CEE 454 HW #2 Solution
4.33 Determine the tabulated and allowable design values for the following members and loading conditions. All members are Select Structural Southern Pine. Bending occurs about the strong axis.
** See Chapter 4 of NDS (not supplement) or Section 4.12 – 4.22 of the textbook for the detail.
- Structural Southern Pine → NDS Supplement Table 4B (p37 ~ )
a. Roof joists are 2*6 at 24 in. o.c. which directly support the roof sheathing. Loads are (D + S).
- Given Condition: (D + S), 2*6, Roof joist
- From the table (p38), the tabulated values are as following
(unit: psi)
Fb / Ft / Fv / Fc / Fc / E2550 / 1400 / 175 / 565 / 2000 / 1800000
- The allowable design values after applying appropriate adjustment factors are as following
Fb’ = Fb *CD * CM * Ct * CF * Cr = 2550*1.15*1.0*1.0*1.0*1.15 = 3372 psi
(Note that this problem is for a joist. → Cr = 1.15)
Ft’ = Ft *CD * CM * Ct * CF= 1400*1.15 = 1610 psi
Fv’ = Fv *CD * CM * Ct = 201 psi
Fc’ = Fc * CM * Ct = 565 psi
Fc’ = Fc * CD * CM * Ct * CF = 2300 psi
E’ = E * CM * Ct = 1800000 psi
b. 4*12 supports (D + L + Lr)
- Given Condition: (D + L + Lr), 4*12
- From the table (p38), the tabulated values are as following
(unit: psi)
Fb / Ft / Fv / Fc / Fc / E1900 / 1050 / 175 / 565 / 1800 / 1800000
- The allowable design values after applying appropriate adjustment factors are as following
Fb’ = Fb *CD * CF = 1900*1.25*1.1 = 2613 psi (Note that CF is applied only to Fb.)
Ft’ = Ft *CD = 1050*1.25 = 1313 psi
Fv’ = Fv *CD = 219 psi
Fc’ = Fc = 565 psi
Fc’ = Fc * CD = 2250 psi
E’ = E = 1800000 psi
c. Purlins in a roof are 2*10 at 4 ft o.c. Loads are (D + Lr)
- Given Condition: (D + Lr), 2*10, spacing = 4 ft → Cr = 1.0
- From the table (p38), the tabulated values are as following
(unit: psi)
Fb / Ft / Fv / Fc / Fc / E2050 / 1100 / 175 / 565 / 1850 / 1800000
- The allowable design values after applying appropriate adjustment factors are as following
Fb’ = Fb *CD = 2050*1.25 = 2563 psi
Ft’ = Ft *CD = 1100*1.25 = 1375 psi
Fv’ = Fv *CD = 219 psi
Fc’ = Fc = 565 psi
Fc’ = Fc * CD = 2313 psi
E’ = E = 1800000 psi
d. Floor beams are 4*10 at 4 ft o.c. Loads are (D + L + W)
- Given Condition: (D + L + W), 4*10, spacing = 4 ft → Cr = 1.0
- From the table (p38), the tabulated values are as following
(unit: psi)
Fb / Ft / Fv / Fc / Fc / E2050 / 1100 / 175 / 565 / 1850 / 1800000
- The allowable design values after applying appropriate adjustment factors are as following
Fb’ = Fb *CD * CF = 2050*1.6*1.1 = 3608 psi (Note that CF is applied only to Fb.)
Ft’ = Ft *CD = 1100*1.6 = 1760 psi
Fv’ = Fv *CD = 280 psi
Fc’ = Fc = 565 psi
Fc’ = Fc * CD = 2960 psi
E’ = E = 1800000 psi
5.11 Given: A 5 1/8* 28.5 24F-1.8E Douglas Fir glulam is used to span 32 ft, carrying a load of (D + S). The load is a uniform load over a simple span, and the beam is supported so that buckling is prevented.
a. Sketch the beam and the cross section. Show calculations to verify the section properties Sx and Ix for the member, and compare with values in NDS supplement Table 1C.
- → agree with the NDS supplement value
- → agree with the NDS supplement value
b. Determine the allowable stresses associated with the section properties in part a. These include Fbx+, F’cx, F’vx, and E’x.
** See Chapter 5 of NDS or Section 5.5 – 5.7 of the textbook
- CD = 1.15, Cv =
- Fbx+ = 2400*1.15*0.88 = 2429 psi
- F’cx = 650 psi
- F’vx = 265*1.15 = 305 psi
- E’x = 1800 ksi
c. Repeat part b except the moisture content of the member may exceed 16%
- From NDS supplement p59, CM for each stresses will be used as following
Fbx+ / F’cx / F’vx / E’x0.8 / 0.53 / 0.875 / 0.833
- Fbx+ = 2429*0.8 = 1943 psi
- F’cx = 650*0.53 = 345 psi
- F’vx = 305*0.875 = 267 psi
- E’x = 1800*0.833 = 1499 ksi
6.1 Given: The roof beam in Fig. 6.A with the following information:
Load: / P = 2 k /Load combination: / D + Lr
Span: / L = 8 ft
Member size: / 4*8
Stress grade and species: / No. 1 DF-L
Unbraced length: / lu = 0
Moisture Content: / MC 19 %
Deflection limit: / Allow. L/360
a. Size category
- Dimension lumber based on the member size
b. Tabulated stresses
- Fb = 1000 psi, Fv = 180 psi, E = 1700 ksi
c. Allowable stresses
- CD = 1.25, CF = 1.3
- Fb’ = 1000*1.25*1.3 = 1625 psi
- Fv’ = 180*1.25 = 225 psi
- E’ = 1700 ksi
- allow = 8*12/360 = 0.267 in
d. Actual stresses
- A = 25.38 in2Sx = 30.66 in3, Ixx = 111.1 in4 from NDS supplement Table 1B
- M = PL/4 = 2*1000*8*12/4 = 48000 in-lb
→ fb = M/Sx = 48000/30.66 = 1566 psi
- fv = psi
- Δ = in
e. Compare the actual and allowable design values and determine if the member is adequate.
- For bending, 1625 > 1566 → OK
- For shear, 225 > 59.1 → OK
- For deflection, 0.267 > 0.195 → OK
6.2 Repeat Prob. 6.2 except the moisture content exceeds 19%
- CM for bending, shear, and E are 0.85, 0.97, and 0.9, respectively
Part c.
- Fb’ = 1625*0.85 = 1381 psi
- Fv’ = 225*0.97 = 218 psi
- E’ = 1700*0.9 = 1530 ksi
Part d.
- Δ = in
Part e
- For bending, 1381 < 1566 → NG
- For shear, 218 > 59.1 → OK
- For deflection, 8*12/360 = 0.267 > 0.217 → OK