THREE-DIMENSIONAL STRESS Machinery's Handbook, 31st Edition
224
A 2 3 --- B = –
2 A 3 27 – -----
D 3 27 = ---
AB 3 ---- C –
=
D
E
F
E – 2 F ----
D 3 = --
G 3 -- cos
A 3 + --
arccos
=
2 H
=
G
H
I
A 3 + --
G 3 -- 120 ° +
A 3 + --
G 3 -- 240 ° +
J = Step 4. Calculate the true maximum shear stress S s (max) using the formula . S S S 05 large small = − ^ h s (max) in which S large is equal to the algebraically largest of the calculated principal stresses I , J , or K and S small is algebraically the smallest. The maximum principal normal stresses and the maximum true shear stress calculated above may be used with any of the various theories of failure. 2 H cos = K 2 H cos Sample Calculations.— The following examples illustrate some typical strength of materials calculations, using both English and metric SI units of measurement. Example 1(a): A round bar made from SAE 1025 low carbon steel is to support a direct tension load of 50,000 lbs. Using a factor of safety of 4, and assuming that the stress concentration factor K = 1, a suitable standard diameter is to be determined. Calculations are to be based on a yield strength of 40,000 psi. Solution: Because the factor of safety and strength of the material are known, the al- lowable working stress s w may be calculated using Equation (1): 40,000 ∕ 4 = 10,000 psi. The relationship between working stress s w and nominal stress σ is given by Equation (2). Since K = 1, σ = 10,000 psi. Applying Equation (9) in the Table of Simple Stresses on page 216 , the area of the bar can be solved for: A = 50,000 ∕ 10,000 or 5 square inches. The next largest standard diameter corresponding to this area is 2 9 ∕ 16 inches. Example 1(b): A similar example to that given in Example 1(a), using metric SI units, is as follows. A round steel bar of 300 meganewtons/meter 2 yield strength is to withstand a direct tension of 200 kilonewtons. Using a safety factor of 4, and assuming that the stress concentration factor K = 1, a suitable diameter is to be determined. Because the factor of safety and the strength of the material are known, the allow able working stress s w may be calculated using Equation (1): 300 ∕ 4 = 75 mega- newtons/meter 2 . The relationship between working stress and nominal stress σ is given by Equation (2). Since K = 1, σ = 75 MN/m 2 . Applying Equation (9) in the Table of Simple Stresses on page 216, the area of the bar can be determined from: , . A kN N 200 200 000 000267m 2 = = 75 MN ⁄ m 2 75,000,000 N ⁄ m 2 = The diameter corresponding to this area is 0.058 meters, or approximately 0.06 m. Millimeters can be employed in the calculations in place of meters, provided the treatment is consistent throughout. In this instance the diameter would be 60 mm. Note: If the tension in the bar is produced by hanging a mass of M kilograms from its end, the value is Mg newtons, where g = approximately 9.81 meters per second 2 . Example 2(a) : What would the total elongation of the bar in Example 1(a) be if its length were 60 inches? Applying Equation (18) on page 218 , . , , , . e 5 157 30 000 000 50 000 60 0019inch # # = = Example 2(b): What would be the total elongation of the bar in Example 1(b) if its length were 1.5 meters? The problem is solved by applying Equation (17) in which F = 200 kilonewtons; L = 1.5 meters; A = π 0.06 2 ⁄4 = 0.00283 m 2 . Assuming a modulus of elasticity E of 200 giganewtons/meter 2 , then the calculation is:
. e 0 00283 200 000 000 000 200 000 15 , , , , . # # =
. 0000530m
=
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