Design Shear Stress Solution - UCFileSpace Tools -...
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DESIGN FOR SHEAR STRESS 1 τ= P S A Guideline for design shear stress τ d based on yield strength in shear S ys or ultimate strength in shear S us 1. Recognize 6000psi 2. Identify P 3. Visualize shear plane dimensions 4. Substitute into equation τ d = P S A 1.Use guideline and material property to find τ d 2. Visualize shear plane 3. Write equation τ d =S us = P S A Solve for τ d given = 6000 psi P S given = 21000 lb (8 inches) x (a inches) τ d = P S A S 6000 lb ¿ 2 = 21000 lb ¿¿ a = 0.438 in τ d = S US = (448 N/mm 2 ) (0.82) (circumference) Χ P S = (circumference)x(thickness) = τ d = P S A S
Transcript of Design Shear Stress Solution - UCFileSpace Tools -...
DESIGN FOR SHEAR STRESS 1
τ=PSAS
Guideline for design shear stress τd based on yield strength in shear Sys or ultimate strength in shear Sus
1. Recognize 6000psi as τd
2. Identify PS
3. Visualize shear plane dimensions
4. Substitute into equation τ d=PSAS
and solve for a
1.Use guideline and material property to find τd
Force must cause material to fail.
2. Visualize shear plane and find AS
3. Write equation τ d=Sus=PSAS
Solve for force PS
τd given = 6000 psi
PS given = 21000 lb
(8 inches) x (a inches)
τ d=PSAS
6000 lb¿2 =
21000 lb¿¿
a = 0.438 in
τd = SUS = (448 N/mm2)(0.82)
(circumference) Χ PS = (circumference)x(thickness) = π∙ d∙ t
τ d=PSAS
(0.82)(448 Nmm2 )=
PSπ (20mm )(8mm)
PS = 184700 N = 184.7 kN
40mm
AyAs
F2
3.6kN
30o
10o
DESIGN FOR SHEAR STRESS 2
1. Draw free-body diagram
2. ΣFx = 0 ΣFy =0 ΣM=0 (three equation, 3 unknowns)
3. Solve for Ax Ay and A
4. Find Guideline for τd 5. Find material property, Sus or Sys
6. Find smallest possible pin diameter to be safe
HW#12
∑ F x=0=F2 ( cos10o )+Ax+3.6kN (cos30o)∑ F y=0=F2 (sin 10o )+A y−3.6 kN (sin 30o)
∑M F2=0=−Ax (25mm )+A y (10mm )−3.6kN ( cos30o ) (55mm )−3.6kN (sin 30o )(50mm)
A x=10.24kNA y=0.544 kNA=√(10.25kN )2+(0.544 kN )2=10.26 kN
τ d=S y8
Sy = 1000 N/mm2
τ d=PSAS
1000 Nmm2
8=10.26×103N
π d2
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DESIGN FOR SHEAR STRESS 3
