Question 11.8: A wood post of rectangular cross section (Fig. 11-40) is con...

(a) Allowable axial load. The allowable load (from Eq. 11-92) is

P_{allow}=F^{\prime}_{c}A=F_{c}C_{P}A

in which F_{c} = 11 MPa and

A=bh=(120  mm)(160  mm)=19.2\times 10^{3}  mm^{2}

To find the stability factor C_{P}, we first calculate the slenderness ratio, as follows:

\frac{L_{e}}{d}=\frac{1.8  m}{120  mm}=15

in which d is the smaller dimension of the cross section. Next, we obtain the ratio Φ from Eq. (11-94):

\phi =\frac{F_{cE}}{F^{\ast}_{c} }=\frac{K_{cE}E}{F_{c}(L_{e}/d)^{2}}=\frac{(0.3)(13  GPa)}{(11  MPa)(15)^{2}}=1.5758

Then we substitute Φ into Eq. (11-95) for C_{P}, while also using c = 0.8, and we obtain

C_{P}=\frac{1+\phi}{2c}-\sqrt{\left[\frac{1+\phi}{2c}\right]^{2}-\frac{\phi}{c}}            (11-95)

C_{P}=\frac{1+1.5758}{1.6}-\sqrt{\left[\frac{1+1.5758}{1.6}\right]^{2}-\frac{1.5758}{0.8}}=0.8212

Finally, the allowable axial load is

P_{allow}=F_{c}C_{P}A = (11 MPa)(0.8212)(19.2 × 10³ mm²) = 173 kN

(b) Maximum allowable length. We begin by determining the required value of C_{P}. Rearranging Eq. (11-92) and replacing P_{allow} by the load P, we obtain the formula for C_{P} shown below. Then, we substitute numerical values and obtain the following result:

C_{P}=\frac{P}{F_{c}A}=\frac{100  kN}{}=0.47348

Substituting this value of C_{P} into Eq. (11-95), and also setting c equal to 0.8, we get the following equation in which Φ is the only unknown quantity:

C_{P}=0.47348=\frac{1+\phi}{1.6}-\sqrt{\left[\frac{1+\phi}{1.6}\right]^{2}-\frac{\phi}{0.8}}

Solving numerically by trial and error, we find

Φ = 0.55864

Finally, from Eq. (11-94), we get

\frac{L}{d}=\sqrt{\frac{K_{cE}E}{\phi F_{c}}}=\sqrt{\frac{(0.3)(13  GPa)}{}}=25.19

and

L_{\max} = 25.19d = (25.19)(120 mm) = 3.02 m

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 100 kN.
(c) Minimum width of square cross section. The minimum width b_{\min} can be found by trial and error, using the procedure described in part (a). The steps are as follows:

1.       Select a trial value of b (meters)
2.      Calculate the slenderness ratio L/d = 2.6/b (nondimensional)
3.      Calculate the ratio Φ from Eq. (11-94):

\phi=\frac{K_{cE}E}{F_{c}(L_{e}/d)^{2}}=\frac{(0.3)(13  GPa)}{(11  MPa)(2.6/b)^{2}}=52.448b^{2}     (nondimensional)

4.      Substitute Φ into Eq. (11-95) and calculate C_{P}  (nondimensional)
5.      Calculate the load P from Eq. (11-92):

P=F_{c}C_{P}A=(11  MPa)(C_{P})(b^{2})=11,000  C_{P}b^{2}(kilonewtons)

6.      Compare the calculated value of P with the given load of 125 kN. If P is less than 125 kN, select a larger trial value for b and repeat steps (2) through (7). If P is larger than 125 kN by a significant amount, select a smaller value for b and repeat the steps. Continue until P reaches a satisfactory value.

Let us take a trial value of b equals to 130 mm, or 0.130 m. Then steps (2) through (5) produce the following results:

L/d = 2.6/b = 20            Φ = 52.448 b² = 0.88637

C_{P} = 0.64791                P = 11,000 C_{P}b^{2} = 120.4 kN

Since the given load is 125 kN, we select a larger value of b, say 0.132 m, for the next trial. Proceeding in this manner with successive trials, we obtain the following results:

b = 0.132 m; P = 126.3 kN            b = 0.131 m; P = 123.4 kN

Therefore, the minimum width of the square cross section is

b_{\min} = 0.132 m = 132 m