✔Homework 4 (Due Monday, October 1)

David Wagner 2007/09/30 18:52

Solve the following problems in your textbook:

✔4.23(0.5 points)

What reference design values for wood, if any, are not subject to adjustment for duration of loading (ASD)? For time effect (LRFD)?

For sawn lumber, Fc, E, and Emin are not adjusted for the ASD load duration factor, while only E and Emin are not adjusted for the LRFD time effect.

✔4.24(0.5 points)

Above what moisture content is it necessary to reduce the reference design values for most species of

  • (a) sawn lumber: 19%
  • (b) glulam: 16%

✔4.27(0.5 points)

Should lumber be pressure-treated if it is to be used in an application where it will be continuously submerged in

  • fresh water? No
  • Salt water? Yes

Explain. When continuously submerged in fresh water, wood does not decay appreciably. However, according to the text, page 4.23, “Pressure treated piles have an extensive record in resisting attack by marine borers.”

✔4.30(1.5 points)

Determine the ASD reference and adjusted design values for the following members and loading conditions. All members are No. 2 Hem-Fir and are fully braced to prevent lateral-torsional buckling. Bending occurs about the strong axis.

a. Roof joists are 2 x 10 at 16 in. o.c. which directly support the roof sheathing. Loads are (D + S).

SymbolReference Design ValueCDCMCtCLCFCfuCiCrCPCTCbAdjusted Design Value
Fb 850 1.15 1 1 1 1.1 1 1 1.15 1237
Ft 525 1.15 1 1 1.1 1 664
Fv 150 1.15 1 1 1 173
Fc⊥ 405 1 1 1 1.19 782
Fc 1300 1.15 1 1 1.0 1 1 1495
E 1.3e6 1 1 1 1.3e6
Emin 0.47e6 1 1 1 1 0.47e6

b. A 6 x 14 carries an equipment load that can be considered a permanent load.

Assuming temperatures 100-125ºF.

C_F=(12/d)^{1/9}=root{9}{12/14}=0.98

SymbolReference Design ValueCDCMCtCLCFCfuCiCrCPCTCbAdjusted Design Value
Fb 375 0.9 1 0.8 1 0.98 1 1 1 265
Ft 325 0.9 1 0.9 0.98 1 258
Fv 135 0.9 1 0.8 1 97
Fc⊥ 405 1 0.8 1 1.00 324
Fc 475 0.9 1 0.8 0.98 1 1 335
E 1.1e6 1 0.9 1 0.99e6
Emin 0.4e6 1 0.9 1 1 0.36e6

c. Purlins in a roof are 4 x 14 at 8 ft o.c. Loads are (D + Lr).

Assuming temperatures 100-125ºF.

SymbolReference Design ValueCDCMCtCLCFCfuCiCrCPCTCbAdjusted Design Value
Fb 850 1.25 1 0.8 1 1.0 1 1 1 850
Ft 525 1.25 1 0.9 0.9 1 532
Fv 150 1.25 1 0.8 1 150
Fc⊥ 405 1 0.8 1 1.10 356
Fc 1300 1.25 1 0.8 0.9 1 1 1170
E 1.3e6 1 0.9 1 1.17e6
Emin 0.47e6 1 0.9 1 1 0.42e6

d. Floor beams are 4 x 6 at 4 ft o.c. Loads are (D + L). High-humidity conditions exist, and the moisture content may exceed 19 percent.

SymbolReference Design ValueCDCMCtCLCFCfuCiCrCPCTCbAdjusted Design Value
Fb 850 1 1.00* 1 1 1.3 1 1 1 1105
Ft 525 1 1.00 1 1.3 1 683
Fv 150 1 0.97 1 1 146
Fc⊥ 405 0.67 1 1 1.10 298
Fc 1300 1 0.80† 1 1.1 1 1 1144
E 1.3e6 0.90 1 1 1.17e6
Emin 0.47e6 0.90 1 1 1 0.42e6
  • *F_b C_F≤ 1150 psi
  • F_c C_F> 750 psi

✔4.31(1.5 points)

Determine the LRFD nominal and adjusted design values and adjusted design resistances for the following members and loading conditions. All members are No. 2 Hem-Fir and are fully braced to prevent lateral-torsional buckling. Bending occurs about the strong axis.

a. Roof joists are 2 x 10 at 16 in. o.c. which directly support the roof sheathing. Loads are 1.2D + 1.6S.

