Cyclic Tensile Loading of Bars

Introduction

Bars are thin structural members. This chapter gives procedures for determining the resistance to fatigue failure of bars under varying loads.

Nomenclature for Bar Fatigue Analysis

3.5 Cyclic Tensile Loading of Bars

The case to be considered now is that for which an alternating axial load is applied to the bar. A diagram of this loading is shown in Figure 3-14. The mean load is designated as P_m, and the alternating component is designated as P_a.

For a brittle homogeneous material, a stress concentration factor must be applied to both the alternating and the mean stress in equations for failure under alternating loads since a static load is affected by stress concentration. Thus, the Soderberg relation becomes

where S_m and S_a are equal to P_m/A and P_a/A, respectively, and n is the factor of safety. In this case, K_e may be taken to be equal to K_t, so that the basic design equation becomes

where the A is either the reduced area at the discontinuity or the full area of the bar as given in Figures 3-1 through 3-12.

For a ductile material, a stress concentration factor need only be applied to the alternating stress in equations for failure since ductile materials under static loading have an effective stress concentration factor of one. In view of this, the Soderberg relation becomes

where f_tm and f_ta are equal to P_m/A and P_a/A, respectively, and n is the factor of safety. Here, A_m is the reduced area of the bar, and A_a may be either the reduced or the full area of the bar as given in Figures 3-1 through 3-12. Thus, the basic design equation for ductile rods under alternating tensile loads is

where K_e is equal to q (K_t − 1) + 1. Here, K_t is the theoretical stress concentration factor as obtained from Figures 3-1 through 3-12, and q may be obtained from Figure 3-15 for steel. For aluminum, magnesium and titanium alloys, fatigue data is scattered to the extent that a value of one is suggested for q.

Since cast iron is insensitive to stress raisers, stress concentration factors need not be applied in any formula governing alternatlng stresses. Thus, the Soderberg equation may be used in its basic form

where f_tm and f_ta are equal to P_m/A and P_a/A, respectively, and A is the reduced area of the rod. Thus, the design equation for cast iron bars under an alternating tensile load becomes

The previous discussion of bars under alternating tensile loads is summarized in Table 3-2.

Table 3-2: Design of Bars Under Cyclic Tensile Load

Material Type	Design Equation	Explanation of Terms
ductile	$$ { 1 \over n } = { P_m \over A_m ~F_{ty} } + { K_e ~P_a \over A_a ~f_{se} } $$	Am = reduced area of section. Ke = q (Kt − 1) + 1 where q is obtained from Figure 3-15 for steel or set equal to 1 for aluminum, magnesium, and titanium alloys, and where Kt is obtained from Figures 3-1 through 3-12. Aa = reduced or full area of section as given in Figures 3-1 through 3-12.
homogeneous brittle	$$ { 1 \over n } = { K_t ~P_m \over A ~F_{ty} } + { K_t ~P_a \over A ~f_{se} } $$	Kt = theoretical stress concentration factor as obtained from Figures 3-1 through 3-12. A = reduced or full area of section as given by Figures 3-1 through 3-12.
cast iron	$$ { 1 \over n } = { P_m \over A ~F_{ty} } + { P_a \over A ~f_{se} } $$	A = reduced area of section.

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3.6 Sample Problem- Bar Under Cyclic Tensile Load

Given: A stepped circular bar with a fillet of 0.02 in. radius is to support a cyclic tensile load that is given as P = 20,000 lbs + 10,000 sinωt lbs. This bar, as shown in Figure 3-16, is made of an annealed steel for which F_ty = 45,000 psi and f_se = 35,000 psi. Use a factor of safety of 1.5.

Find: The diameter d.

Solution: From Section 3.5, the basic design equation for a ductile material is

$$ { 1 \over n } = { P_m \over A_m ~F_{ty} } + { K_e ~P_a \over A_a ~f_{se} } $$

P_m, P_a, and n are equal to 20,000 lbs, 10,000 lbs. and 1.5, respectively. By checking the explanation of terms in Section 3.5 and checking the figure referred to there, we find that both A_a and A_m are the reduced area π d²/4. This explanation of terms also tells us that K_e = q (K_t − 1) + 1, where q is given by Figure 3-15. Consulting this figure, q = 0.89 and thus K_e = 0.89 (K_t − 1) + 1.

The above values may now be substituted into the basic equation to obtain

$$ { 1 \over 1.5 } = { 20000 \over {\pi d^2 \over 4} (45000) } + { [0.89 (K_t - 1) + 1] ~10000 \over {\pi d^2 \over 4} (35000) } $$

Simplifying this equation gives

$$ 0.66 = { 0.605 + 0.323 K_t \over d^2 } $$

Try d = 1.33. Since d/D and r/d are equal to 0.870 and 0.0601, respectively, K_t = 1.75 according to Figure 3-1. Substituting this in the above equation gives 0.6 = 0.6. Thus the guess for diameter is correct. If it were not, other diameters would have to be chosen to see if one satisfies the equation.

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