HP 48GII

Equation Reference 5-57
Solution:
σ x1=18933.9405_kPa, σ y1=821.0595_kPa, τ x1y1= -686.2151_kPa.
Mohr's Circle (14, 4)
Equations:
S I
N
2
θ
p 1
⋅ ( )
τ
x y
σ
x
σ
y
–
2
- - - - - - - - - - - - - - - - - - - - -
⎝
⎠
⎛
⎞
2
τ
xy
2
+
-
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - ----------------------
=
σ
1
σ
x
σ
y +
2
-- - - - - - - - - - - - - - - - - - - - -
σ
x
σ
y
–
2
-
- - - - - - - - - -----------
ρ
xy
2
++ =
σ
1
σ
2 +
σ
x
σ
y+=
θ
p 2
θ
p 1 90 + =
τ
max
σ
1
σ
2
–
2
---------------------
=
θ
s
θ
p 1 45
–
=
σ
avg
σ
x
σ
y+
2
----------------------
=
Example:
Given:
σ x=-5600_psi, σ y=-18400_psi, τ xy=4800_psi.
Solution:
σ 1= -4000_psi, σ 2= -20000_psi, θ p1=18.4349_°, θ p2=108.4349_°, τ max=8000_psi,
θ s= -26.5651_°, σ avg= -12000_psi.
Waves (15)
Variable
Description
β
Sound level
λ
Wavelength
ω
Angular frequency
ρ
Density of medium
B
Bulk modulus of elasticity
f
Frequency
I
Sound intensity
k
Angular wave number
s
Longitudinal displacement at x and t