Question 10.13: (a) Determine the member stiffness of the joint as shown in ......

The stiffness of the bolt can be written as

k_b=\frac{A_d A_t E}{A_d l_t+A_t l_d}

The area of the unthreaded portion of the bolts is

A_d=\frac{\pi d^2}{4}

Therefore, the area of the unthreaded portion of the bolt (M12) is given as

A_d=\frac{\pi \times(12)^2}{4}=113 \mathrm{~mm}^2

The tensile stress area for metric threads for a diameter of 12 mm and assuming coarse pitch, is given as (Table 10.2)

A_t=84.3 \mathrm{~mm}^2

The length of the grip is given as

L_G= sum of the thickness of the plates

L_G=20+25=45 \mathrm{~mm}

The height of the hexagonal head of the bolt for a 12 mm size bolt and assuming regular hexagonal nut, is given as

H=0.65 d=0.65 \times 12=7.8 \mathrm{~mm}

From the standard value (refer to Table 10.6), H = 7.95 mm

The total length of the bolt is given as

Total length =L_G+H

Therefore, the total length of the bolt is given as

Total length = 45 + 7.95 = 52.95 mm

The length of a standard bolt that is nearest to the value obtained above is

L = 60 mm

The thread length for a bolt length of 60 mm and a bolt diameter of 12 mm can be written as

L_T=2 d+6

Therefore, the thread length is given as

L_T=(2 \times 12)+6=30 \mathrm{~mm}

The length of the unthreaded portion in the grip is L_d=L-L_T

Therefore, the length of the unthreaded portion in the grip can be written as

l_d=60-30=30 \mathrm{~mm}

The length of the threaded portion in the grip is l_t=L_G-L_d

Therefore, the length of the threaded portion in the grip is given as

l_t=45-30=15 mm (total thickness is 45 mm)

The modulus of elasticity for the bolts made of steel is E = 207 GPa

Therefore, the stiffness of the bolts is given as

k_b=\frac{A_d A_t E}{A_t l_d+A_d l_t}

k_b=\frac{113 \times 84.3 \times 207 \times 10^3}{84.3 \times 30+113 \times 15}=466.82 \mathrm{~kN} / \mathrm{mm}

As in this problem the two members are made of different materials, the member stiffness needs to be calculated using the procedure outlined below.

Since the members do not have an identical thickness, three regions (frustum’s) need to be considered to calculate the overall member stiffness. These three frustums are shown in the figure below.

The stiffness of the member in each frustum assuming a half-apex angle \alpha=30^{\circ} can be written as

k \approx E d\left(\frac{0.702+0.654(d / L)}{1-0.12(d / L)}\right)

and as per Sighley

k=\frac{0.5774 \pi E d}{\ln \frac{(1.155 t+D-d)(D+d)}{(1.155 t+D+d)(D-d)}}

Upper frustum

The thickness of the member in the upper frustum is given as

l=t=20 \mathrm{~mm}

The modulus of elasticity of the member material in the upper frustum (steel) E = 207 GPa

The dimension for the upper frustum can be written as

D=d_w=1.5 d

Therefore, the dimension D for the upper frustum is given as

d=D=1.5 \times 12=18 \mathrm{~mm}

Therefore, the stiffness of the member in the upper frustum is given as

k_1 \approx 207 \times 10^3 \times 18\left(\frac{0.702+0.654(18 / 20)}{1-0.12(18 / 20)}\right)=5391 \mathrm{~kN} / \mathrm{mm}

As per Shigley

k_1=\frac{0.5774 \times \pi \times 207 \times 10^9 \times 12 \times 10^{-3}}{\ln \left(\frac{(1.155 \times 20)+18-12)(18+12)}{(1.155 \times 20)+18+12)(18-12)}\right)}=4470 \times 10^6 \mathrm{~N} / \mathrm{m}

Central frustum

The thickness of the member in the central frustum is given as

l=t=2.5 \mathrm{~mm}

The modulus of elasticity of the member material in the central frustum (gray cast iron FG 300), E = 90 to 113 GPa

Let E = 100 GPa

The dimension, D, for the central frustum can be written as

D=d_w+2\left(20 \tan 30^{\circ}\right)

Therefore, the dimension D for the central frustum is given as

D=18+\left(2 \times 20 \tan 30^{\circ}\right)=41.09 \mathrm{~mm}

Therefore, the stiffness of the member in the central frustum is given as

k=\frac{0.5774 \pi E d}{\ln \frac{(1.155 t+D-d)(D+d)}{(1.155 t+D+d)(D-d)}}

k_2=\frac{0.5774 \pi \times 100 \times 10^3 \times 12}{\ln \left\{\frac{(1.155 \times 2.5+41.09-12)(41.09+12)}{(1.155 \times 2.5+41.09+12)(41.09-12)}\right\}}=52229.2 \mathrm{~kN} / \mathrm{mm}

Lower frustum

The thickness of the member in the lower frustum is given as

l=t=22.5 \mathrm{~mm}

The modulus of elasticity of the member material in the lower frustum (gray cast iron), E = 100 GPa

The dimension D=d_w

Therefore, the dimension D for the lower frustum is given as D = 18 mm

Therefore, the stiffness of the member in the lower frustum is given as

k_3=\frac{0.5774 \pi \times 100 \times 10^3 \times 12}{\ln \left\{\frac{(1.155 \times 22.5+18-12)(18+12)}{(1.155 \times 22.5+18+12)(18-12)}\right\}}=2074 \mathrm{~kN} / \mathrm{mm}

The net stiffness of the member can be written as

k_m=\left(\frac{1}{k_1}+\frac{1}{k_2}+\frac{1}{k_3}\right)^{-1}

Therefore, the net stiffness of the member is given as

k_m=\left(\frac{1}{4470}+\frac{1}{52229.2}+\frac{1}{2074}\right)^{-1}=1379.3 \mathrm{~kN} / \mathrm{mm}

The stiffness constant of the joint, C=\frac{k_b}{k_b+k_m}

C=\frac{466.82}{466.82+1379.3}=0.253

The load factor guarding against over-proof loading is obtained as follows.

We have F_b=F_i+\frac{k_b}{\left(k_b+k_m\right)} F=F_i+C F

If N is the factor of safety guarding against the proof load failure

F_b=\sigma_p A_t

and \sigma_p A_t=F_i+N C F

or N=\frac{\sigma_p A_t-F_i}{C F}=\frac{650 \times 84.3-0.9 \times 650 \times 84.3}{0.253 \times 66267}=2.124

where the proof strength for ISO class 9.8 bolts is 650 MPa.

\sigma_p=650 \mathrm{~MPa}

The proof load is then F_p=A_t S_p=84.3 \times 650=54.795 \mathrm{~kN}

The preload force F_i=0.9 F_p=0.9 \times 54.795=49.3155 \mathrm{~kN}

The external load acting on each bolt is given as

F=\frac{\text { effective sealing area } \times \text { presssure }}{\text { number of bolts }}

The effective sealing diameter is specified as 150 mm.

Therefore, the external load acting on each bolt is given as

F=\frac{\frac{\pi}{4} \times 150^2 \times 3}{8}=6.627 \mathrm{~kN}

The load factor guarding against joint separation is given as

N=\frac{F_i}{(1-C) F}=\frac{49.3155}{(1-0.253) \times 6.627}=9.962