Problem 1010 and Problem 1011 | Investigation of timber reinforced by two steel channels | Strength of Materials Review at MATHalino

Problem 1010
A pair of C250 × 30 steel channels are securely bolted to wood beam 200 mm by 254 mm, as shown in Fig. P-1010. From Table B-2 in Appendix B, the depth of the channel is also 254 mm.) If bending occurs about the axis 1-1, determine the safe resisting moment if the allowable stresses σ_s = 120 MPa and σ_w = 8 MPa. Assume n = 20.

Problem 1011
In Problem 1010, determine the safe resisting moment if bending occurs about axis 2-2.

Solution

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Relevant data for C250 × 30 steel channel
$I_x = 32.7 \times 10^6 ~ \text{mm}^4$

$I_y = 1.16 \times 10^6 ~ \text{mm}^4$

$x = 15.3 ~ \text{mm}$

$\text{Area} = 3780 ~ \text{mm}^2$

$\text{Depth} = 254 ~ \text{mm}^2$

$\text{Flange width} = 69 ~ \text{mm}^2$

Equivalent wood area
$A_w = 200(254) = 50\,800 ~ \text{mm}^2$

$A_s = 2(n \times \text{Area}) = 2(20 \times 3780) = 151\,200 ~ \text{mm}^2$

$A = A_w + A_s = 50\,800 + 151\,200 = 202\,000 ~ \text{mm}^2$

Solution 1010: Bending occurs at section 1-1

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$I_{1-1} = I_w + 2nI_x = \dfrac{200(254^3)}{12} + 2(20)(32.7 \times 10^6)$

$I_{1-1} = 1\,581\,117\,733 ~ \text{mm}^4$

Measured from Section 1-1, the distance of extreme wood fiber and extreme steel fiber are equal. In this case, no need to investigate both materials.

$\dfrac{f_{bs}}{n} = \dfrac{120}{20} = 6 ~ \text{MPa}$

$f_{bw} = 8 ~ \text{MPa}$

Since f_bs / n < f_bw, steel is critical.

$\dfrac{f_{bs}}{n} = \dfrac{Mc}{I_{1-1}}$

$6 = \dfrac{M(127)}{1\,581\,117\,733}$

$M = 74\,698\,475.57 ~ \text{N}\cdot\text{mm}$

$M = 74.7 ~ \text{kN}\cdot\text{m}$ answer

Solution 1011: Bending occurs at section 2-2

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$I_{2-2} = I_w + 2n[ \, I_y + \text{Area}(100 + x)^2 \, ]$

$I_{2-2} = \dfrac{254(200^3)}{12} + 2(20)[ \, (1.16 \times 10^6) + 3780(100 + 15.3)^2 \, ]$

$I_{2-2} = 2\,225\,799\,741 ~ \text{mm}^4$

Since $\left( \dfrac{f_{bs}}{n} = 6 ~ \text{MPa} \right) \lt (f_{bw} = 8 ~ \text{MPa})$ and the extreme steel fiber is farther than extreme wood fiber from Section 2-2, steel is critical. In this case, no need to investigate the wood.
$\dfrac{f_{bs}}{n} = \dfrac{Mc}{I_{2-2}}$

$6 = \dfrac{M(100 + 69)}{2\,225\,799\,741}$

$M = 79\,022\,476.01 ~ \text{N}\cdot\text{mm}$

$M = 79.02 ~ \text{kN}\cdot\text{m}$ answer

From the above results, this section is stronger in bending at Section 2-2.

Problem 1010 and Problem 1011 | Investigation of timber reinforced by two steel channels

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