In step 3, we applied loads in three directions, but each member still bent in one plane at a time. The rafter bent about its strong axis and the short beam about its weak axis. In this step, we'll load one member in two directions at once. The model is a 4 m column fixed at its base with a horizontal point load at its top along X, another along Y, and the two together as a combination. We'll see how Structure Bench draws and tabulates bending in two planes, check the moment and the deflection in each plane by hand, and then find the most highly stressed fibre in the section, where the stresses from the two planes add.
The model for this step (A 4 m UC 203 x 203 x 46 column fixed at its base, under 10 kN along X, 5 kN along Y, and both together) isn't available yet.
1.0 The idea
A member in a 3D frame has two bending planes. Each one has its own second moment of area, its own bending moment diagram and its own shear force diagram. Structure Bench names them by the local axis the member bends about. Mzโ is bending about local z, which for a beam with its web vertical is the strong axis, and is bending about local y, the weak axis. A member bending in both planes at the same time is in .
Myโ
biaxial bending
The important thing about biaxial bending in a linear analysis is that the two planes don't interact. In each plane, the shear is the slope of the moment diagram, dMzโ/dx=Vyโ and dMyโ/dx=Vzโ, and the deflection comes from integrating that plane's curvature, M/EI, with that plane's own I. For a section that is symmetric about both of its axes, such as a UB, UC, SHS, RHS or CHS, a load along one axis produces no moment, no shear and no deflection about the other. (Channels and angles are the exception, and we'll come back to them in section 4.) So, a column loaded along X and along Y is really two cantilever problems that happen to share a member, and we can solve them one at a time and add the answers. In other words, superposition works across the planes as well as across load cases.
Where the two planes do meet is in the stress. Bending about the strong axis puts its extreme stress on the flange faces, and bending about the weak axis puts its extreme stress at the flange tips. A corner of a flange is both at once, so the normal stresses from the two planes simply add there, and the worst fibre of a biaxially bent section is always at a corner.
The 3D results reflect all of this. Every member end reports six actions (N, Vyโ, Vzโ, T, Myโ and Mzโ), every support reports six reactions, and the two moment diagrams are drawn in their own planes, each with its own switch.
2.0 The model
The column is a single frame member, 4 m tall, from a node at the origin to a node 4 m up. The section is a UC 203 x 203 x 46 from the catalogue in S355 steel, with A = 58.7 cmยฒ, Izโ = 4,570 cmโด about the strong axis and Iyโ = 1,550 cmโด about the weak axis. The base is Fixed, with all six tick-boxes ticked, and the top is free, so the column is a cantilever standing on end.
The column has no roll angle, so it takes the column default from step 2. Local y runs along global X, which means the web runs along X too. So, a load along X acts in the plane of the web and bends the column about its strong axis, and a load along Y bends it about its weak axis. There are two load cases, Along X with 10 kN at the top and Along Y with 5 kN at the top, and a combination, Along X + along Y, which adds them.
Things to try
Press Load the model. The Result set opens on Along X. Switch on Bending moment. Its two chips, Mz and My, light up with it, and a straight line is drawn from 40.0 kNm at the base to zero at the top, in the plane of the load. Click My off and nothing changes, because this case has no weak-axis moment. Click Mz off and the diagram disappears. Switch it back on, switch on Shear force, and hover just above the base. The probe card reads Mz 40.0 kNm and Vy -10.0 kN, and the shear is the same all the way up (its sign gives the direction and its magnitude is the load). Hover the base support, and you'll see a reaction of 10.0 kN along X and a moment of 40.0 kNm about Y, with the other four components zero.
Pick Along Y. Now the diagram is drawn in the other plane, with 20.0 kNm at the base, and it belongs to the My chip. Mz shows nothing. The reaction is 5.0 kN along Y and 20.0 kNm about X. Click FRONT on the view cube, then orbit a quarter turn, and you'll see the two cases bend the column in two planes at right angles.
