At a point in a stressed body, the known stresses are $\sigma_{x}$ = 40 MPa (T), $\sigma_{y}$ = 20 MPa (C), $\sigma_{z}$ = 20 MPa (T), $\tau_{x y}$ = +40 MPa, $\tau_{yz}$ = 0, and $\tau_{zx}$ = +30 MPa. Determine:
(a) the normal and shear stresses on a plane whose outward normal is oriented at angles of 40°, 75°, and 54° with the x, y, and z axes, respectively.
(b) the principal stresses and the absolute maximum shear stress at the point.

Question

At a point in a stressed body, the known stresses are [latex]\sigma_{x}[/latex] = 40 MPa (T), [latex]\sigma_{y}[/latex] = 20 MPa (C), [latex]\sigma_{z}[/latex] = 20 MPa (T), [latex] au_{x y}[/latex] = +40 MPa, [latex] au_{yz}[/latex] = 0, and [latex] au_{zx}[/latex] = +30 MPa. Determine:
(a) the normal and shear stresses on a plane whose outward normal is oriented at angles of 40°, 75°, and 54° with the x, y, and z axes, respectively.
(b) the principal stresses and the absolute maximum shear stress at the point.

Accepted Answer

The known stresses are

[latex]\begin{array}{lll}\sigma_{x}=40 MPa & \sigma_{y}=-20 MPa & \sigma_{ z }=20 MPa \ au_{x y}=40 MPa & au_{y z}=0 MPa & au_{z x}=30 MPa\end{array}[/latex]

(a) The plane of interest is defined by its direction cosines:

[latex]l=\cos \left(40^{\circ} ight)=0.7660 \quad m=\cos \left(75^{\circ} ight)=0.2588 \quad n=\cos \left(54^{\circ} ight)=0.5878[/latex]

The three orthogonal components of the resultant stress are:

[latex]\begin{aligned}&S_{x}=\sigma_{x} \cdot l+ au_{x y} \cdot m+ au_{z x} \cdot n=(40)(0.7660)+(40)(0.2588)+(30)(0.5878)=58.63 MPa \&S_{y}= au_{x y} \cdot l+\sigma_{y} \cdot m+ au_{y z} \cdot n=(40)(0.7660)+(-20)(0.2588)+(0)(0.5878)=25.47 MPa \&S_{z}= au_{z x} \cdot l+ au_{y z} \cdot m+\sigma_{z} \cdot n=(30)(0.7660)+(0)(0.2588)+(20)(0.5878)=34.74 MPa\end{aligned}[/latex]

The normal component [latex]\sigma_{n}[/latex] of the resultant stress is

[latex]\begin{aligned}\sigma_{n} &=S_{x} \cdot l+S_{y} \cdot m+S_{z} \cdot n \&=(58.63)(0.7660)+(25.47)(0.2588)+(34.74)(0.5878) \&=71.92 MPa =71.9 MPa ( T )\end{aligned}[/latex]

The shear stress [latex] au_{n t}[/latex] on the oblique plane can be obtained from the relation [latex]S^{2}=\sigma_{n}^{2}+ au_{n t}^{2}[/latex].

[latex]S^{2}=S_{x}^{2}+S_{y}^{2}+S_{z}^{2}=(58.63)^{2}+(25.47)^{2}+(34.74)^{2}=5,292.40[/latex]

and thus;

[latex] au_{n t}=\sqrt{\left(S^{2}-\sigma_{n}^{2} ight)}=\sqrt{5,292.40-(71.92)^{2}}=10.95 MPa[/latex]

(b) The principal stresses can be obtained from the roots of the cubic equation [Eq. (12.27)]

[latex]\sigma_{p}^{3}-I_{1} \sigma_{p}^{2}+I_{2} \sigma_{p}-I_{3}=0[/latex]

The three invariants have values of

[latex]\begin{aligned}I_{1} &=\sigma_{x}+\sigma_{y}+\sigma_{z} \&=(40)+(-20)+(20)=40 \I_{2} &=\sigma_{x} \sigma_{y}+\sigma_{y} \sigma_{z}+\sigma_{z} \sigma_{x}- au_{x y}^{2}- au_{y z}^{2}- au_{z x}^{2} \&=(40)(-20)+(-20)(20)+(20)(40)-(40)^{2}-(0)^{2}-(30)^{2}=-2,900 \I_{3} &=\sigma_{x} \sigma_{y} \sigma_{z}+2 au_{x y} au_{y z} au_{z x}-\left(\sigma_{x} au_{y z}^{2}+\sigma_{y} au_{z x}^{2}+\sigma_{z} au_{x y}^{2} ight) \&=(40)(-20)(20)+2(40)(0)(30)-\left[(40)(0)^{2}+(-20)(30)^{2}+(20)(40)^{2} ight]=-30,000\end{aligned}[/latex]

therefore, Eq. (12.27) is

[latex]\sigma_{p}^{3}-(40) \sigma_{p}^{2}+(-2,900) \sigma_{p}-(-30,000)=0[/latex]

The three roots of this cubic equation are the principal stresses:

[latex]\begin{aligned}\sigma_{p 1} &=73.8 MPa ( T ) \\sigma_{p 2} &=9.41 MPa ( T ) \\sigma_{p 3} &=43.2 MPa ( C )\end{aligned}[/latex]

The absolute maximum shear stress at the point is found from

[latex] au_{ abs \max }=\frac{\sigma_{\max }-\sigma_{\min }}{2}=\frac{73.8-(-43.2)}{2}=58.5 MPa[/latex]

Question 12.91: At a point in a stressed body, the known stresses are σx = 4...

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