Homework 5 (Due Monday, October 8)

— David Wagner 2007/10/06 10:21

Solve the following problems in your textbook: 6.1 and 6.2

Problem 6.1

a. Dimension Lumber
Fb Fv E
b. Reference Value 1000 180 1.7e6
CD 1.25 1.25 –
CM 1.0 1.0 1.0
Ct 1.0 1.0 1.0
CL 1.0 – –
CF 1.3 – –
Cfu 1.0 – –
Ci 1.0 1.0 1.0
Cr 1.0* – –
CP – – –
CT – – 1.0
Cb – – –
c. Adjusted ASD 1625 225 1.7e6
F'b F'v E'

*Member Spacing not given.

fbfvΔ
d. Stresses 1566 psi 59 psi 0.2 in

M={PL}/4={{400+1600}*8*12}/4=48000 in-lb f_b = M/S = 48000/30.66=1566 psi < 1625✔

V=P/2={400+1600}/2=1000 lb; f_v=3V/2A = 3*1000/{2*25.38}=59.1 psi <225✔

L/360=8*12/360=0.27 in, L/240=8*12/240=0.40 in. Delta = {PL^3}/{48EI} = {2000*8^3*12^3}/{48*1.7e6*111.1} =0.195 in < 0.27 in✔

e. ASD Adequate?Yes
Reference Value 1000 180 1.7e6
KF 2.16/0.85 2.16/0.75 –
FbnFvnE
f. Nominal LRFD 2541 518 1.7e6
CM 1.0 1.0 1.0
Ct 1.0 1.0 1.0
CL 1.0 – –
CF 1.3 – –
Cfu 1.0 – –
Ci 1.0 1.0 1.0
Cr 1.0* – –
CP – – –
CT – – 1.0
Cb – – –
φ 0.85 0.75 –
λ 0.8 0.8 –
g. Adjusted LRFD 2246 311 1.7e6
F'bn F'vn E'

*Member Spacing not given.

  • {M prime}_{n} = S{F prime}_{bn} = 30.66*2246=68862 in-lb
  • {V prime}_{n} = {2/3}A{F prime}_{bn} = {2/3}*25.38*311=5262 lb
M'nV'n
h. Adjusted Resistance 68862 in-lb 5262 lb

i. Factored moment and shear (LRFD), and actual deflection:

  • P=1.2*400+1.6*1600=3040 lb
  • M_u={PL}/4={{3040}*8*12}/4=72960 in-lb > 68862✘
  • V_u=P/2={3040}/2=1520 lb <5262✔

L/360=8*12/360=0.27 in, L/240=8*12/240=0.40 in.

Delta = {PL^3}/{48EI} = {2000*8^3*12^3}/{48*1.7e6*111.1} =0.195 in < 0.27 in✔

MuVuΔ
i. 72960 in-lb 1520 lb 0.195 in
j. LRFD Adequate?No

Problem 6.2

Repeat 6.1 with moisture exceeding 19%.

a. Dimension Lumber
Fb Fv E
b. Reference Value 1000 180 1.7e6
CD 1.25 1.25 –
CM 0.85* 0.97 0.9
Ct 1.0 1.0 1.0
CL 1.0 – –
CF 1.3 – –
Cfu 1.0 – –
Ci 1.0 1.0 1.0
Cr 1.0† – –
CP – – –
CT – – 1.0
Cb – – –
c. Adjusted ASD 1381 218 1.53e6
F'b F'v E'
  • *Fb CF > 1150
  • †Member Spacing not given.
fbfvΔ
d. Stresses 1566 psi 59 psi 0.2 in

M={PL}/4={{400+1600}*8*12}/4=48000 in-lb f_b = M/S = 48000/30.66=1566 lb > 1381✘

V=P/2={400+1600}/2=1000 lb; f_v=3V/2A = 3*1000/{2*25.38}=59.1 lb <218✔

L/360=8*12/360=0.27 in, L/240=8*12/240=0.40 in. Delta = {PL^3}/{48EI} = {2000*8^3*12^3}/{48*1.53e6*111.1} =0.216 in < 0.27 in✔

e. ASD Adequate?No

Part f.

Reference Value 1000 180 1.7e6
KF 2.16/0.85 2.16/0.75 –
FbnFvnE
f. Nominal LRFD 2541 518 1.7e6
CM 0.85 0.97 0.9
Ct 1.0 1.0 1.0
CL 1.0 – –
CF 1.3 – –
Cfu 1.0 – –
Ci 1.0 1.0 1.0
Cr 1.0* – –
CP – – –
CT – – 1.0
Cb – – –
φ 0.85 0.75 –
λ 0.8 0.8 –
g. Adjusted LRFD 1909 302 1.53e6
F'bn F'vn E'

*Member Spacing not given.

  • {M prime}_{n} = S{F prime}_{bn} = 30.66*1909=58530 in-lb
  • {V prime}_{n} = {2/3}A{F prime}_{bn} = {2/3}*25.38*302=5110 lb
M'nV'n
h. Adjusted Resistance 58530 in-lb 5110 lb

i. Factored moment and shear (LRFD), and actual deflection:

  • P=1.2*400+1.6*1600=3040 lb
  • M_u={PL}/4={{3040}*8*12}/4=72960 in-lb > 58530 ✘
  • V_u=P/2={3040}/2=1520 lb <5110✔

L/360=8*12/360=0.27 in, L/240=8*12/240=0.40 in.

Delta = {PL^3}/{48EI} = {2000*8^3*12^3}/{48*1.53e6*111.1} =0.217 in < 0.27 in✔

MuVuΔ
i. 72960 in-lb 1520 lb 0.217 in
j. LRFD Adequate?No

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