Consider Figure 4.11(c) and take σx = 350 kg/cm² and σy = −700 kg/cm² (compressive). Find out the value of Φ defining the plane on which no normal stress acts. What is the magnitude of shear stress on that plane?

Question

Consider Figure 4.11(c) and take [latex] \sigma_x=350 kg / cm ^2 ext { and } \sigma_y=-700 kg / cm ^2 [/latex] (compressive). Find out the value of [latex] \phi [/latex] defining the plane on which no normal stress acts. What is the magnitude of shear stress on that plane?

Accepted Answer

We know that

[latex] \sigma_{ ext {average }}=\frac{1}{2}\left(\sigma_x+\sigma_y ight) [/latex]

Here, [latex] \sigma_x=350 kg / cm ^2 ext { and } \sigma_y=-700 kg / cm ^2 [/latex]. Therefore,

[latex] \begin{aligned} \sigma_{ ext {avcrage }} & =\frac{1}{2}(350-700)=-175 kg / cm ^2 \ au_{\max } & =\frac{1}{2}\left(\sigma_x-\sigma_y ight)=\frac{1}{2}(350+700)=525 kg / cm ^2 \end{aligned} [/latex]

Now, [latex] \sigma_n [/latex] is normal stress given by

[latex] \sigma_n=\sigma_{ ext {average }}+ au_{\max } \cos 2 \phi=-175+525 \cos 2 \phi [/latex]

Equating [latex] \sigma_n ext { with } heta=0^{\circ} ext {, we get }-175+525 \cos 2 \phi=0 ext {. Therefore, } \phi=35.26^{\circ} [/latex].

Thus, the plane on which normal stress does not act is making an angle [latex] \phi=35.26^{\circ} [/latex], shear stress on the same plane is given by

[latex] au_n= au_{\max } \sin 2 \phi [/latex]

Putting [latex] au_{\max }=525 kg / cm ^2 ext { and } \phi=35.26^{\circ}, ext { we get } [/latex]

[latex] au_n=525\left(\sin 2 imes 35.26^{\circ} ight)=494.9 kg / cm ^2 [/latex]

Therefore, value of [latex] \phi [/latex] defining the angle is 35.26° and shear stress is 494.9 kg/cm².

Question 4.3: Consider Figure 4.11(c) and take σx = 350 kg/cm² and σy = −7...

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