Question

Required information A 1.75-in. steel plate and a 1-in gray cast iron plate are compressed with one bolt and nut. The bolt is in-13 UNC. A 12 N American Standard plain washer is used under the head of the bolt and under the nut. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Determine the bolt stiffness. The bolt stiffness is Mib/in.

Also find, the length of the bolt rounded up to nearest 1/4 inches. Please box all your answers. thank you

Answer 1

Given 1.75 in Steel Plate lin care iron Bolt % in 13 UNC A N American Standard washes as Grip length l=ts ttc + 2 t ts- thickness of Steel Place; Ес - thickness of cast iron Washer thickness tacle A-32 basic dimensions of american Standard plain face of nut. t = 0.045 in la 1.75 +1 + 2 C 0.0953 d = 2.94 in Nut height w 16 Fasteners length L + H = 2.94 + 1 16 - 3.3115 L = 3.5 Suitable length of bolt - 3.5

b) Bolt stiffness Ad. At. E kb Ad. It t Atild Theread length LT = 2d+! = 2 (0.5) + 1 4 Ly= 1.25 in Length of unthreaded portion grip ld = L-LT - 3.5 -1.25 in ld = 2.25in length of threaded portion in grip lt = 2.94 -2.25 -0.69 Area o unthreaded portion Ad = ㅈ 4 . [o.s3² Ad 0.1963 in 2 Tonsile stress 2 area As = 0.14 19 in [ hom table 8-2 diameters and area of warse pitch )

0.1963 X 0.1419 x 30 Kb = [0.1963x0.69] + [0.1419x2.25) 1.83 mebslin (C) Material stiffness: Frustum of the material. orsin K → 10.095" 1:25 Steel 0.5 cast iron Member Steffness for each Steel Washer frustum 0.5774 ñ Eid en [1.155++-d ] [D+d) [1.155t + D+d) Tod] Inner diameter 0.53) in t table A-32] D 0.75 t = 0.095 in

0.5774.7. 30 x 0.531 en [1.15560.095) +0.75 -0.531] [0.75+0.531) [1.155 [0.095)+ 0.75 +0.531] (0.75-0.521) Ki 89.20 Mebfli Member Satness for top plate and top steel ditum : d=0.54 D- 1.25 + 2 10.095) tan 30 2 D=0.73 t = 1.25 i 0.5774xX 30 0.5 K2 = tu.155(1.25) +0.73-0.5) (0.73+0.5) en [1.155(1.25) 40.7370.5) [0.73-0.5) K2= 22.524 Mlbflin Member Stiliness for top plate and lower steel feustum d = o.sin D=0.73 + 2(1) tan 30 t=0.5in D=1.889

K3 = 0.5774 X X 30 X 0.5 [ 1.155X0.5 +1.889 -0.5] [1.8897015) [1.155 X0.5 +1-889 +0.5) (1.889-0.5) K3= 206.96 Mebf/ Member stiffness for cast iron Frustum: d = 0.5 in t-lú D=0.78 in E14.5 Mpsi 0.5774 x T x 14.5 X 0.5 [1-155x0.5 +0-73-t.5) (0.73+0.5) en [1155x0.5 +0.73 +0.5) (0.73-0.5) kq = 11.63 Mobilin - km = 2 + + + نا ki ke + + + 89.2 22.524 206.96 11.63 km 6.397 mebflin