Pick the combination, Along X + along Y. Both diagrams are drawn at once, each in its own plane, with the same 40.0 and 20.0 kNm at the base, and the Member forces table lists 20.00 kNm under |My| and 40.00 kNm under |Mz| for the one member. Switch on Deformed shape and look at the Displacements table. The top moves 22.229 mm along X and 32.770 mm along Y, exactly the values from the two cases on their own.
With the combination still selected, switch on Member stress and select the ฯ total chip. Zoom in on the base, where both moments are largest, and hover the four corners of the flanges. One corner is the most highly compressed point on the section, and the diagonally opposite corner carries the largest tension. The stresses at the other two corners are lower, because at those corners the stresses from the two planes have opposite signs. Compare what the card reads at the worst corner with the 220 MPa of the hand check below.
If you're a member, press Copy to my projects: your copy opens in the editor. Pick the column and, in its properties, set roll (ยฐ) to 45, then press Solve and pick the combination. The moments at the base are now 42.4 kNm of Mz and 14.1 kNm of My. These come from the same loads, resolved onto a section that has turned. The top moves 33.06 mm along X and only 0.29 mm along Y, even though the load has a Y component. Once the section's axes aren't lined up with the load, the column doesn't deflect in the direction of the load.
Set the roll back to 0 and add a third load case, Axial, with a nodal load at the top of Fz -300 kN. In the Load combinations section press + combination and build one with all three cases at a factor of 1.0, then solve and pick it. The two moments haven't moved, the base reaction gains 300.0 kN upwards, and the top drops 0.973 mm. Switch on Member stress with the ฯ total chip again. Every fibre has gained the same compression, so the line of zero stress has moved off the centroid towards the tension corner.
3.0 What you'll see
Two planes, one column
The solver
By hand
Moment at the base, Along X (Mz)
40.0 kNm
PL=10ร4=40 kNm
Moment at the base, Along Y (My)
20.0 kNm
PL=5ร4=20 kNm
Base reaction, Along X
-10.0 kN along X, -40.0 kNm about Y
the load and its moment, reversed
Base reaction, Along Y
-5.0 kN along Y, 20.0 kNm about X
the same
Top deflection, Along X
22.23 mm
PL3/3EIzโ=22.23 mm
Top deflection, Along Y
32.77 mm
PL3/3EIyโ=32.77 mm
Top rotation, Along X then Along Y
8.336 mrad, 12.289 mrad
PL2/2EI=8.34 and 12.29 mrad
Moments and top deflections, the combination
the four values above, unchanged
superposition
Rolled 45ยฐ, the combination
42.4 kNm of Mz, -14.1 kNm of My
(40+20)/2โ=42.4, kNm
With 300 kN of axial load, the top drops by
-0.97 mm
NL/EA=0.97 mm
Corner stress at the base, the combination
read it from the ฯ card
Mzโ/Wzโ+M MPa
The reactions carry the sign of their direction, as the table prints them, so each base reaction acts against its load. The rolled column's negative My says which face that moment stretches.
The two cases are textbook cantilevers, and the combination is simply the two added together. Notice that the Y load is half the X load but deflects the top half as much again, 32.8 mm against 22.2, because the weak axis has barely a third of the strong axis's second moment of area. The same thing happens in the stress, and this is the key lesson of this step. The 20 kNm about the weak axis produces a larger stress at the corner than the 40 kNm about the strong axis does. If you need a refresher on why one section has two very different second moments of area, take a look at this EngineeringSkills tutorial, The Second Moment of Area.
Hand check: each plane on its own, then the corner
The moments. Each horizontal load at the top of a cantilever puts a moment of PL at the base, growing linearly from zero at the free end. Along X,
Mzโ=PxโL=10ร4=40ย kNm
and along Y,
Myโ=PyโL=5ร4=
The solver reads 40.0 and 20.0 kNm at the base, each in its own plane, and the shear in each plane is the load, 10 kN and 5 kN, constant over the height.
