CIV-421 · Lesson 3 of 12 · 21 min
Timber columns and combined loading
Timber columns carry the pavilion roof, the floors of the annex and the walls of the salt shed. Like steel columns, they fail by crushing when they are short and by buckling when they are slender, with a smooth transition in between. Unlike steel, wood has a low modulus and variable stiffness, so even quite stocky timber columns lose a large part of their crushing strength to buckling. This lesson designs the interior columns of the pavilion and then a salt shed post that carries roof load and bending from the salt pile at the same time.
Crushing and buckling
A very short wood column fails by crushing parallel to grain, with fibers folding into a characteristic diagonal kink band. A long one fails by elastic buckling at the Euler load, which depends on the stiffness and the slenderness but not on the strength. Real columns sit in between, and their failure load is reduced by the crookedness of the piece, variations in stiffness along its length, knots and the slow creep of wood under load.
The NDS combines these effects in the column stability factor \(C_P\), from the Ylinen equation. For a rectangular column, the slenderness ratio is \(l_e/d\), the effective length over the dimension in the direction of buckling, and it may not exceed 50. The adjusted compression design value is
Here \(F_c^*\) is the reference compression design value multiplied by all applicable factors except \(C_P\), \(E'_{min}\) the adjusted stability modulus, and \(c\) a constant that reflects how straight and uniform the product is: 0.8 for sawn lumber, 0.85 for round timber poles and 0.9 for glulam and structural composite lumber. As \(l_e/d\) grows, \(F_{cE}\) falls and \(C_P\) approaches the Euler value.
Effective length
The effective length is the unbraced length multiplied by a factor that depends on the end conditions, exactly as in steel design. Timber connections are rarely fixed: bolted and screwed connections allow some rotation, and wood bearing on steel seats acts as a pin. Columns are therefore normally designed as pinned at both ends, with the effective length equal to the actual length between points of lateral support. Bracing in one direction only, such as girts or wall sheathing, reduces the effective length about one axis, so each axis is checked separately with its own length and dimension.
Stud walls
Studs in the annex walls are columns too, braced about their weak axis by the sheathing nailed to them. A 2x6 stud 10 ft tall has a slenderness of \(120/5.5 = 21.8\) about its strong axis and is held by the sheathing about the weak one. Studs share load like joists, so the repetitive member factor applies to their bending design values, and walls with gypsum and wood sheathing on both faces work as composite assemblies. The annex studs at 16 in. carry the second floor and roof with ample reserve; the critical studs are those beside window openings, where the jack studs carry the header reactions.
The pavilion interior columns
Each interior column of the pavilion supports the ends of two glulam roof beams, an ASD reaction of 10.6 kips from each, and is 15 ft tall from the slab to the beam seat. The columns use the same 24F-V4 Douglas fir-larch stock as the beams, with \(F_c = 1650\) psi and \(E_{y,min} = 0.85 \times 10^6\) psi, so the plant can cut them from the same production run. A 6-3/4 in. width matches the beams, so the steel saddle at the top can be simple.
Worked problem
Pavilion interior column
Select a glulam column for \(P = 21.5\) kips including self-weight, pinned at both ends, \(l_e = 15\) ft, D + S (\(C_D = 1.15\)).
Show solution
- Try 5-1/8 x 6: \(l_e/d = 180/5.125 = 35.1\), \(F_{cE} = 0.822(850{,}000)/35.1^2 = 566\) psi, \(F_c^* = 1650(1.15) = 1898\) psi:
- \[C_P = 0.287 \qquad P_{allow} = 1898(0.287)(30.75) = 16.7\ \text{kips} < 21.5\]
- Try 6-3/4 x 7-1/2: \(l_e/d = 180/6.75 = 26.7\):
- \[F_{cE} = \frac{0.822(850{,}000)}{26.7^2} = 983\ \text{psi} \qquad \frac{F_{cE}}{F_c^*} = 0.518\]
- \[C_P = \frac{1.518}{1.8} - \sqrt{\Big(\frac{1.518}{1.8}\Big)^2 - \frac{0.518}{0.9}} = 0.475 \qquad P_{allow} = 1898(0.475)(50.6) = 45.6\ \text{kips} \ge 21.5\]
Answer: Use 6-3/4 x 7-1/2 columns at 47 percent of their allowable load. The 5-1/8 in. width fails because \(C_P\) drops to 0.29 at a slenderness of 35. The extra capacity is useful: it leaves room for a future solar array on the roof and gives a robust column in a public building where a vehicle or a fire could damage one face.
The example shows the most important habit in timber column design: width matters far more than area. Going from 130 to 175 mm (5-1/8 to 6-3/4 in.) increases the area by about two thirds but nearly triples the resistance, because the slenderness falls from 35 to 26 and the stability factor almost doubles.
Bearing at the column ends
At the top, the beams bear on a steel saddle, and the load passes into the column end grain. End-grain bearing is strong: the end-grain bearing design value \(F_c^*\) applies without the stability factor, and a full-width steel plate spreads the load. At the base, the column stands on a steel shoe that lifts it at least 1 in. above the slab, so water and cleaning fluids cannot wick into the end grain, which absorbs water much faster than side grain.
