Question

Problem 8-11 An M14x2 hex-head bolt with a nut is used to clamp together two 15-mm steel plates. (a) Determine a suitable length for the bolt, rounded up to the nearest 5 mm. (1) Determine the bolt stiffness. (c) Determine the stiffness of the members. (d) Assume that this bolt joint will not be permanent. Determine the pre-load. (c) How much turn is needed to develop such a pre-load? (d) How much torque is needed to produce such a pre-load? (e) What is the maximal external separation load allowed before yielding? (f) What is the maximal external separation load allowed before separation?

Answer 1

Given bolt has M14x2 hexagonal bead bolt And thicknees ē t = 15 muy @ Frow Specification of bolt, we get the bolt diameter = d = 14 wong - Frony table dingensions of hexagonal nut, we get the height = H = 12.8 msus Grip length = b = plate thickness = 15+15 = 30 m ... Bolt length = t = 0+H = 30 + 12.8 = 4.2.8 muy Suitable length nearest sumy is [= 45 way - we have threaded length for usetric series . for L = 125 mong is L = 2d t6 ming = 2114) + 6 = 34 may from table of coarse pitch & fine pitch threads at nonsinal dia of 14 wins, we get Tensile stress area = At = 115 mm2 unthreaded Portion area = Ad = I d ² = 1 x 14 1 = 153.9 may length of unthreaded Portion grip = ld=2-Lq = 45-34 = llmuy Threaded portion length in grip = b = lrld = 30-11 = 19 mm - we know for sted, E = 207689

- Adatt ... Bolt stiffness =k6= - Aaltt Ald 153.94 X 1157 201 (153.94x19) +(115x11) k=874.62 MN/us @ Merber stiffness = kus = 0 0.57749 Ed 26 r.o.57748to5d7 - 0.5774 ITX 2076 14 051742+ 2.5d 3) King =- 26 to 0.5774 x 30 + 0.5x14 T0.57748 30 + 2.5x14 2 km = 3116.45 MN/W) - (d) If bolt was not promanant, they Preload =P = 1.9 kml = 1.943116.4 x 15 x 103 MN TP = 88.8 MN)