Question

Need step by step detailed solution plz.

The figure shows a steel bearing block that is to be bolted to a steel ceiling joist with 20 mm thick flanges, supporting a gravity load of 40 kN. M24 ISO 8.8 coarse threaded bolts are used with 4.6-mm-thick steel washers under the bolt head and nut. The dimension A, marked in the figure is 22mm. (a) Find the wrench torque required if the fasteners are lubricated during assembly and the joint is to be permanent. (b) Determine the factors of safety guarding against yielding, overload, and joint separation (c) If the joint is oversized, and it is decided to use only two bolts instead of four, recalculate the factors in part (b) and comment.

Answer 1

Solution Any (a) : - bolt (coarse threaded! For 924, Iso 8,8 grade Proof strength Sp = 600 MPa Tensile stress area Q 353 mm2 Preload tension for permanent connections: - Fi= 0.90 At SP Fi = 0.9 X353 X 600 fi = 190620 N a The wrench tarque required : T KEid where R = Tarque coettrient = 0.18 (for lubricated assembly 0:18 X190620 x(946163) T 823.4784 N-M AM(6) :- The helle diameter, D=1.5d = 1.5X24 36mm Young's modulus of steel, E = 207 X1OS MPa stiffness of the washer as follows: Ki= 0.57747 XEd (1) Anr(1.155++ D-d) (D+d) 1 (1.155t + Ord) (D. d) 0.57747 X 207 X103x24 Kts n. (4.155X4.6+36-24) 66 +24) (1.115X4.6+36+24) (36-24) K1=3.1998107 07 N/mm

helly diameter for the steel bearing block, D = 1.5dt at dan 30' 0 = (1,5X20 + 2(4.6) Han 30 - 91.31mm. Stiftness of the steel bearing block, (quation ) K2 0.5774X* X 207 X103 x24 16.155 X22 +41.31-24) (41.31+24) an 4(4.155X22+41.31+24) (42.31–24) Y = 1.5679 X107 N/mm the stiffness for the steel Joist :- K3= 0.5774 TX 2078103424 in S. (1.555 X20741.31-24) ( 41.31 +24) (1.155 *20 +41.34+24) (41.31-24) K3 = 1.65367 x 107 N/mm calculate the combined stiffness for the system; 1 1 2 1 ka K₃ km 1 1 1.5679X107 3.199x107 km 1.65367 x 107 + K1 2 + ot km = 5.354 X 106 N/mm Calculate the thickness of all materials squeezed between face of bolt & face of nuts l = 2 X 4,6 + 22+ 20 51.2 mm

bolt, obtain the height H of the H = 21.5mm (for Men bolt) Total length of the batt: - L = eth 51.2+ 21.5 L= 72.7 mm Prefferred size, L= 80 mm The threaded length, LT adt6 LT 2 (29)+6 = 54 mm Length of anthreaded portion in grip, la L-LT 80-54 - 26 mm Length of threaded portion in grip, Ita l-ld 51.2 26 lt = 25.2 mm nominal area of the bott (Ad) :- Ad= tlu d2 = /GX (2412 - 452.39 mm2 Calculate the bolt :- kb the stiffness of Ad At E Ad lt + Atld (452.39)(353)X207 X108 (452.39)(25.2)+(353)(26) ko = 1.60638106 N/mm Calculate the forest constant (C): –

C = Kb Kb+ km 1.6063 X 106 с (1.6063 X109) +(6.854X106) 0.2307 --> Calculate the factor factor of safety against yielding, Ptotal = 40 KN 40000 N load / bolt p= 40000 10000 N SN=4 =no. of 2 bolts www.ot? op Sp At ep & Ei (600)(353) (0.23)(10000)+(1.90620) np= inp=1.0978 calulate the factar of safety against loading, NL Spat - Ei ср a = (600) (353) — (3.96 620) 0.2307 X10000 NL: 9.177 Factor of safety Fi P (1-1) against Joint separation :- đ90620 20000(4 - 0.230775 No= no 24.78

only two Ans(): If the Joint is. oversized, and and it is decided to use bolts instead of u, N=2 Plotal 0000 = 20000 N N 2 -> Factor of safety against yielding, Sp At (P+Fi (600) (353) 0.2307) (20000) +1906.20 1.0848 Factar of safety against overload, NL SpAt -Fi СР 600) (353)-110620 = (0.2307) (90000) DL 34.388 Factor of safety safety against Joint separation :- Fi 1 90620 P(1-0) 20000 (1-0.2307) no no = 12.39 Here, we com observe that by using a boods instead of 4 balts, it does not affect sauch on the yielding factor but largely affects the sately against overload & Toint separation. [almost half of the previous] factor of

Thank you.