Question 11.9: A wood post of rectangular cross section (Fig. 11-41) is con...
A wood post of rectangular cross section (Fig. 11-41) is constructed of Douglas fir lumber having a compressive design stress F_{c} =1600 psi and modulus of elasticity E =1900 ksi. The length of the post is L, and the cross-sectional dimensions are b and h. The supports at the ends of the post provide pinned-end conditions, so the length L becomes the effective length L_{e}. Also, buckling is free to occur about either principal axis of the cross section. Note: Since the post is made of sawn lumber, the constant c equals 0.8.
(a) Determine the allowable axial load P_{allow} if L = 7 ft, b = 4.75 in., and h = 6.25 in.
(b) Determine the maximum allowable length L_{max} if the axial load P = 40 kips, b = 4.75 in., and h = 6.25 in.
(c) Determine the minimum width b_{min} of the cross section if the column is square, P = 38 kips, and L = 8.5 ft.
Use a four-step problem-solving approach. Combine steps as needed for an efficient solution.
Part (a): Allowable axial load.
1, 2, 3. Conceptualize, Categorize, Analyze: The allowable load from Eq. (11-96) is
P_{\text {allow }}=F_{c}^{\prime} A=F_{c} C_{P} Ain which F_{c} = 1600 psi and
A=b h=(4.75 \text { in. })(6.25 \text { in. })=29.6875 \text { in }^{2}To find the stability factor C_{P}, first calculate the slenderness ratio as
\frac{L_{e}}{d}=\frac{7 ft }{4.75 in.}=17.684in which d is the smaller dimension of the cross section. Next, obtain the ratio Φ from Eq. (11-98):
\phi=\frac{F_{c E}}{F_{c}^{*}}=\frac{0.822 E}{F_{c}\left(L_{e} / d\right)^{2}}=\frac{0.822(1900 ksi )}{(1600 psi )(17.684)^{2}}=3.121Then substitute Φ into Eq. (11-99) for C_{P} while also using c = 0.8 and obtain
C_{P}=\frac{1+\phi}{2 c}-\sqrt{\left[\frac{1+\phi}{2 c}\right]^{2}-\frac{\phi}{c}} (11-99)
C_{P}=\frac{1+3.121}{1.6}-\sqrt{\left(\frac{1+3.121}{1.6}\right)^{2}-\frac{3.121}{0.8}}=0.9234. Finalize: Finally, the allowable axial load is
P_{\text {allow }}=F_{c} C_{P} A=1600 psi (0.923)\left(29.6875 in ^{2}\right)=43.8 kipsPart (b): Maximum allowable length.
1, 2, 3. Conceptualize, Categorize, Analyze: Begin by determining the required value of C_{P} . Rearrange Eq. (11-96) and replace P_{allow} by the load P to obtain the formula for C_{P} shown here. Then substitute numerical values and obtain
C_{P}=\frac{P}{F_{c} A}=\frac{40 kips }{1600 psi \left(29.6875 in ^{2}\right)}=0.84211Substituting this value of C_{P} into Eq. (11-99) and also setting c equal to 0.8 gives the equation in which Φ is the only unknown quantity:
C_{P}=0.84211=\frac{1+\phi}{1.6}-\sqrt{\left[\frac{1+\phi}{1.6}\right]^{2}-\frac{\phi}{0.8}}Solve numerically by trial-and-error to find
\phi=1.74039Finally, use Eq. (11-98) to get
\frac{L}{d}=\sqrt{\frac{0.822 E}{\phi F_{c}}}=\sqrt{\frac{0.822(1900 ksi )}{1.74039(1600 psi )}}=23.683and
L_{\max }=23.683 d=23.683(4.75 \text { in. })=9.37 ft4. Finalize: Any larger value of the length L will produce a smaller value of C_{P} and, hence, a load P that is less than the actual load of 40 kips.
Part (c): Minimum width of square cross section.
1, 2. Conceptualize, Categorize: The minimum width b_{min} can be found by trial-and-error using the procedure described in part (a). Follow these steps.
i. Select a trial value of b (feet).
ii. Calculate the slenderness ratio L / d = 8.5 / b (nondimensional).
iii. Calculate the ratio Φ from Eq. (11-98):
\phi=\frac{0.822 E}{F_{c}\left(L_{ e } / d\right)^{2}}=\frac{0.822(1900 ksi )}{1.6 ksi \left(\frac{8.5}{b}\right)^{2}}=13.51 b^{2} (nondimensional)
iv. Substitute Φ into Eq. (11-99) and calculate C_{P} (nondimensional).
v. Calculate the load P from Eq. (11-96):
P=F_{c} C_{P} A=(1600 psi )\left(\frac{144 in ^{2}}{ ft ^{2}}\right)\left(C_{P}\right) b^{2}=230.4 C_{P} b^{2} (kips)
vi. Compare the calculated value of P with the given load of 38 kips. If P is less than 38 kips, select a larger trial value for b and repeat steps (ii) through (v). If P is larger than 38 kips by a significant amount, select a smaller value for b and repeat the steps. Continue until P reaches a satisfactory value.
3, 4. Analyze, Finalize: Take a trial value of b equal to 5 in., or 0.417 ft. Then steps (ii) through (v) produce
\frac{L}{d}=\frac{8.5}{b}=\frac{8.5}{0.417}=20.384 \quad \phi=13.51 b^{2}=13.51\left(0.417^{2}\right)=2.349C_{P}=0.891 \quad P=230.4 C_{P} b^{2}=230.4(0.891)\left(0.417^{2}\right)=35.7 kips
Since the given load is 38 kips, select a larger value of b, say 5.2 in., for the next trial. Proceeding in this manner with successive trials leads to
b=5.2 \text { in., } P=38.97 kipsb=5.15 \text { in., } P=38.13 kips
Therefore, the minimum width of the square cross section is
b_{\min }=5.15 in.