Wood Design – Tutorial #3

Trevor Winchester

Wood Design – Tutorial #3

Problem 1

Use design aids to select a joist for the following conditions. Completed detailed calculations to confirm the actual values. Consider moment, shear, and deflection. Given information:

l = 4 m (length of joist)
S = 300 mm (spacing between joists)
D = 0.15 kPa (dead load, includes self-weight)
L = 2.4 kPa (live load)
Joist is simply supported
Limit deflections to l/180 from all loads
Wet service conditions
Case 2 system factor

Solution:

Determine the conditions to select the joist (i.e., required moment and shear resistance and stiffness):

Factored uniformly distributed load:

$w_f=1.25D+1.5L$

$w_f=(1.25\times0.15kPa+1.5\times2.4kPa)\times0.3m=1.14kN/m$

Factored shear:

$V_f=\frac{w_fl}{2}=\frac{1.14kN/m\times4m}{2}=2.28 kN$

Factored moment:

$M_f=\frac{w_fl^2}{8}=\frac{1.14kN/m\times(4m)^2}{8}=2.28 kNm$

Unfactored (service) uniformly distributed load:

$w=D+L$

$w=(0.15kPa+2.4kPa)\times0.3m=0.765kN/m$

Deflection requirements due to total service loads:

$\bigtriangleup\leq\frac{l}{180}$

$\frac{5wl^4}{384\timesE_sI}=\frac{5\times0.765N/mm\times(4000mm)^4}{384\times(E_sI)_r_e_q}\le\frac{4000mm}{180}$

$(E_sI)_r_e_q\ge 115\times10^9Nmm^2$

Note: The deflection requirements are outlined in Cl. 5.4. Here, only the limit for all service loads is considered. The sustained loads (i.e., dead loads) will not govern since they are significantly less than the live loads. The limit of L/360 is not a code requirement but is based on best practices.

Limit due to live loads only:

$w_l=L$

$w_l=2.4kPa\times0.3m=0.72 kN/m$

$\bigtriangleup\leq\frac{l}{360}$

$\frac{5wl^4}{384E_sI}=\frac{5\times0.72N/mm\times(4000mm)^4}{384(E_sI)_r_e_q}\le\frac{4000mm}{360}$

$(E_sI)_r_e_q\ge 216\times10^9Nmm^2$

Thus, deflection due to live load only would govern if considered.

The above values are our design requirements for shear, moment and deflection. However, to use the design tables, we must correct for the wet service conditions. The checklist for selecting joists is on Pg. 37. On Pg. 38, the corrections for wet service conditions are provided:

$K_S_b = 0.84 for M_r$

$K_S_v = 0.96 for V_r$

$K_S_E = 0.94 for E_sI$

Note: The same values for wet service condition are available in Table 6.10 of CSA O86, Pg. 57.

Apply the corrections to our design values gives the minimum requirements we can use to select a joist from the tables:

$V_r_t\ge \frac{V_f}{K_S_v}=\frac{2.28 kN}{0.96}=2.38 kN$

$M_r_t\ge \frac{M_f}{K_S_b}=\frac{2.28 kNm}{0.84}=2.71 kNm$

$E_sI_t\geq\frac{(E_sI)_r_e_q}{K_S_E}=\frac{115\times10^9Nmm^2}{0.94}=122\times10^9Nmm^2$

Select a SPF No.1/No.2 38×184 (2×8) (based on Case 2 system factor) (Pg. 43):

$V_r=10.6 kN>V_r_t=2.38 kN$

$M_r=3.83kNm>M_r_t=2.71 kNm$

$E_sI=187\times10^9Nmm^2>E_sI_t=122\times10^9Nmm^2$

Note: From the table both moment and deflection governed the selection.

Detailed calculations can now be completed to determine the actual values of the above three criteria and show that they exceed the requirements. We expect the actual resistance and stiffness to be lower than what is shown above due to the wet service conditions.

