Page 296 - Mechanical Behavior of Materials

Section 7.6 Discussion of the Basic Failure Criteria 297

An alternative is load factor design, where each load expected in actual service is multiplied
by a load factor Y, which is similar to a safety factor. The loads so increased are used to determine
stresses, so that the failure condition is analyzed. For yield and fracture criteria, effective stresses
¯ σ calculated with the increased loads are then equated to the material’s failure stress, such as the
yield strength, σ o , or the ultimate tensile strength, σ ut , as appropriate. Load factors Y may differ
for different sources of loading to reflect circumstances such as different uncertainties in the actual
values of the various load inputs. Ideally, the Y values would be based on statistical analysis of loads
measured in service.
We will use the following nomenclature: The value of a load P expected in actual service is
denoted P, and the load factor for this load is Y P . Then the increased (factored) load used in analysis
ˆ
of the failure condition is P f = Y P P.
ˆ
Example 7.6
Consider the situation of Ex. 7.3(b), where a solid shaft of diameter d is made of AISI 1020 steel
(as rolled) and is subjected in service to an axial force P = 200 kN and a torque T = 1.50 kN·m.
What diameter is required if load factors of Y P = 1.60 and Y T = 2.50 are required for the axial
force and torque, respectively?

Solution The factored loads for analysis of the failure condition are

ˆ
P f = Y P P = 1.60(200,000 N) = 320,000 N
6
6
T f = Y T T = 2.50(1.50 × 10 N·mm) = 3.75 × 10 N·mm
ˆ
Then we modify the Ex. 7.3 solution, proceeding similarly, except for employing P f and T f .
The maximum shear stress yield criterion thus gives

¯ σ S = MAX (|σ 1 − σ 2 | , |σ 2 − σ 3 | , |σ 3 − σ 1 |) = |σ 1 − σ 2 |

2 2 2

2P f 16T f 4 2 8T f
¯ σ S = 2 + = P + = σ o = 260 MPa
f
πd 2 πd 3 πd 2 d
where the effective stress is now equated directly to the yield strength of the AISI 1020 steel,
as the stresses have already been increased by load factors. Substituting P f and T f and solving
iteratively gives d = 55.5mm (Ans.).
Comment If the preceding solution is repeated with Y P = Y T = 2.00, the same answer
(d = 54.1 mm) as in Ex. 7.3 is obtained. As employed here, load factors that all have the same
value are mathematically equivalent to safety factors.

7.6.3 Stress Raiser Effects

Engineering components necessarily have complex geometry that causes stresses to be locally
elevated—for example, holes, fillets, grooves, keyways, and splines. Such stress raisers are often