At a point in a bracket the stresses on two mutually perpendicular planes are 120 N/mm2 and
60 N/mm2
both tensile. The shear stress across these planes is 30 N/mm2
. Find using the Mohr's
stress circle i) Principal stresses at the point, ii) Maximum shear stress and iii) resultant stress on
a plane inclined at 60° to the axis of the major principal stress.
Answers
Answer:
The following three points calculation can be defined as follows:
Explanation:
\begin{gathered}\sigma_x= 100 \frac {N}{mm^2}\\\\\sigma_y= 50 \frac {N}{mm^2}\\\\T_{xy}=80 \frac {N}{mm^2}\\\\\theta =20^{\circ}\end{gathered}
σ
x
=100
mm
2
N
σ
y
=50
mm
2
N
T
xy
=80
mm
2
N
θ=20
∘
The resultant stress, principle stress, and maximum shear stress status are determined by the analytic method which can be defined as follows:
\to \sigma_\theta=\frac{\sigma_x+\sigma_y}{2}+\frac{\sigma_x-\sigma_y}{2} \cos 2 \theta + T_{xy} \sin 2\theta→σ
θ
=
2
σ
x
+σ
y
+
2
σ
x
−σ
y
cos2θ+T
xy
sin2θ
=\frac{\100+50}{2}+\frac{100-50}{2} \cos 40+80\sin 40=
2
\100+50
+
2
100−50
cos40+80sin40
=145.57 \ Mpa=145.57 Mpa
\to T_\theta=\frac{\sigma_x-\sigma_y}{2}\sin 2\theta-T_{xy}\cos 2 \theta→T
θ
=
2
σ
x
−σ
y
sin2θ−T
xy
cos2θ
\begin{gathered}=\frac{\100-50}{2}\sin 40-80\cos 40\\\\=-45.21 \ Mpa\end{gathered}
=
2
\100−50
sin40−80cos40
=−45.21 Mpa
\begin{gathered}\to \sigma_R=\sqrt{\sigma_B^2+T_\theta^2}\\\end{gathered}
→σ
R
=
σ
B
2
+T
θ
2
\begin{gathered}=\sqrt{145.57^2-45.21^2}\\\\=152.43 \ Mpa\end{gathered}
=
145.57
2
−45.21
2
=152.43 Mpa
\begin{gathered}\to \tan \theta= \frac{T_\theta}{\sigma_\theta}\\\\ \theta= \tan^{-1} \frac{T_\theta}{\sigma_\theta}\\\\\end{gathered}
→tanθ=
σ
θ
T
θ
θ=tan
−1
σ
θ
T
θ
\begin{gathered}= \tan^{-1} \frac{ -45.21}{145.57}\\\\=17.25^{\circ}\end{gathered}
=tan
−1
145.57
−45.21
=17.25
∘
Calculating Maximum shear stress:
T_{max}=\sqrt{(\frac{\sigma_x-\sigma_y}{2})^2+T_{xy}}T
max
=
(
2
σ
x
−σ
y
)
2
+T
xy
\begin{gathered}=\sqrt{25^2 +80^2}\\\\=83. 81 \ Mpa\end{gathered}
=
25
2
+80
2
=83.81 Mpa
\theta_s=\frac{1}{2} \tan^{-1}\frac{\sigma_x-\sigma_y}{2T_{xy}}θ
s
=
2
1
tan
−1
2T
xy
σ
x
−σ
y
=8.677^{\circ}=8.677
∘
\begin{gathered}\sigma_{avg}=75\\\\\sigma_{p1}=158.81 \ Mpa\\\\\sigma_{p2}= -8.81 \ Mpa\\\\\sigma_{\theta\ p1} =36.62^{\circ}\end{gathered}
σ
avg
=75
σ
p1
=158.81 Mpa
σ
p2
=−8.81 Mpa
σ
θ p1
=36.62
∘
Scale 1 MPa = cm in the mohr circle
Its location with maximum shear stress container and main aircraft are interconnected