A=13.88 in² ; S=21.39 in³

SymbolReference Design ValueCMCtCLCFCfuCiCrCPCTCbKFφxλAdjusted Design ValueA or SAdjusted Design Resistance
Fb 850 1 1 1 1.1 1 1 1.15 2.54 0.85 0.8 1858 21.3939.7 kip-in
Ft 525 1 1 1.1 1 2.70 0.80 0.8 998 13.8813.9 k
Fv 150 1 1 1 2.88 0.75 0.8 259 13.883.6 k
Fc⊥ 405 1 1 1 1.19 2.08 0.90 0.8 723 13.8810.0 k
Fc 1300 1 1 1 1 xx1x 2.40 0.90 0.8 2246 13.8831.2 k
E 1.3e6 1 1 1 1.3e6
Emin 4.7e6 1 1 1 1 1.76 0.85 0.71e6

b. A 6 x 16 carries an equipment load that can be considered storage using 1.2D + 1.6L.

A=85.25 in² ; S=220.2 in³ ; assume temperature < 100ºF

C_F=(12/d)^{1/9}=root{9}{12/16}=0.97

SymbolReference Design ValueCMCtCLCFCfuCiCrCPCTCbKFφxλA justed Design ValueA or SAdjusted Design Resistance
Fb 375 1 1 1 0.97 1 1 1 2.54 0.85 0.7 550 220.2 121.1 kip-in
Ft 325 1 1 0.97 1 2.70 0.80 0.7 477 85.25 40.6 k
Fv 135 1 1 1 2.88 0.75 0.7 204 85.25 17.4 k
Fc⊥ 405 1 1 1 1 2.08 0.90 0.7 532 85.25 45.3 k
Fc 475 1 1 0.97 1 1 2.40 0.90 0.7 697 85.25 59.4 k
E 1.1e6 1 1 1 1.1e6
Emin 0.4e5 1 1 1 1 1.76 0.85 0.6e6

c. Purlins in a roof are 4 x 14 at 8 ft o.c. Loads are 1.2D + 1.6Lr.

Assume fairly hot.

A=46.38 in² ; S=102.4 in³

SymbolReference Design ValueCMCtCLCFCfuCiCrCPCTCbKFφxλAdjusted Design ValueA or SAdjusted Design Resistance
Fb 850 1 0.8 1 1 1 1 1 2.54 0.85 0.8 1175 102.4 120 kip-in
Ft 525 1 0.9 0.9 1 2.70 0.80 0.8 735 46.38 34.1 k
Fv 150 1 0.8 1 2.88 0.75 0.8 207 46.38 9.6 k
Fc⊥ 405 1 0.8 1 1.1 2.08 0.90 0.8 535 46.38 54.8 k
Fc 1300 1 0.8 0.9 1 1 2.40 0.90 0.8 1617 46.38 75.0 k
E 1.3e6 1 0.9 1 1.17e6
Emin 0.47e6 1 0.9 1 1 1.76 0.85 0.63e6

d. Floor beams are 4 x 6 at 4 ft o.c. Loads are 1.2D + 1.6L with L from occupancy. High-humidity conditions exist, and the moisture content may exceed 19 percent.

