The composite beam is composed of two steel plates (E_{s}=200 GPa) and a wood core (E_{w}=11GPa) ,which are prefectly bonded , as shown in figure. Given the allowable bending stress for the steel (sigma _{all})s=120 MPa ,and for the wood (sigma

(a) To determine the maximum moment that can be applied to the beam about the x-axis, we need to find the section where the bending stress is maximum. Considering the beam as a rectangular section, the distance of the center of gravity of the steel plates from the neutral axis is {eq}d_1=3,cm {/eq} and the distance of the center of gravity of the wood core from the neutral axis is {eq}d_2=2.5,cm {/eq}. Assuming uniform stress distribution within each material, the maximum bending stress occurs at the top of the steel plates and the bottom of the wood core. Therefore, we have: - For the steel plates: {eq}sigma_s = \frac{My}{I_{s}} = \frac{M(d_1)}{frac{1}{12}bh^3} {/eq} - For the wood core: {eq}sigma_w = \frac{My}{I_{w}} = \frac{M(d_2)}{frac{1}{12}bh^3} {/eq} where {eq}M {/eq} is the bending moment, {eq}y {/eq} is the distance from the neutral axis to the extreme fiber, {eq}I_s {/eq} and {eq}I_w {/eq} are the moments of inertia of the steel plates and wood core, respectively, {eq}b {/eq} is the width of the composite beam, and {eq}h {/eq} is the total height of the composite beam. Since the allowable bending stress for the steel is {eq}sigma_{all,s}=120,MPa {/eq}, we can write: {eq}frac{M(d_1)}{frac{1}{12}bh^3} \leq sigma_{all,s} {/eq} Rearranging, we get: {eq}M \leq \frac{sigma_{all,s}frac{1}{12}bh^3}{d_1} = \frac{120times10^6timesfrac{1}{12}times10times(3+2.5)^3}{3} \approx boxed{7414,Ncdot m} {/eq} Therefore, the maximum moment that can be applied to the beam about the x-axis is \approx imately {eq}7414,Ncdot m {/eq}. (b) To determine the maximum moment that can be applied to the beam if the moment is applied about the y-axis instead of the x-axis, we need to calculate the new bending stress distribution and find the section where the maximum bending stress occurs. The new moment of inertia of the composite beam about the y-axis is {eq}I_y = \frac{1}{12}bh^3 = \frac{1}{12}times(10+2)times(2.5+3)^3 = 1787.5,cm^4 {/eq}. Assuming uniform stress distribution within each material, the maximum bending stress occurs at the left of the steel plates and the right of the wood core. Therefore, we have: - For the steel plates: {eq}sigma_s = \frac{My}{I_{y}} = \frac{M(d_1)}{I_y} {/eq} - For the wood core: {eq}sigma_w = \frac{My}{I_{y}} = \frac{M(d_2)}{I_y} {/eq} where {eq}M {/eq}, {eq}d_1 {/eq}, {eq}d_2 {/eq}, {eq}b {/eq}, and {eq}h {/eq} are the same as before. Since the allowable bending stress for the wood is {eq}sigma_{all,w}=24,MPa {/eq}, we can write: {eq}frac{M(d_2)}{I_y} \leq sigma_{all,w} {/eq} Rearranging, we get: {eq}M \leq \frac{sigma_{all,w}I_y}{d_2} = \frac{24times10^6times1787.5}{2.5} \approx boxed{4.26times10^8,Ncdot mm} {/eq} Note that we converted the units to N·mm for consistency with the units of the moments of inertia. Alternatively, we could have used the units of N·m by dividing the result by 1000. Therefore, the maximum moment that can be applied to the beam about the y-axis is \approx imately {eq}4.26times10^8,Ncdot mm {/eq}.