Combined bending and axial load
Many timber members carry axial load and bending together: wall studs under wind, top chords of trusses, posts with eccentric beam seats and, in the salt shed, the posts that support the roof while resisting the push of salt against the wall lining. Axial compression amplifies the bending moment as the member deflects, just as in steel beam-columns.
The NDS interaction equation for compression with bending about one axis has a squared axial term and an amplifier on the bending term:
\(F_c'\) is the compression design value considering buckling about either axis, \(f_{b1}\) the bending stress, and \(F_{cE1}\) the critical buckling value in the plane of bending. The bending design value \(F_{b1}'\) includes \(C_L\) if the compression edge is unbraced.
The salt shed posts
The salt shed walls are lined with heavy timber boards that retain the salt pile. Behind the lining, glulam posts at 4 ft on center carry the lining and part of the roof. Each post is 16 ft from the floor slab to the eave tie, pinned at both ends, and the lining boards brace it about its weak axis at 4 ft. Owen estimates the ASD roof load on one post as 9.0 kips and the salt pressure, applied to the post as a uniform load, as 0.42 kip/ft. Stored salt stays in the shed for months, so the team uses \(C_D = 1.0\) rather than the snow value. The shed is enclosed, and with a protective finish the supplier accepts dry-service values; for an open-sided shed they would not.
# Lesson 3: glulam columns and a salt shed post in combined loading (NDS, ASD)
import math
Fc, Emin = 1650.0, 0.85e6 # 24F-V4 DF used as a column: Fc parallel, Ey,min (psi), approx.
def CP(Fc_star, le_d, Emin_=Emin, c=0.9):
FcE = 0.822 * Emin_ / le_d**2
a = FcE / Fc_star
return (1 + a) / (2 * c) - math.sqrt(((1 + a) / (2 * c))**2 - a / c), FcE
# Interior pavilion column: two beam reactions of 10.6 kips (D + S), height 15 ft, CD = 1.15
P = 2 * 10.6 + 0.3
print(f"Interior column, le = 15 ft, P = {P:.1f} kips")
for b, d in [(5.125, 6.0), (6.75, 7.5), (6.75, 9.0)]:
Fcs = Fc * 1.15
cp, FcE = CP(Fcs, 180 / min(b, d))
Pa = Fcs * cp * b * d / 1000
print(f" {b}x{d}: le/d = {180/min(b,d):.1f}, FcE = {FcE:.0f} psi, CP = {cp:.3f}, Pa = {Pa:.1f} kips")
# Salt shed wall post: 5-1/8 x 10-1/2, 16 ft between floor and eave tie, D + S + H, CD = 1.0
b, d, L = 5.125, 10.5, 192.0
P, w = 9.0, 0.42 * 1000 / 12 # kips and lb/in. (salt pressure on the lining, service)
M = w * L**2 / 8 # lb-in.
A, Sx = b * d, b * d**2 / 6
Fb_ = 2400 * 1.0 * min(1.0, (21 / 16)**0.1 * (12 / d)**0.1 * (5.125 / b)**0.1)
Fcs = Fc * 1.0
cpx, FcEx = CP(Fcs, L / d, 0.95e6) # strong axis, Ex,min
cpy, FcEy = CP(Fcs, 48 / b) # lining braces the weak axis at 4 ft
Fc_ = Fcs * min(cpx, cpy)
fc, fb = P * 1000 / A, M / Sx
ratio = (fc / Fc_)**2 + fb / (Fb_ * (1 - fc / FcEx))
print(f"\nSalt shed post {b}x{d}: Fc' = {Fc_:.0f} psi (CP,x {cpx:.3f}, CP,y {cpy:.3f}), fc = {fc:.0f} psi")
print(f" M = {M/12000:.2f} kip-ft, fb = {fb:.0f} psi, Fb' = {Fb_:.0f} psi, FcE1 = {FcEx:.0f} psi")
print(f" (fc/Fc')^2 + fb/[Fb'(1 - fc/FcE1)] = {(fc/Fc_)**2:.3f} + {fb/(Fb_*(1-fc/FcEx)):.3f} = {ratio:.2f}")Interior column, le = 15 ft, P = 21.5 kips 5.125x6.0: le/d = 35.1, FcE = 566 psi, CP = 0.287, Pa = 16.7 kips 6.75x7.5: le/d = 26.7, FcE = 983 psi, CP = 0.475, Pa = 45.6 kips 6.75x9.0: le/d = 26.7, FcE = 983 psi, CP = 0.475, Pa = 54.7 kips Salt shed post 5.125x10.5: Fc' = 1426 psi (CP,x 0.864, CP,y 0.975), fc = 167 psi M = 13.44 kip-ft, fb = 1713 psi, Fb' = 2400 psi, FcE1 = 2335 psi (fc/Fc')^2 + fb/[Fb'(1 - fc/FcE1)] = 0.014 + 0.769 = 0.78
Worked problem
Salt shed post
Check a 5-1/8 x 10-1/2 24F-V4 post, \(l_e = 16\) ft, \(P = 9.0\) kips, \(w = 0.42\) kip/ft, bending about the strong axis, weak axis braced at 4 ft, \(C_D = 1.0\).