A=19.25 in² ; S=17.65 in³

SymbolReference Design ValueCMCtCLCFCfuCiCrCPCTCbKFφxλAdjusted Design ValueA or SAdjusted Design Resistance
Fb 850 1 1 1 1.3 1 1 1 2.54 0.85 0.8 1909 17.65 33.7 kip-in
Ft 525 1 1 1.3 1 2.70 0.80 0.8 1179 19.25 22.7 k
Fv 150 0.97 1 1 2.88 0.75 0.8 251 19.25 4.8 k
Fc⊥ 405 0.67 1 1 1.1 2.08 0.90 0.8 448 19.25 8.6 k
Fc 1300 0.8 1 1.1 1 1 2.40 0.90 0.8 1977 19.25 38.0 k
E 1.3e6 0.9 1 1 1.17e6
Emin 0.47e6 0.9 1 1 1 1.76 0.85 0.64e6

✔4.35(1.5 points)

Estimate the total shrinkage that will occur in a four-story building similar to the one in Example 4.3. Floor joists are 2 x 10's instead of 2 x 12's. The initial moisture content can be taken as 19 percent, and the final moisture content is assumed to be 9 percent. All other information is the same as in Example 4.3.

SV = 0.002, Delta_{MC}=19-9=10 %

S_{floor}=SV d Delta_{MC} = 0.002*9.25*10=/m>0.185 in

<m>S_{plate}=SV d Delta_{MC} =0.002*1.5*10=0.03 in

S = Sigma S = 3*S_{floor} + (4+8)*S_{plate}=3*0.185+12*0.03 =0.555+.36=0.915

Total S = 0.915 in. ≈ 1 in.

✔5.1(0.5 points)

What is the usual thickness of lamination used to fabricate glulam members from

a. Western Species: 1½” b. Southern Pine: 1 3/8”

c. Under what conditions would thinner laminations be used?

Thinner laminations are used to construct sharply-curved members.

✔5.2(0.5 points)

What are the usual widths of glulam members fabricated from

a. Western Species: 3 1/8, 5 1/8, 6 3/4, 8 3/4, 10 3/4 b. Southern Pine: 3, 5, 6 3/4, 8½, 10½

✔5.11(1.5 points)

Given: 5 1/8 x 28.5 24F-1.8E Douglas Fir glulam is used to span 32 ft, carrying an ASD load combination 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.

S_x = {bh^2}/6 = {5.125*28.5^2}/6 =693.8 approx693.8 in table 1C

693.8 in3

I_x = {bh^3}/12 = {5.125*28.5^3}/12 =9886.6 approx9887 in table 1C

9886.6 in4

b. Determine the adjusted ASD values associated with the section properties in part a. These include F+'bx F'ux and F'x.

Assume normal temperatures.

C_V = root{10}{{21*12*5.125}/{Ldb}} =root{10}{1291.5/{32*5.125*28.5}}root{10}{0.2763}=0.8793 approx0.88 <1✔

SymbolReference Design ValueCDCMCtCLCVCfuCcCPCbAdjusted Design Value
Fbx+ 2400 1.15 1 1 1 0.88 1 1 2429
Fbx- 1450 1.15 1 1 1 0.88 1 1 1467
Fc⊥x 650 1 1 650
Fvx 265 1.15 1 1 1 305
Ex 1.8e6 1 1 1.8e6
Ex~min 0.93e6 1 1 0.93e6
Fby- 1450 1.15 1 1 1 1 1 1 1668
Fc⊥y 560 1 1 1 560
Fvy 230 1.15 1 1 265
Ey 1.6e6 1 1 1.6e6
Ey~min 0.83e6 1 1 0.83e6
Ft 1100 1.15 1 1 1.1 1392
Fc 1600 1.15 1 1 1.0 1 1840
Eaxial 1.7e6 1.15 1 1 1.0 1.96e6

c. Repeat part b except the moisture content of the member may exceed 16 percent.

