Location of principal plane and their magnitude

Consider a rectangular element ABCD having unit thickness subjected to normal and shear stresses as mentioned below,
$\sigma _{x}=Tensile\: stress\: on\: face\: AD\: and\: BC$
$\sigma _{y}=Tensile\: stress\: on\: face\: AB\: and\: DC$
$q=Shear\: stress\: on\: all\: faces$
Let consider an inclined plane BE at inclination of θ w. r. t. face BC. This plane subjected to normal stresses σ_x and σ_y. Plane BE also subjected to shear stresses ‘q’.
Now, consider wedge BCE. As shown in fig. 6.

$\sigma _{x}=Tensile\: stress\: on\: face\: BC$
$\sigma _{y}=Tensile\: stress\: on\: face\: EC$
$q=Shear\: stress\: on\: faces\: BC\: and\: EC$
$\sigma _{n}=Normal\: stress\: on\: face\: BE$
$\sigma _{t}=Tangential\: stress\: on\: face\: BE$
This stresses are converted in to forces by using relation;
$\sigma =\frac{Force}{Area}$
$Force=\sigma \times Area$
$\therefore Tensile\: force\: on \: face\: BC=\sigma _{x }\cdot\: BC\cdot 1=\sigma _{x }\cdot\: BC$
Here to find the area; we know one dimension BC and other dimension as unit.
Similarly,
$Shear\: force\: on \: face\: BC=q\cdot\: BC$
$Tensile\: force\: on \: face\: EC=\sigma _{y }\cdot\: EC$
$Shear\: force\: on \: face\: EC=q\cdot\: EC$
$Normal\: force\: on \: face\: BE=\sigma _{n }\cdot\: BE$
$Tangential\: force\: on \: face\: BE=\sigma _{t }\cdot\: BE$
Consider X-axis to the plane BE and resolve all forces.
For face BC
Inclination of σ_x∙BC force w.r.t. X-axis (BE plane) = (90-θ).
X component =σ_x∙BC∙cos⁡(90-θ)=σ_x∙BC∙sinθ
Y component = σ_x∙BC∙sin⁡(90-θ)=σ_x∙BC∙cosθ
Inclination of q∙BC force w.r.t. X-axis (BE plane)=θ.
X component=q∙BC∙cos⁡θ
Y component = q∙BC∙sin⁡θ
For face EC
Inclination of σ_y∙EC force w.r.t. X-axis (BE plane) =θ.
X component=σ_y∙EC∙cos⁡θ
Y component = σ_y∙EC∙sin⁡θ
Inclination of q∙EC force w.r.t. X-axis (BE plane) = (90 - θ).
X component=q∙EC∙cos⁡(90-θ)=q∙EC∙sinθ
Y component = q∙EC∙sin⁡(90-θ)=q∙EC∙cosθ
Now,

Find tangential stress σ_t

$\sum F_{x}=0$
$\leftarrow =\rightarrow$
$\sigma _{y}\cdot ECcos\theta +q\cdot BCcos\theta +\sigma _{t}\cdot BE$ $=q\cdot ECsin\theta +\sigma _{x}BCsin\theta$
Diving by BE on both side
$\sigma _{y}\left (\frac{EC}{BE} \right )cos\theta +q\left (\frac{BC}{BE} \right )cos\theta+\sigma _{t}$ $=q \left (\frac{EC}{BE} \right )sin\theta +\sigma _{x}\left (\frac{BC}{BE} \right )sin\theta$

$\sigma _{y}\cdot sin\theta\cdot cos\theta +q\cdot cos\theta\cdot cos\theta+\sigma _{t}$ $=q \cdot sin\theta\cdot sin\theta +\sigma _{x}\cdot cos\theta\cdot sin\theta$

$\sigma _{y}\cdot sin\theta\cdot cos\theta +q\cdot cos^{2}\theta +\sigma _{t}$ $=q \cdot sin^{2}\theta +\sigma _{x}\cdot cos\theta\cdot sin\theta$

$\sigma _{t}=q \cdot sin^{2}\theta +\sigma _{x}\cdot cos\theta\cdot sin\theta$ $-\sigma _{y}\cdot sin\theta\cdot cos\theta -q\cdot cos^{2}\theta$

$\sigma _{t}=q \left (sin^{2}\theta - cos^{2}\theta \right ) + sin\theta \cdot cos \theta \left (\sigma _{x} - \sigma _{y}\right )$

$\sigma _{t}=sin\theta \cdot cos \theta \left (\sigma _{x} - \sigma _{y}\right )-q \left (cos^{2}\theta - sin^{2}\theta \right )$

$\sigma _{t}=\frac{2}{2} \cdot sin\theta \cdot cos \theta \left (\sigma _{x} - \sigma _{y}\right )-q cos2\theta$

$\sigma _{t}=\frac{\left (\sigma _{x} - \sigma _{y}\right )}{2} \cdot 2 \cdot sin\theta \cdot cos \theta -q cos2\theta$

∴ $\sigma _{t}=\frac{\left (\sigma _{x} - \sigma _{y}\right )}{2} \cdot sin2\theta -q cos2\theta$

