46 sections, each 5 m high. The rafter is a UB 406 x 178 x 60, modelled as two members (rafter-1 and rafter-2), one either side of a node at its midspan. The rafter is rigidly connected to the columns at the eaves, so moment is transmitted around each eaves joint and into the columns.
There are three load cases:
a dead (or permanent) load of 10 kN/m on the rafter,
an imposed (or variable) load of 15 kN/m on the rafter,
a wind load of 8 kN/m on the left-hand column, pushing the frame to the right.
These are arranged into two ULS combinations:
ULS: 1.35 G + 1.5 Q, with gravity loads only,
ULS: 1.35 G + 1.5 W, with the wind.
Notice that the wind combination deliberately has no imposed load in it. The imposed load on a roof represents maintenance access, and Eurocode 1 (EN 1991-1-1, 3.3.2) says it need not be applied together with wind. In other words, nobody is going to be up on the roof doing maintenance in a storm! A floor's imposed load would be different: it would be included alongside the wind, at a reduced combination value. So, before you combine loads on your own structure, take a moment to check what each imposed load actually represents.
The wind load is on the generous side. A line load of 8 kN/m works out at about 1 kN/m² of wind pressure on an 8 m bay (the spacing between neighbouring frames). We've chosen it so that the wind combination gives a clear result.
2.0 A routine for checking stresses
When you open a frame of your own, it's tempting to switch on a stress layer, look for the biggest number and move on. A routine makes sure you look at the right load state, in the right places, and that you don't forget what the stress results leave out. Here it is, and then we'll run it on this frame.
Check every combination, one at a time. The member stress view doesn't paint envelopes. An envelope collects the maximum at every point from different load states, and those maximums never happen together. A peak stress has to come from one real set of forces acting at the same time, so we step through the combinations one by one.
Start with U*. It mutes everything below yield, so on a large model you can see straight away whether anything is overstressed, and where.
Then switch to U and find the hotspots. Now every part of the frame is coloured by its utilisation, and the most heavily utilised regions stand out. These are where the detailed checks go.
Read σ and τ at each hotspot, one chip at a time. Then check one number by hand. If the hand check doesn't agree, stop and find out why before going any further.
Place a stress plane where the formulas are weakest: under high shear and anywhere an open section carries torque. This is where the beam formulas behind the member stress view are least reliable.
List what the stress check leaves out. That means the frame's overall stability, every member in compression, every compression flange without restraint, and every connection. None of these show up in a stress result, so each one is checked separately.
Run the routine
Press Load the model, switch on Member stress and select the U* chip. Step through both ULS combinations in the result set selector, looking for anything overstressed.
Switch to U and find the most heavily utilised point in each combination: the legend in the bottom-left corner of the canvas names the member and the position of the peak. Then step through the σ, τ and σvM​ chips, hovering at that point each time.
Check the hotspot under the wind combination by hand, using the axial force and bending moment at that point. (The hand check in section 3.0 works through it.)
If you're a member, copy the model and solve your copy. Open the Stresses tab, press New section and click on the right-hand column, then press New section again to stop placing planes. The column runs up from its base, so in the planes table set the new plane's x (m) to 5 m to put it right at the top. Pick the wind combination and click the plane's row: the analysis runs, and you can compare its von Mises peak with the painted value.
In your copy, try adding a wind case blowing the other way, or swapping the columns for a lighter section, and run the routine again. Notice whether the hotspot moves, and whether anything is now overstressed under U*.
3.0 What you'll see
What you should see
Combination
Hotspot
Forces there
Von Mises
U
1.35 G + 1.5 Q
Rafter, midspan
12.9 kN (the frame's thrust), 223.4 kN·m
211.8 MPa
0.60
1.35 G + 1.5 Q
Top of each column
144.0 kN, 64.6 kN·m
168.3 MPa
0.47
1.35 G + 1.5 W
Top of the right-hand column
72.8 kN, 111.6 kN·m
260.7 MPa
0.73
Nothing turns magenta under U*, so nothing in the frame reaches yield under either combination. A stress plane at the top of the right-hand column gives a von Mises peak of 261 MPa, against the painted 260.7 MPa.
Under gravity alone, the rafter's midspan governs. The axial force there is the frame's thrust (the inward horizontal reaction at each pinned base, carried across the rafter as compression), and it's small, so almost all of the stress at midspan comes from the sagging moment.
At the top of each column, the axial force is half of the total vertical load. Under the gravity combination, that total is (1.35 × 10 + 1.5 × 15) × 8 = 288 kN, and half of it is the 144 kN in the table. The moment at the top of the column is the thrust multiplied by the 5 m column height, because a pinned base carries no moment.
