Question 10.12: A steel bracket is bolted to steel ceiling joists by two bol......

Bolt preload is usually specified between 75% and 90% of proof stress. Hence assuming

0.75 F_p \leq F_i \leq 0.9 F_p

where F_p=\text { proof load }=A_t \times \sigma_p=258 \times 600=154.8 \mathrm{~kN}

\sigma_p=600 \mathrm{~MPa}

d=20 \mathrm{~mm}

A_t=258 \mathrm{~mm}^2

0.75 \times 154.8 \leq F_i \leq 0.9 \times 154.8

116.1 \leq F_i \leq 139.32 \mathrm{~kN}

Let F_i be 125 kN.

If friction is normal and the same for both,

\left(M_t\right)_{\mathrm{tot}}=0.2 F_i d=0.2 \times 125 \times 20=500 \mathrm{~N} \cdot \mathrm{m}

In order to determine how the load is shared between the bolts and joint members, it is necessary to examine their stiffness. As the joint is metal to metal, using the following relation, the stiffness can be obtained.

k_c=\frac{A_c E_c}{L}=\frac{F}{\delta} \approx E d\left(\frac{0.702+0.654(d / L)}{1-0.12(d / L)}\right)

\approx 207 \times 20\left(\frac{0.702+0.654(20 / 24)}{1-0.12(20 / 24)}\right)=5736.2 \mathrm{~kN} / \mathrm{mm}

where L = 48/2 = 24 mm and E = 207 GPa.

(L is assumed as ½ of the grip length)

\frac{1}{\left(k_c\right)_{\mathrm{eff}}}=\frac{1}{k_{c_1}}+\frac{1}{k_{c_2}}=\frac{1}{5736.2}+\frac{1}{5736.2} (assuming equally shared)

Nut height = 0.9 d = 0.9 × 20 = 18 mm

Half of the nut length = 9 mm

Assuming 2 exposed threads of length = 2p = 2 × 2 = 4 mm

L_t=2 d+6=2 \times 20+6=46 \mathrm{~mm}

Length of bolt =L_G+H+2 p

L=48+18+4=70 \mathrm{~mm}

l_d= length of unthreaded portion =L-L_t=70-46=24 \mathrm{~mm}

l_t= length of threaded portion =L_G-l_d=48-24=24 \mathrm{~mm}

Area of shank, A_d=\frac{\pi}{4} d^2=\frac{\pi}{4} \times 20^2=314.16 \mathrm{~mm}^2

Area of threaded part, A_t=258 \mathrm{~mm}^2

k_b=\frac{A_t A_d E}{A_t l_d+A_d l_t}=\frac{258 \times 314.16 \times 207 \times 10^3}{258 \times 24+314.16 \times 24}=1221.8 \mathrm{~kN} / \mathrm{mm}

Hence, for joint assembly the stiffness is

\frac{1}{k_e}=\frac{1}{k_c}+\frac{1}{k_b}

\frac{1}{k_e}=\frac{1}{2868.1}+\frac{1}{1221.8}

k_e=856.8 \mathrm{~kN} / \mathrm{mm}

C=\frac{k_b}{k_b+k_c}=\frac{1221.8}{1221.8+2868.1}=0.299

Shared load when 40 kN is applied

F_b=F_i+\frac{k_b}{\left(k_c+k_b\right)} F=125+\frac{k_e}{k_c} F=125+\frac{856.8}{2868.1} \times 20 (load shared equally; hence 20 kN is taken in the computation)

F_b=130.98 \mathrm{~kN}=131 \mathrm{~kN}

F_c=F_i-(1-C) F=125-(1-0.299) \times 20=110.98 \mathrm{~kN}