This is shear stress on inclined plane.
(Note: cos²⁡θ-sin²⁡θ = cos2θ ; 2∙sin⁡θ∙cos⁡θ = sin2θ)

Now,

Find normal stress σ_n

$\sum F_{y}=0$
$\uparrow =\downarrow$
$\sigma _{n}\cdot BE$ $=\sigma _{x} BC \cdot cos\theta +qBC\cdot sin\theta$ $+\sigma _{y}EC\cdot sin\theta +qEC\cdot sin\theta$
Diving by BE on both side,
$\sigma _{n}$ $=\sigma _{x} \cdot \left (\frac{BC}{BE} \right )\cdot cos\theta +q\cdot \left (\frac{BC}{BE} \right )\cdot sin\theta$ $+\sigma _{y}\cdot \left (\frac{EC}{BE} \right )\cdot sin\theta +q\cdot \left (\frac{EC}{BE} \right )\cdot sin\theta$

$\sigma _{n}$ $=\sigma _{x} \cdot cos\theta \cdot cos\theta +q\cdot cos\theta \cdot sin\theta$ $+\sigma _{y}\cdot sin\theta \cdot sin\theta +q\cdot sin\theta \cdot sin\theta$

$\sigma _{n}$ $=\sigma _{x} \cdot cos^{2}\theta +q\cdot cos\theta \cdot sin\theta$ $+\sigma _{y}\cdot sin^{2}\theta +q\cdot sin\theta \cdot sin\theta$

$\sigma _{n}$ $=\sigma _{x} \cdot cos^{2}\theta +\sigma _{y}\cdot sin^{2}\theta + 2q\cdot sin\theta \cdot cos\theta$

But,
$cos^{2}\theta =\frac{1+cos2\theta }{2};\: sin^{2}\theta =\frac{1-cos2\theta }{2}$ and $2 \cdot sin\theta \cdot cos \theta = sin2\theta$

$\sigma _{n}$ $=\sigma _{x} \left (\frac{1+cos2\theta }{2} \right ) +\sigma _{y}\left (\frac{1-cos2\theta }{2} \right ) + q\cdot sin2\theta$

$\sigma _{n}$ $=\frac{\sigma _{x}}{2} + \frac{\sigma _{x}cos2\theta }{2} + \frac{\sigma _{y}}{2} -\frac{ \frac{\sigma _{y}}{2} cos2\theta }{2} + q\cdot sin2\theta$

$\sigma _{n}$ $=\frac{\sigma _{x}}{2} + \frac{\sigma _{y}}{2} + \frac{\sigma _{x}cos2\theta }{2} -\frac{ \frac{\sigma _{y}}{2} cos2\theta }{2} + q\cdot sin2\theta$

∴ $\sigma _{n}$ $=\left( \frac{\sigma _{x} + \sigma _{y}}{2} \right) + \left( \frac{\sigma _{x} - \sigma _{y} }{2} \right) \cdot cos2\theta + q\cdot sin2\theta$

This is normal stress on inclined plane.

Resultant stress σ_r

$\therefore \sigma _{r}=\sqrt{\left (\sigma _{n} \right )^{2}+\left (\sigma _{t} \right )^{2}}$

Resultant stress σ_r

$\Phi =tan^{-1}\frac{\sigma_{t}}{\sigma_{n}}$

Location of Principal Plane

We know that principal plane is obtained at no shear stress therefor put σ_t = 0 in formula
$\sigma _{t}=\frac{\left (\sigma _{x} - \sigma _{y}\right )}{2} \cdot sin2\theta -q cos2\theta$

$0=\frac{\left (\sigma _{x} - \sigma _{y}\right )}{2} \cdot sin2\theta -q cos2\theta$

$q cos2\theta=\frac{\left (\sigma _{x} - \sigma _{y}\right )}{2} \cdot sin2\theta$

$\frac{2q}{\left (\sigma _{x} - \sigma _{y}\right )}=\frac{sin2\theta}{cos2\theta}$

$\frac{2q}{\left (\sigma _{x} - \sigma _{y}\right )}=tan2\theta$

Value of tan 2θ may be +ve or –ve.
Assume +ve value at 2θ1 and –ve value at 2θ2. To find this value form a fig. as below from above tan2θ value.

From above fig. diagonal of triangle
$=\sqrt{\left (2q \right )^{2}+\left (\sigma _{x} - \sigma _{y} \right )^{2}}$

Now, $sin2\theta _{1}=\frac{2q}{\sqrt{4q^{2}+\left ( \sigma _{x}-\sigma _{y} \right )^{2}}}$

$cos2\theta _{1}=\frac{\sigma _{x}-\sigma _{y}}{\sqrt{4q^{2}+\left ( \sigma _{x}-\sigma _{y} \right )^{2}}}$