Show solution
- \[M = \frac{0.42(16)^2}{8} = 13.4\ \text{kip·ft} \qquad f_b = \frac{13.4(12{,}000)}{94.2} = 1713\ \text{psi} \qquad F_b' = 2400\ \text{psi}\]
- Compression: strong axis \(l_e/d = 18.3\), \(C_P = 0.864\); weak axis \(l_e/d = 9.4\), \(C_P = 0.975\); so \(F_c' = 1426\) psi and \(f_c = 9000/53.8 = 167\) psi.
- \[F_{cE1} = \frac{0.822(950{,}000)}{18.3^2} = 2335\ \text{psi}\]
- \[\Big(\frac{167}{1426}\Big)^2 + \frac{1713}{2400(1 - 167/2335)} = 0.014 + 0.769 = 0.78 \le 1.0\]
Answer: The post works at 0.78. Bending dominates and the axial term is almost negligible, which is typical of wall posts. The amplifier adds about 8 percent to the bending term. The post is adopted with a stainless steel base shoe on the concrete curb.
Tension members
Timber tension members, such as truss bottom chords and the eave ties of the salt shed, are designed for \(f_t = T/A_n \le F_t' = F_tC_DC_MC_tC_FC_i\) on the net area at the connections. Tension parallel to grain is strong in clear wood but very sensitive to knots and slope of grain, so tension values are lower than bending values. The connections, not the member, usually govern; lesson 5 deals with them. Tension perpendicular to grain is never relied on.
Built-up and spaced columns
Several pieces of lumber nailed or bolted together can form a column, as at the jack studs beside a window. Unless the connection is designed to make the pieces act together, a built-up column is weaker than a solid one of the same size, because the pieces can slip relative to one another as the column bends. Both standards give rules for nailed and bolted built-up columns, with reduction factors that recognize this partial action. Glulam avoids the issue because its laminations are glued.
Common mistakes
MistakeChecking slenderness only about the axis with the larger moment.
InsteadCheck buckling about both axes with their own unbraced lengths and dimensions; the weaker one governs the axial resistance.
MistakeChoosing a narrow column with plenty of area.
InsteadSlenderness depends on the least dimension; widening the column is the most efficient way to gain strength.
MistakeOmitting the \(1 - f_c/F_{cE1}\) term in the interaction equation.
InsteadAxial compression magnifies bending as the member deflects; include the amplifier with the Euler load in the plane of bending.
MistakeSetting column bases directly on the slab.
InsteadLift the end grain above the slab on a steel shoe so it cannot absorb water.
Summary and key terms
- Columns: \(F_c' = F_c^*C_P\) with \(C_P\) from the Ylinen equation, \(c = 0.8\) sawn and 0.9 glulam, and \(l_e/d \le 50\).
- Effective lengths are normally the unbraced lengths, with pinned ends, checked separately about each axis.
- Combined loading: \((f_c/F_c')^2 + f_{b1}/[F_{b1}'(1 - f_c/F_{cE1})] \le 1.0\).
- The pavilion interior column is 6-3/4 x 7-1/2, 45.6 kips allowable against 21.5 kips; the salt shed post, 5-1/8 x 10-1/2, has an interaction ratio of 0.78.
- Width, not area, is the most efficient way to raise column strength.
- Slenderness ratio
- The effective length divided by the cross-section dimension in the direction of buckling.
- Column stability factor
- CP, the reduction in compression design value for buckling.
- Effective length
- The length between points of zero moment in the buckled shape, normally the unbraced length for timber.
- Beam-column
- A member carrying axial compression and bending together.
- Moment amplifier
- The factor that increases the bending effect because of axial load acting on the deflected shape.
- End grain
- The cut surface across the grain at the end of a member.
- Built-up column
- A column made of several pieces fastened together, acting only partly as one piece.
- Eave tie
- A tension member at the eaves that resists the outward thrust of a roof frame.
Practice problems
Work each problem on paper first, then reveal the answer to check.
Find \(C_P\) for a 6-3/4 x 9 glulam column 15 ft long.
Show answer
\(l_e/d = 26.7\) as before, so \(C_P = 0.475\); the allowable load rises to 54.7 kips only because the area grows, since the width is unchanged.
Why is the axial term squared in the timber interaction equation?
Show answer
Tests show timber beam-columns lose less capacity to axial load than a linear equation predicts, especially at low axial loads.
A stud is braced by sheathing about its weak axis. What slenderness governs its compression design?
Show answer
The strong-axis slenderness, the full stud height over its depth.
If the salt pressure on the post doubled to 0.84 kip/ft, would 5-1/8 x 10-1/2 still work?
Show answer
No: the bending term would be about 1.54; a 6-3/4 x 12 post or closer spacing would be needed.
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