Question

4、a bracket is shown in the figure, the cross - section area of the steel bar ab is a₁ = 6cm²; the cross - section area of wooden bar bc is a₂ = 300cm², knowing the allowable stress of steel is σ=140mpa, the allowable tensile stress of wood is σₗ=8mpa, and the allowable compressive stress of wood is σᵣ=4mpa. try determine the allowable load p of the bracket.

Explanation:

Step1: Calculate the angle $\theta$

$\tan\theta=\frac{2.2}{1.4}$, so $\theta = \arctan(\frac{2.2}{1.4})\approx 57.53^{\circ}$

Step2: Analyze the force - equilibrium at point B

In the x - direction: $N_1 - N_2\cos\theta=0$, so $N_1 = N_2\cos\theta$. In the y - direction: $N_2\sin\theta=P$.

Step3: Determine the allowable load based on the steel bar

The stress formula is $\sigma=\frac{N}{A}$. For the steel bar AB, $\sigma_{max1}=\frac{N_1}{A_1}\leq[\sigma]$. Since $N_1 = N_2\cos\theta$ and $N_2=\frac{P}{\sin\theta}$, then $N_1=\frac{P\cos\theta}{\sin\theta}$. Substituting into the stress - allowable formula: $\frac{P\cot\theta}{A_1}\leq[\sigma]$. Given $A_1 = 6\times10^{- 4}m^2$ and $[\sigma]=140\times10^{6}Pa$, we have $P_1\leq[\sigma]A_1\tan\theta=140\times10^{6}\times6\times10^{-4}\times\tan57.53^{\circ}\approx1.27\times10^{5}N$.

Step4: Determine the allowable load based on the tensile strength of the wooden bar

For the wooden bar BC in tension, $\sigma_{max2}=\frac{N_2\cos\theta}{A_2}\leq[\sigma_t]$. Since $N_2=\frac{P}{\sin\theta}$, then $\frac{P\cot\theta}{A_2}\leq[\sigma_t]$. Given $A_2 = 300\times10^{-4}m^2$ and $[\sigma_t]=8\times10^{6}Pa$, we have $P_2\leq[\sigma_t]A_2\tan\theta=8\times10^{6}\times300\times10^{-4}\times\tan57.53^{\circ}\approx3.64\times10^{5}N$.

Step5: Determine the allowable load based on the compressive strength of the wooden bar

For the wooden bar BC in compression, $\sigma_{max3}=\frac{N_2\sin\theta}{A_2}\leq[\sigma_c]$. Since $N_2=\frac{P}{\sin\theta}$, then $\frac{P}{A_2}\leq[\sigma_c]$. Given $A_2 = 300\times10^{-4}m^2$ and $[\sigma_c]=4\times10^{6}Pa$, we have $P_3\leq[\sigma_c]A_2=4\times10^{6}\times300\times10^{-4}=1.2\times10^{5}N$.

Answer:

The allowable load $P$ of the bracket is $120kN$.