Question

Question 2: For the connection shown in Figure determine whether the bolts are adequate to support the applied load under combined loading. The bolts are 6M20 (A325) in standard holes. Assume that each bolt takes an equal amount of shear and tension Pu 600 kN d 50 mm 80 mm 80 mm 30°

Answer 1

Given: The figure is shown below, Pu= 600 KN 50 mm 80 mm LA 80 mm Each bolt takes an equal amount of shear and tension. The angle 600 sin is 30°. Calculation:

From the above figure the number of bolts are 6. Since, Fx = 400 N/mm² and F, = 250 N/mm2 Draw the free body diagram for force,

600 sine 600 KN → 600 cose From above the vertical force that is shear force can be calculated as, F, = 600 sin 30° = 600x- = 300 KN

So load cause by each bolt can be calculated as, 300 FByeah balt== = 50 kN And similarly the horizontal force that is tensile force can be calculated as,

F, = 600 cos 30° = 600x3 = 519.6 kN So tensile force created in each bolt 519.6 Tensile force in each bolt = 86.6 kN From the table, Minimum tensile strength is 827 N/mm2. The formula to calculate shear strength of bolts can be given as, Vad = 4s in the Substitute all the value in above equation.

0.75*(20)2}x400 13x1.25 0.75* (20)2 x 400 V3x1.25 94200 2.165 = 43510.39 N =43.51 KN And formula to calculate Vat is given as, V = 2.5kde La Here, Take the minimum of the three value of k.

(1). 5.-0.25=32 -0.25 =0.96 = 0.75 Substitute all the value in equation of Vd Vas = 2.5x0.75 x 22x10x400 = 412.5 x 400 = 132000 N =132 KN Strength of bolt is given as, Strength of bolt =min Vds and Vób). So, strength of bolt is given as, Strength of bolt=min Vds and Våb}

=min 45.3 kN and 132 kN) = 45.3 kN (not acceptable) And so the joint is not safe. The formula to calculate the strength of bolt can be given as, tensile strengths of bolt = 0.94,bfy = 0.9x 244.91 x 400 = 88167.6 kN =88.16 kN The above value should be less than the f, 'mo Agb

se poate 4.5 = 250 x 1,35 (20) = 89204.545 kN =89.20 KN Here, 88.16 1.25 = 70.528 Since, 86.6 > 70.528, so this also not safe in tension. For a high strength friction grip bolt, 73x1.25 Viso = 0.78*7*(20) 783744 75x1.25x4 783744 13x1.25x4 783744 8.66 = 90501.61 N =90.50 KN

And again, tensile strength can be calculated as, Tensile strength = 0.9x 244.92 x 800 =176.34 kN Since, 176.34 kN > 86.6 kN (Acceptable} The above equation is safe in tension.