Check moment resistance (Cl. 6.5.3):

fb = 11.8 MPa (Table 6.4, Structural Joist + Plank)

KD = 1.0 (standard term) (Cl. 5.3.2 & Table 5.1)

KH = 1.4 (Cl. 5.3.5, Cl. 6.4.4 & Table 6.12, Case 2)

KSb = 0.84 (wet service condition) (Cl. 5.3.3, Cl. 6.4.2 & Table 6.10)

KT = 1.0 (no treatment) (Cl. 5.3.4, Cl. 6.4.3 & Table 6.11)

$F_b=f_b(K_DK_HK_S_b_KT)$

$F_b=11.8 MPa\times(1.0\times1.4\times0.84\times1.0)=13.9 MPa$

KZb = 1.2 (Cl. 5.3.6, Cl. 6.4.5 & Table 6.13)

KL = 1.0 (See below discussion)

The selected beam has a d/b (184 mm / 38 mm) ratio of 4.84, which means additional lateral support is required (Cl. 6.5.3.2). If we assume that a direct connection is provided to the decking, then adequate support is provided and KL = 1.0.

$S=\frac{1}{6}bh^2=\frac{1}{6}38mm\times(184mm)^2=214\times10^3mm^3$

$M_r=φF_bSK_Z_bK_L$

$M_r=0.9\times13.9MPa\times214\times10^3mm3^\times1.2\times1.0=3.2kNm\ge Mf=2.28 kNm$

Therefore, the section satisfies the moment requirements.

Check shear resistance (Cl. 6.5.4.3, no notch):

fv = 1.5 MPa (Table 6.4, Structural Joist + Plank)

KD = 1.0 (standard term) (Cl. 5.3.2 & Table 5.1)

KH = 1.4 (Cl. 5.3.5, Cl. 6.4.4 & Table 6.12, Case 2)

KSv = 0.96 (wet service condition) (Cl. 5.3.3, Cl. 6.4.2 & Table 6.10)

KT = 1.0 (no treatment) (Cl. 5.3.4, Cl. 6.4.3 & Table 6.11)

$F_v=f_v(K_DK_HK_S_v_KT)$

$F_v=1.5 MPa\times(1.0\times1.4\times0.96\times1.0)=2.02 MPa$

$K_Z_v$

= 1.2 (Cl. 5.3.6, Cl. 6.4.5 & Table 6.13)

$V_r=\phiF_v\frac{2A_n}{3}K_z_v$

$V_r=0.9\times2.02MPa\times\frac{2}{3}\times38 mm\times184 mm\times1.2=10.2 kN\ge V_f=2.28 kN$

Therefore, the section satisfies the shear requirements.

Check stiffness requirement (Cl. 5.4.1):

E = 9500 MPa (Table 6.4, Structural Joist + Plank)

KSE = 0.94 (wet service condition, Table 6.10)

KT = 1.0 (no treatment)

$E_s=E(K_S_EK_T)$

$E_s=9500 MPa (0.94\times1.0)=8930MPa$

$I=\frac{1}{12}bh^3=\frac{1}{12}38 mm\times(184 mm)^3=19.7\times10^6mm^4$

Therefore:

$E_sI=8930MPa\times19.7x10^6mm^4=176\times10^9Nmm^2 \ge (E_sI)_r_e_q=115\times10^9Nmm^2$

Thus, the deflection requirements are satisfied, and the selected section is adequate.

Problem 2

An existing structure is being renovated to accommodate new manufacturing equipment. The intended use requires the installation of significant additional mechanical equipment. The existing structure was designed to support a dead load of 1.25 kPa and a live load of 4.8 kPa. The loads for the renovated structure are D = 4.0 kPa and L = 3.6 kPa.

The existing structure contains 241 mm x 343 mm HF SS beams with a tributary width of 2 m and length of 5.0 m. Determine if the existing beams are adequate to resist the applied loads required for the renovated structure. If the beam is not adequate, determine what is the maximum dead load that can be applied given a live load of 3.6 kPa.

Consider moment resistance. Assume that adequate bracing is provided.

