Two steel members are joined by two M 12 × 1.5, class 5.8 bolts. The external load applied to a bolted joint fluctuates between 0 and 12 kN. The endurance strength of the bolt material in reversed axial loading is 176 MPa. If the ratio of the combined stiffness of the member to bolts is 3, determine
(a) The minimum initial tightening load that should be applied to prevent separation.
(b) The factor of safety guarding against fatigue loading using Goodman criterion.
For M12 × 1.5, class 5.8 bolts
A_t=88.1 \mathrm{~mm}^2, \sigma_p=380 \mathrm{~MPa}, \sigma_u=520 \mathrm{~MPa}
assuming rolled threads, k_f=2.2
Given \frac{k_b}{k_c}=\frac{1}{3}
where k_c is the combined stiffness and k_b is the bolt stiffness.
C=\frac{k_b}{k_b+k_c}=\frac{k_b}{k_b+3 k_b}=\frac{1}{4}=0.25
(a) We have F_b=F_i+\frac{k_b}{\left(k_c+k_b\right)} F
Load to cause joint separation
F_b=F
and the above equation becomes
F=F_i+\frac{k_b}{\left(k_c+k_b\right)} F
or F_i=F\left\{1-\frac{k_b}{\left(k_c+k_b\right)}\right\}
Hence to prevent separation
F_i>F\left\{1-\frac{k_b}{\left(k_c+k_b\right)}\right\}
F_i>12(1-0.25)=9 \mathrm{~kN}
(b) F_b=F_i+\frac{k_b}{\left(k_c+k_b\right)} F
\left(F_b\right)_{\max }=9+\frac{1}{4} \times 12=12 \mathrm{~kN}
\left(F_b\right)_{\min }=9+\frac{1}{4} \times 0=9 \mathrm{~kN}
\left(F_b\right)_m=\frac{9+12}{2}=10.5 \mathrm{~kN}
\left(F_b\right)_a=\frac{12-9}{2}=1.5 \mathrm{~kN}
\sigma_m=\frac{\left(F_b\right)_m}{A_t}=\frac{10.5 \times 1000}{88.1}=119.18 \mathrm{~MPa}
\sigma_a=\frac{\left(F_b\right)_a}{A_t}=\frac{1.5 \times 1000}{88.1} 17.02 \mathrm{~MPa}
S_e=\frac{176}{2.2}=80 \mathrm{~MPa}
\sigma_i=\frac{F_i}{A_t}=\frac{9000}{88.1}=102.16 \mathrm{~MPa}
S_m=\frac{\sigma_u\left(S_e+\sigma_i\right)}{S_e+\sigma_u}=\frac{520(80+102.16)}{80+520}=157.87 \mathrm{~MPa}
S_a=S_m-\sigma_i
S_a=157.87-102.16=55.712 \mathrm{~MPa}
Factor of safety against fatigue failure is (C = 1)
N=\frac{S_a}{\sigma_a}=\frac{55.712}{17.02}=3.273
Alternatively,
N=\frac{\sigma_u A_t-F_i}{\frac{C F}{2}\left(1+\frac{\sigma_u}{S_e}\right)}=\frac{520 \times 88.1-9000}{\frac{0.25 \times 12000}{2}\left(1+\frac{520}{80}\right)}=3.272
Factor of safety in the absence of preload (C = 1)
N_0=\frac{\sigma_u A_t}{\frac{F}{2}\left(1+\frac{\sigma_u}{S_e}\right)}=\frac{520 \times 88.1}{\frac{12000}{2}\left(1+\frac{520}{80}\right)}=1.018
It is clear that the presence of pre load is beneficial as it increases the factor of safety.