The shaft is supported by bearings at \(A\) and \(B\) that exert force components only in the \(x\) and \(z\) directions on the shaft. If the allowable normal stress for the shaft is \(\sigma_{\text {allow }}=15 \mathrm{ksi}\), determine to the nearest \(\frac{1}{8}\) in. the smallest diameter of the shaft that will support the loading. Use the maximum distortion energy theory of failure.

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N = F', 100 lb 6 in. x 8 in. 12 in. 2 in. F=300 lb 10 in. Fy = 300 lb 4 in. B 6 in