Question

A cantilevered cylindrical shaft of length 1.6m and diameter 100 mm is loaded by the force vector F = -3; -5j + 10k kN applied 20 mm above the cross-sectional centroid on the end of the shaft, as shown. Note that the drawing does not correctly indicate the direction of the vector F (the direction of the vector F is given by the vector expression above), but it does correctly indicate the location that the vector is applied. 800 mm - 10.20.1600)mm 100 mm Determine the states of stress at the four points (A, B, C, and D) located in a plane at the midpoint of the shaft's length. Express those stress states on stress cubes. Clearly label the coordinate system and all of the stresses on each of the four cubes.

Answer 1

Solution i a za f=- zî 5] took KN 8oomm -TB A -1.6m 16mm y -324 face-1 ſ face-1 Planeter Cd loo mm) tace - 2 Face- ay-Plane - >ge Foc) BKN Tu 1971 ffy 45kn F = 10kr (Tension loon i. I OY Forces, i 1) Fr Bending on nz plance (reutral only on yz) L Torston - ( Torsten = 3x20 = 60 Nm) Plance Constant & Fy 3) Fr — - Lada Creab shun shuns conse Bending on y z plane (Neutral ands) on az) simple Tenslon of Tokn fone.

1 ) 2 nd 4 64 Stress due to (fe) force on all points (ABCO) :- 4 as Bending due to fire i 5 5) a bending pagelten My moment constant a Fore) & Distance (Two) 2.5" y = distance of fretume flbrue (Yz diz) (8) = Normal kanting Strees (m) = (Fr.) x 800 a 3x 800 (M = 2400 NM) = 8630 x 103 rimm =) Put all values in eno : Î - j aim (65) - Qubota 012. = 32x2409 X103 32.42460/ x103 71d4 . a m.mm SC 7430 Klo3 - 32 18h everyone их оə2 32 +24 (охл. 24.46 TGV = 24.46 Noron e or mpal - -24.46 N/mm 2 or mpa = - At A :- (stury) = +65) (Tensile, upper fibre) At Be C and D (struy) o C plene in Mantral plona) N y co

(b) Torsion due to for Torsion en . Ty Torque (Nomm] = Box13 mm J Polar moment of Inertia =df 32 (als as sheer shrum ed destina Ri from centru Ra Radhy of circle (shatt) fatting all values is ez ® her top (0) = (602.3) x16 37 Td 3 Gitreme Cal = (60x103) x16 - 1646 ne X (100) 3 - Toon = 0.30 N/mma to Enburne = 0.3 almos- i At A, B, C a fol tesla = (tele - (Tos) c = 0.3 Mpa (N/mm2) los), = o Con Shaft Center). 2) Strugs due to fy on all points (AB Co) :- L (fy) only creab bending on y z plane 4 Neutral anis on noz. plane 37

(- g) M= (fy)x 800 (disture) M (moment) = 5x105x qoxias m = 4000 NiM = 4000 x103 Nonna Patting all valuy in binding equation m ely 4000 x103) x dla = (bely Lynn 7 Nimm To 164 32x 4000 x103 = 32x4 = TX (10o3 A = 40.74 N/mm2 Tooly = 40024 Nimm2 or mpa] 124 Nimma or mpa At A i At B i Strugs dur (fyle=0 (on Neutral plane) Strue due to (fy) = -(toly = -40174 (compune) ne (Tenslon) Ate & (Фотре-.) (Strunon due to by = -40,74 Strus due les fy = (toly (Tension 40174 Mpa. tu Mpa Ato strus - o

1 3 Stress due fz on all poinly o (FZ) = Tensile load passes tough centures (7) - 1.273 (Tenside] (Tensile-Stress) mpa .. Stous on all points (A, B, GD) ds same that is equal to (1.273 mpa) 0,= 11273 mpa. # Cammulative Struges ; - 1 At point A : dia to fu Total Stress = (tot tos} of) 1 = (24.46, 0.37 1. {N/mm2 or mpa. . 2 At point B. 8 = [24:46,0-3, 1024 ] Nimma - Total strugs = E-Goly (6.13 573 = (-40, 14, 013]1.3]/mm2 or mpa 3) At point co- = (-40074, 0.3, 1.27] Nimma Total Strass = {(foly, tes} or] P = [+40.74, 0.3] 12/11mm? or mps 4 At Point D :- (+40.74 0.3, 1:27) W/mm2 Total Stress =(0, of 8q) = Co, o, 1027] wl mm2 2

* Stress maton (1) Point Al Гар буу Тыа 1 1, | + 5 - To gi? ( 11 1:ја | to41 4 L, •2х ог, ба с 2 2 A 4 fokszor (25) в Со оо. о- Elu 3 от 1 По о. 3 от | 0.3 o o Loo 2517 J теті - ч, Ч J, 22 o iss I z 25.M/ (2\ ради Point - , (1) (о 0.3 от Тә-3 o o Т Lo o -35.ча аг - Чо23 ( — — 35.42 Mp4 (3) Point - д: 1 д: 1 о Ca 2 Чо?+ (+22 - 42 ми | o o 4 | 1 (4) point-D Гco1 - o o on Cool• 12 |