The deflections. The tip of a cantilever under a point load moves PL3/3EI, and here each plane has its own EI. With E = 210 GPa, the strong axis has kNmยฒ, so along X,
ฮดxโ=3EI
which is the solver's 22.229 mm. The weak axis has EIyโ=210ร106ร1, kNmยฒ, so along Y,
ฮดyโ=3EI
against the solver's 32.770 mm. The tip rotation is PL2/2EI in each plane: 10ร16/(2ร9,597)= rad, the 8.336 mrad in the Displacements table, and rad, the 12.289 mrad.
The corner. The normal stress from bending is M/W, where W=I/(h/2) is the elastic section modulus. From the catalogue, h = 203.2 mm and b = 203.6 mm, so,
Wzโ=10.164,570โ
and the two bending stresses at the base are,
WzโM
At the corner where both planes compress, the stress is 88.9+131.3=220.3 MPa of compression, and at the diagonally opposite corner the same 220.3 MPa of tension. This is the hand-calculated value. The member stress view samples the section on its own grid, so hover the corner at the base and read what the card prints there, then compare the two.
The rolled column. Rolling the section by 45ยฐ changes neither the loads nor the moment they produce at the base, only the pair of axes we resolve that moment onto. The base moment has a 40 kNm component from the X load and a 20 kNm component from the Y load, at right angles to each other. The rolled axes sit at 45ยฐ to both, so each component contributes cos45ยฐ of itself to each axis,
Mzโ=2
against the solver's 42.426 and 14.142. The deflection works the same way in reverse. We resolve the load onto the rolled axes, work out the deflection from each component in its own plane, and add the two movements back together. Because the weak plane is three times as flexible, the two movements nearly cancel along Y and add up along X, which is why the top moves 33.06 mm along X and only 0.29 mm along Y.
The axial load. A 300 kN compression on a 58.7 cmยฒ section adds 300,000/5,870=51.1 MPa of compression to every fibre, so the corners now read 220.3+51.1=271.4 MPa of compression and 220.3 MPa of tension, with the zero line nearer the tension corner. The column shortens by,
ฮด=EANLโ=
the 0.973 mm in the Displacements table. The moments don't change at all, because in a first-order analysis the axial load acts on the undeflected column.
4.0 What the two planes don't share
Uncoupled planes, doubly symmetric sections
The two bending planes and torsion are solved independently of each other at element level, with the section's Izโ and Iyโ taken about the local axes and the shear centre taken to sit at the centroid. That's exact for a section that is symmetric about both of its axes (every UB, UC, SHS, RHS and CHS in the catalogue), because its principal axes are the local axes and its shear centre really is at its centroid. It's wrong for a section that isn't, and the catalogue has two kinds that aren't: the PFC channels and the L angles. A real channel's shear centre sits outside its web, so a load parallel to the web acting through its centroid twists it as well as bending it. A real angle's principal axes are inclined to its legs, so a load along one leg bends it about both principal axes. The solver uses the listed properties about the local axes and reports no twist and no coupling, and the same goes for a custom rectangle standing in for either. So a channel or an angle in a 3D model is a member to check by hand.
Remember too that the member stress view is based on elastic beam theory. The 220 MPa at the corner is a number you need to check against the steel's yield strength yourself.
5.0 What to take from this step
A 3D member bends in two planes, and Structure Bench reports them separately. Mzโ and Myโ each have their own diagram, their own chip under the Bending moment layer, their own column in the Member forces table and their own entry on the probe card. Within a linear analysis the planes don't interact, so each one can be checked with the formulas you already know, using that plane's second moment of area. A combination is then the two added together.
The planes meet at the corners of the section, where the normal stresses add. Because the weak axis has the smaller section modulus, a modest weak-axis moment can produce a larger stress at a corner than a bigger strong-axis moment. So, when a member bends in both planes, check the stress at a corner, not just on a flange face. If the section is rolled, resolve the loads onto the section's own axes before you reach for a formula. If you've worked through the stress analysis pathway, you've already met this column from the stress side. Here, we've approached it from the internal actions associated with the stress.
Next, we'll look at the third rotation a 3D member has, torsion.