SymbolReference Design ValueCDCMCtCLCVCfuCcCPCbAdjusted Design Value
Fbx+ 2400 1.15 0.8 1 1 0.88 1 1 1943
Fbx- 1450 1.15 0.8 1 1 0.88 1 1 1174
Fc⊥x 650 0.53 1 345
Fvx 265 1.15 0.875 1 1 267
Ex 1.8e6 0.833 1 1.5e6
Ex~min 0.93e6 0.833 1 0.78e6
Fby- 1450 1.15 0.8 1 1 1 1 1 1334
Fc⊥y 560 0.53 1 1 297
Fvy 230 1.15 0.875 1 231
Ey 1.6e6 0.833 1 1.3e6
Ey~min 0.83e6 0.833 1 0.69e6
Ft 1100 1.15 0.8 1 1.1 1113
Fc 1600 1.15 0.73 1 1.0 1 1343
Eaxial 1.7e6 1.15 1 1 1.0 1.96e6

✔5.12(1.5 points)

Repeat problem 5.11 using LRFD. The applicable load combination is (1.2D + 1.6S).

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.

S_x = {bh^2}/6 = {5.125*28.5^2}/6 =693.8 approx693.8 in table 1C

693.8 in3

I_x = {bh^3}/12 = {5.125*28.5^3}/12 =9886.6 approx9887 in table 1C

9886.6 in4

b. Determine the adjusted [LRFD] values associated with the section properties in part a. These include F+'bx F'ux and F'x.

Assume normal temperatures.

C_V = root{10}{{21*12*5.125}/{Ldb}} =root{10}{1291.5/{32*5.125*28.5}}root{10}{0.2763}=0.8793 approx0.88 <1✔

SymbolReference Design ValueCMCtCLCVCfuCcCPCbKFφxλAdjusted Design Value
Fbx+ 2400 1 1 1 0.88 1 1 2.54 0.85 0.8 3650
Fbx- 1450 1 1 1 0.88 1 1 2.54 0.85 0.8 2205
Fc⊥x 650 1 1 2.08 0.90 0.8 975
Fvx 265 1 1 1 2.88 0.75 0.8 458
Ex 1.8e6 1 1 1.8e6
Ex~min 0.93e6 1 1 1.76 0.85 0.8 1.1e6
Fby- 1450 1 1 1 1 1 1 2.54 0.85 0.8 2506
Fc⊥y 560 1 1 1 2.08 0.90 0.8 840
Fvy 230 1 1 2.88 0.75 0.8 397
Ey 1.6e6 1 1 1.6e6
Ey~min 0.83e6 1 1 1.76 0.85 0.8 0.10e6
Ft 1100 1 1 1.1 2.70 0.80 0.8 2091
Fc 1600 1 1 1.0 1 2.40 0.90 0.8 2465
Eaxial 1.7e6 1 1 1.0 1.36e6

c. Repeat part b except the moisture content of the member may exceed 16 percent.

SymbolReference Design ValueCMCtCLCVCfuCcCPCbKFφxλAdjusted Design Value
Fbx+ 2400 0.8 1 1 0.88 1 1 2.54 0.85 0.8 2920
Fbx- 1450 0.8 1 1 0.88 1 1 2.54 0.85 0.8 1764
Fc⊥x 650 0.53 1 2.08 0.90 0.8 517
Fvx 265 0.875 1 1 2.88 0.75 0.8 401
Ex 1.8e6 0.833 1 1.5e6
Ex~min 0.93e6 0.833 1 1.76 0.85 0.8 0.93e6
Fby- 1450 0.8 1 1 1 1 1 2.54 0.85 0.8 2004
Fc⊥y 560 0.53 1 1 2.08 0.90 0.8 445
Fvy 230 0.875 1 2.88 0.75 0.8 348
Ey 1.6e6 0.833 1 1.3e6
Ey~min 0.83e6 0.833 1 1.76 0.85 0.8 0.83e6
Ft 1100 0.8 1 1.1 2.70 0.80 0.8 1672
Fc 1600 0.73 1 1.0 1 2.40 0.90 0.8 2018
Eaxial 1.7e6 1 1 1.0 1.4e6

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