Question 10.14: A frame as shown in the figure below is required to fix on t......

(a) The proof strength of class 5.8 steel is 380 MPa. The nominal load for each of the two bolts is 5 kN. A factor of safety of 2.5 is sufficient where the human life and the damage of other equipment are not involved. Hence assuming the same condition, a slightly higher factor of safety, say 3, is used in the present design.

Design over load for each bolt is 5 × 3 = 15 kN

Neglecting the stress concentration factor, we can write

\sigma_t=\frac{F}{A_t}

380=\frac{15000}{A_t}

A_t=39.47 \mathrm{~mm}^2

The appropriate standard size of the bolt is \mathrm{M} 10 \times 1.5 \text { with } A_t=58 \mathrm{~mm}^2

(b) The torque required for tightening a bolt can be written as

T=K F_i d

The torque coefficient for lubricated bolts is given as (Table 10.10)

K = 0.18

The preload on the bolts for a permanent connection, can be written as

F_i=0.90 F_p

The proof load on the bolts can be written as

F_p=A_t \sigma_p

The tensile stress area for metric M10 × 1.5 bolts, A_t=58 \mathrm{~mm}^2

Therefore, the proof load on the bolts is given as

F_P=380 \times 58=22.04 \mathrm{~kN}

Therefore, the preload on the bolts is given as

F_i=0.9 \times 22.04=19.836 \mathrm{~kN}

The tightening torque on the bolts is given as

T=0.18 \times 19.836 \times 10=35.71 \mathrm{~kN} \cdot \mathrm{mm}

(c) The stiffness of the bolt, 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 bolts is given as

A_d=\frac{\pi \times 10^2}{4}=78.54 \mathrm{~mm}^2

The height of the hexagonal head of the bolt for a 10 mm size bolt and assuming regular hexagonal nut, is given as

H=0.65 d=0.65 \times 10=6.5 \mathrm{~mm}

The total length of the bolt is given as

\text { total length }=L_G+H

Therefore, the total length of the bolt is given by

Total length = 45 + 6.5 = 51.5 mm

The length of a standard bolt that is nearest to the value obtained above is given as L = 60 mm

The thread length, L_T=2 D+6

L_T=2 \times 10+6=26 \mathrm{~mm}

The length of the unthreaded portion in the grip, l_d=L-L_T

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

l_d=60-26=34 \mathrm{~mm}

The length of the threaded portion in the grip, l_t=L_G-l_d

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

l_t=45-34=11 \mathrm{~mm}

The modulus of elasticity for the bolts assuming they are made of steel, E = 207 GPa Therefore, the stiffness of the bolt is given as

k_b=\frac{78.54 \times 58 \times 207 \times 10^3}{58 \times 34+78.54 \times 11}=332.5 \mathrm{~kN} / \mathrm{mm}

Assuming the steel surface as flat and smooth one, the stiffness are assumed to be proportional to their area.

Hence

\frac{k_b}{k_b+k_m}=\frac{A_d}{A_d+A_m}

A_m=A_c=d^2+0.68 d l+0.065 l^2 [Eq. (10.56)], l is grip length

A_c=d^2+0.68 d l+0.065 l^2 (10.56)

\frac{k_b}{k_b+k_m}=\frac{78.54}{78.54+537.625}=0.128

Therefore, the bolts carry 12.8% of the external load and the members carry 87.2% of the external load.

The load carried by the bolts can be written as F_b=C F+F_i

The total external load is specified to be 10 kN. Assuming the load is equally shared by the two bolts, the external load on each bolt is 5 kN.

Therefore, the load carried by each bolt is given as

F_b=0.128 \times 5+19.836=20.476 \mathrm{~kN}

The load carried by the members, F_m=(1-C) F-F_i

Therefore, the load carried by the members for each bolt is given as

F_m=(1-0.128) \times 5-19.836=-15.476 \mathrm{~kN}