$sin2\theta _{2}=sin\left(2\theta_{1}+180^{0}\right)=-sin2\theta_{1}$

$cos2\theta _{2}=cos\left(2\theta_{1}+180^{0}\right)=-cos\theta_{1}$

θ₁and θ₂are the location of principal stresses

Find magnitude of principal stress

Principal stress is normal stress acting on principal plane.
Now, to find value of principal stress put values of sin⁡2θ₁,cos⁡2θ₁,sin⁡2θ₂ and cos2θ₂ in σ_n formula,
$\sigma _{n}$ $=\left( \frac{\sigma _{x} + \sigma _{y}}{2} \right) + \left( \frac{\sigma _{x} - \sigma _{y} }{2} \right) \cdot cos2\theta + q\cdot sin2\theta$
σ_n1 is the value of magnitude of principal stress at θ₁.
$\sigma _{n1}=\left( \frac{\sigma _{x} + \sigma _{y}}{2} \right)$ $+ \left( \frac{\sigma _{x} - \sigma _{y} }{2} \right) \cdot \frac{\sigma _{x}-\sigma _{y}}{\sqrt{4q^{2}+\left ( \sigma _{x}-\sigma _{y} \right )^{2}}} + q\cdot \frac{2q}{\sqrt{4q^{2}+\left ( \sigma _{x}-\sigma _{y} \right )^{2}}}$

$\sigma _{n1}=\left( \frac{\sigma _{x} + \sigma _{y}}{2} \right)$ $+ \frac{\left( \sigma _{x} - \sigma _{y} \right)^{2} }{2\sqrt{4q^{2}+\left ( \sigma _{x}-\sigma _{y} \right )^{2}}} + \frac{4q^{2}}{2\sqrt{4q^{2}+\left ( \sigma _{x}-\sigma _{y} \right )^{2}}}$

$\sigma _{n1}=\left( \frac{\sigma _{x} + \sigma _{y}}{2} \right) + \frac{\left( \sigma _{x} - \sigma _{y} \right)^{2} + 4q^{2}}{2\sqrt{4q^{2}+\left ( \sigma _{x}-\sigma _{y} \right )^{2} }}$

$\sigma _{n1}=\left( \frac{\sigma _{x} + \sigma _{y}}{2} \right)$ $+ \frac{\left( \sigma _{x} - \sigma _{y} \right)^{2} + 4q^{2}}{2} \cdot \frac{1}{\sqrt{4q^{2}+\left ( \sigma _{x}-\sigma _{y} \right )^{2} }}$

$\sigma _{n1}=\left( \frac{\sigma _{x} + \sigma _{y}}{2} \right)$ $+ \frac{\left( \sigma _{x} - \sigma _{y} \right)^{2} + 4q^{2}}{2} \cdot \frac{1}{\left(4q^{2}+\left ( \sigma _{x}-\sigma _{y} \right )^{2} \right)^{\frac{1}{2}}}$

$\sigma _{n1}=\left( \frac{\sigma _{x} + \sigma _{y}}{2} \right)$ $+ \frac{\left ( \left ( \sigma _{x} -\sigma _{y}\right )^{2}+4q^{2} \right )\cdot \left ( 4q^{2} +\left ( \sigma _{x} -\sigma _{y}\right )^{2}\right )^{-\frac{1}{2}}}{2}$

$\sigma _{n1}=\left( \frac{\sigma _{x} + \sigma _{y}}{2} \right) + \frac{\left ( \left ( \sigma _{x} -\sigma _{y}\right )^{2}+4q^{2} \right )^{\frac{1}{2}}}{2}$

$\sigma _{n1}=\left( \frac{\sigma _{x} + \sigma _{y}}{2} \right) + \frac{\sqrt{\left ( \sigma _{x} -\sigma _{y}\right )^{2}+4q^{2}} }{2}$

$\sigma _{n1}=\left( \frac{\sigma _{x} + \sigma _{y}}{2} \right) + \sqrt{\frac{\left ( \sigma _{x} -\sigma _{y}\right )^{2}+4q^{2} }{4}}$

∴ $\sigma _{n1}=\left( \frac{\sigma _{x} + \sigma _{y}}{2} \right) + \sqrt{\left ( \frac{\sigma _{x}-\sigma _{y}}{2} \right )^{2}+q^{2}}$

Similarly we can find, σ_n2 is the value of magnitude of principal stress at θ₂. We get

∴ $\sigma _{n2}=\left( \frac{\sigma _{x} - \sigma _{y}}{2} \right) + \sqrt{\left ( \frac{\sigma _{x}-\sigma _{y}}{2} \right )^{2}+q^{2}}$

Value of σ_n1 is maximum hence it is called as ‘Major Principal Stress’ and σ_n2 is minimum hence it is called as ‘Minor Principal Stress’.
There are different cases for finding out location of principal plane and magnitude of principal stress depending upon stress acting on body.
Above derivation for when two mutually perpendicular stresses and shear stress is acting on body. When any one of them is absent then that value is taken as zero and when compressive in nature that taken as –ve.

Vinayak Hulwane

I share my knowledge in civil engineering and try to makes it helpful to all studying and practicing civil engineers. Help to improve it.

2 comments

TRANCE YTJune 14, 2023 at 6:36 PM
Can you explain me in easy notes
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TRANCE YTJune 14, 2023 at 6:36 PM
Please explain
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Location of principal plane and their magnitude

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