Once the wind blows, the frame sways to the right and the right-hand eaves joint picks up most of the extra moment. As a result, the top of the right-hand column becomes the most heavily utilised point in the frame, at 0.73. This is despite the wind combination carrying less than half the vertical load of the gravity one: 1.35 × 10 × 8 = 108 kN, against 288 kN.
This is a typical result for a portal frame: the top of the column is where the largest moments in the frame meet the column's axial force. So, when you check a portal frame of your own, don't assume the combination with the most load governs - look at the eaves under every combination that includes lateral load.
Hand check for the top of the right-hand column under wind
The forces. At the top of the right-hand column, under the wind combination, the analysis gives N = 72.75 kN of compression and M = 111.64 kN m.
The section. From step 2, the UC 203 x 203 x 46 has A = 58.7 cm2 and Wz​ = 449.8 cm3. We'll work in N and mm, so that's an area of 5,870 mm² and a section modulus of 449,800 mm³.
The normal stress. The axial force is spread uniformly over the section, while the bending stress is largest at the extreme fibres. On the compression face, the two add together, so the normal stress there is,
σ=−5,87072,750​−
On the tension face, the bending stress changes sign but the axial stress doesn't, so the axial compression reduces it: −12.39+248.20=+235.81 MPa. Therefore, the compression face governs.
The von Mises stress. At the extreme fibre, the only shear is the flange's 4.8 MPa of horizontal shear flow. Combining it with the normal stress (σvM​=σ2+3τ) brings the von Mises stress to 260.7 MPa, so , the value painted in the model. Notice how little the shear adds: the normal stress on its own was 260.59 MPa.
4.0 What the routine leaves out
The last item in the routine matters as much as the other five. For this frame, a utilisation of 0.73 tells us the steel doesn't yield under either combination. It doesn't tell us the frame is adequate - it's the first of several checks, and none of the others can be read from the stress results. Here's what's still to check.
The frame's overall stability. Eurocode 3 measures how close a frame is to buckling sideways as a whole by αcr​, the factor the loads would have to grow by to reach that point. For a single-storey frame, it gives a quick estimate: push the frame sideways with a load H, read the sway δ, and take αcr​=(H/V)(h/δ), with V the total vertical load and h the height.
You can do that in a copy of the model. A 10 kN horizontal load at the left-hand eaves sways this frame 25.4 mm. Under the gravity combination, the total vertical load is the 288 kN we worked out above, so αcr​ is about (10/288)(5/0.0254)=6.8. Under the wind combination, which carries far less vertical load, it's about 18. (The exact values are a little lower, about 6 and 16; the estimate treats the columns as stiffer than pinned-base columns are.)
Below 10, the code requires the extra moments that develop as the frame sways to be allowed for. So, under the gravity combination, the eaves moments would grow a little before any member check starts.
The columns in compression. Each column has to be checked for flexural buckling about both axes, and for the combination of axial force and bending in a member that can buckle. A stress check only compares the stress with yield, and as we've seen, a slender column can buckle well before any fibre yields. If you'd like a refresher on how this is done, take a look at this EngineeringSkills tutorial on steel column design.
The rafter's lateral-torsional buckling. Under gravity, the top flange is in compression at midspan and the bottom flange is in compression near the eaves. How the rafter is restrained - by purlins on the top flange, and stays to the bottom flange - sets its buckling resistance. So, don't assume the purlins restrain the rafter along its whole length - near the eaves, it's the bottom flange that needs restraint. We cover lateral-torsional buckling in more detail in this EngineeringSkills tutorial on steel beam design.
The eaves connections. We've modelled them as perfectly rigid joints. The real connection, with its bolts, welds and any haunch, has to be designed to deliver that. If the real connection is more flexible than the model assumes, the moments and sway in the real frame won't be the ones our model reports.
None of these checks appear anywhere in the stress results, which is why the list belongs in the routine. Without it, it's very easy to see a utilisation below one and assume the job is done.
5.0 What to take from this pathway
Across eight steps, we've followed the forces in a member all the way to the stresses in its fibres. We've worked out the normal stress from axial force and bending, and the shear stress from shear force and torque. We've combined them into a von Mises stress and compared it against yield as a utilisation, and we've used stress planes as a finite element check on all of it. We've also seen where the stresses on their own stop being enough, and which checks have to follow. As with any analysis technique, it's practice that makes it second nature.
Well done for getting to the end! The very best thing you can do now is open a frame of your own, run the routine, and check one number by hand.