HP 48GII

5-56 Equation Reference
Equations:
Example:
Given: P=40000_lbf, L=1_ft, A=3.14159265359_in^2, E=10E6_psi,
Solution:
δ=0.0153_in, ∈ =0.0013, σ =12732.3954_psi.
Shear Stress (14, 2)
Equations:
Example:
Given: L=6_ft, r=2_in, J=10.4003897419_in^4, G=12000000_psi,
τ =12000_psi.
Solution: T=5200.1949_ft
∗ lbf, φ =2.0626_°, γ =5.7296E–2_°.
Stress on an Element (14, 3)
Stresses and strains are positive in the directions shown.
Equations:
Example:
Given:
σ x=15000_kPa, σ y=4755_kPa, τ xy=7500_kPa, θ =30_°.
σ
E
∈⋅
=
∈
δ
L
- - -
=
σ
P
A
----
=
τ
G
γ ⋅
=
γ
r
φ
⋅
L
- - - - - - - - -
=
τ
T r
⋅
J
----------
=
τ
x 1 y 1
σ
x
σ
y
–
2
- - - - - - - - - - - - - - - - - - - - -
⎝
⎠
⎛
⎞
–
S I
N
2
θ
⋅ ( ) ⋅
τ
x y+
σ
y
⋅
=
σ
x 1
σ
x
σ
y +
2
- - - - - - - - - - - - - - - - - - - - - -
σ
x
σ
y
–
2
- - - - - - - - - - - - - - - - - -- - -
+ C O S 2
θ
⋅ ( ) ⋅
τ
xy SI
N
2
θ
⋅( )⋅
+
=
σ
x 1
σ
y 1 +
σ
x
σ
y + =