Consider shear resistance

Consider deflection resistance (limit deflection due to total loads to L/180)

Solution:

Part a)

Factored uniformly distributed load:

$w_f=1.25D+1.5L$

$w_f=(1.25\times4.0 kPa+1.5\times3.6 kPa)2.0 m=20.8 kN/m$

Factored moment:

$M_f=\frac{w_fl^2}{8}=\frac{20.8 kN/m\times(5.0 m)^2}{8}=65.0 kNm$

Check moment resistance (Cl. 6.5.3):

Since the long term loads exceed the standard term loads, Cl. 5.3.2.2 applies for determining the load duration factor, KD.

$K_D=1.0−0.50 log(PL/Ps) \ge 0.65$

$K_D=1.0−0.50 log(4.0 kPa/3.6 kPa) =0.98\ge 0.65$

fb = 16.8 MPa (Table 6.6, Beam and Stringer)

KD = 0.98 (see above) (Cl. 5.3.2.2)

KH = 1.0 (no system) (Cl. 5.3.5, Cl. 6.4.4 & Table 6.12)

KSb = 1.0 (dry service condition) (Cl. 5.3.3, Cl. 6.4.2 & Table 6.10)

KT = 1.0 (no treatment) (Cl. 5.3.4, Cl. 6.4.3 & Table 6.11)

$_b=f_b(K_DK_HK_S_bK_T)$

$F_b=16.8 MPa(0.98\times1.0\times1.0\times1.0)=16.5 MPa$

KZb = 1.0 (Cl. 5.3.6, Cl. 6.4.5 & Table 6.13)

KL = 1.0 (assumed)

$S=\frac{1}{6}bh^2=\frac{1}{6}241 mm∙(343 mm)^2=4,726\times10^3mm^3$

$M_r=\phiF_bSK_Z_bK_L$

$M_r=0.9\times16.5 MPa\times4,726\times10^3mm^3∙1.0∙1.0=70.2 kNm>M_f=65.0 kNm$

Therefore, the section satisfies the moment requirements.

Part b)

Factored shear:

$V_f=\frac{w_fl}{2}=\frac{20.8 kN/m∙5m}{2}=52.0 kN$

Check shear resistance (Cl. 6.5.4.3, no notch):

fv = 1.2 MPa (Table 6.6, Beam and Stringer)

KD = 0.98 (see above) (Cl. 5.3.2)

KH = 1.0 (Cl. 5.3.5, Cl. 6.4.4 & Table 6.12, no system)

KSv = 1.0 (dry service condition) (Cl. 5.3.3, Cl. 6.4.2 & Table 6.10)

KT = 1.0 (no treatment) (Table 6.11)

$F_v=f_v(K_DK_HK_S_vK_T)$

$F_v=1.2 MPa(0.98\times1.0\times1.0\times1.0)=1.18 MPa$

KZv = 1.0 (Table 6.13)

$V_r=\phiF_v\frac{2A_n}{3}K_z_v$

$V_r=0.9\times1.18 MPa\times\frac{2}{3}\times241 mm\times343 mm\times1.0=58.5 kN\geqV_f$

Therefore, the section satisfies the shear requirements.

Part c)

Unfactored (service) uniformly distributed load:

$w=D+L$

$w = (4.0 kPa+3.6 kPa)2.0m = 15.2 kN/m$

E = 11500 MPa (Table 6.6, Beam and Stringer)

KSE = 1.0 (dry service conditions)

KT = 1.0 (no treatment)

$E_s=E(K_S_EK_T)$

$E_s=11,500 MPa$

Solve the second moment of area:

$I=\frac{1}{12}bh^3$

$I=\frac{1}{12}241 mm\times(343 mm)^3=810\times10^6mm^4$

Deflection requirements due to total service loads:

$\bigtriangleup\leq\frac{l}{180}$

$\bigtriangleup=\frac{5wl^4}{384E_sI}$

$\frac{5wl^4}{384E_sI}=\frac{5\times15.2N/mm(5000mm)^4}{384\times11,500 MPa\times810\times10^6mm^4}\leq\frac{5000mm}{180}$

$13.3 mm\leq27.8mm$

Therefore, the section satisfies the deflection requirements.

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Problem 1

Solution:

Problem 2

Solution:

Part a)

Part b)

Part c)

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