Question

A solid circular cantilevered beam is subjected to, at the free end, a downward transverse force P, a torque T and a centric (axial) load F. Given that its diameter is 60 mm, T = 0.1P Nm and F = 10P N.

(a) Determine the largest value of the load P that can be applied if the allowable stresses are 100 MPa in tension and 60 MPa in shear on a section at 120 mm from the free end. Take into account all stresses involved, do not ignore any.

(b) Conclude on the distribution of stresses along the beam.

Answer 1

The solution is shown below

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Nm Given Information. P-Transverse force, T = 0.1P = Torque. F = Axial force = lop N. y d=60mm 1200 lop TOP (1) Bending T= stress due to P at 120mm My P(120) x y da 64 = 1.886 3 X10 4y xp. at y=30 To= is max. 1.886 3 X10 4 x 30 xp = 5.66 X10-3p. (in) Maximum stress. Axial À = [equal along y'dietion] = 3.548103 N/mm² - o до Р Ida Td2 Trans verse Shear a = 4X (P) X (R² - y2) 34 - is Maximum at y=o P 4 - TTR² Shear due to torsion. T > Ts - TY - TÜ) Td4 TR4 32 2 2TY TR4 2x0oIP (4) T[30)4 7.86X108 (P) Ú) =

Let us find where tt = combined shear (q + Ts] is maximum. 4P [p²-g] 7 2Tly) 3124 ro-24] + 2[o.DP ) tpЧ IT OF 4 P 31124 L TR4 0.2P 4P Cf2y) = Элец y = 184 (3) x0,2). 0.075 mm which is almost near Neatral axis]. I 0.075 Let us find total stoels at 0, 6 and ② At o (i) To = Tmax = 5.66610 3 p. 2 Ja = 3154 x 10 3 P. total - Sa+b Total = 9, 2x10 P.U = 9, 2010 3p. (1) Shear at Transverse shear =0 Torsional shear = 16T – 164O.IP TT do U (30) Total = 10 89410 5p. = 1089 yio bp. bet At 2) (1) b = 108863410 3 yp = 10886 3X103 C0.075] P. = 1.415 x1644 p. Talex 3.54x10 3p (equal eve Total = 3.9558 10 3p.

(1) Tat 6.015) = 4P, CR² - y2] Эty py 4.712 710 Up. 16 Ty = 5.894109, = To at 0.075 = " TRY (which is negligable ! T + Ts - Trotol = 4.712 810-4.p. A (3) Ta = 3.54x10 3p I total = 3.54x10-3p. ] cao T = 48 CR² - y2] [is ri ] Tp = 0 ; T t = (4=o ] 3TR4 = 4 T 8 TIR2 = 16 X10 p. find principal Let us Strees at all locations. 2 2 oy=o 5,2 = Extry + Soxtrogy? + Try? here, Tx= Total, Trey-Tritos Major -75, = x + (x + tugo At Tato = 9.2 X10 3p + ( 19. 2x10 3p)² + (11898107) 4.6410 p + 4.6X10-3p Stress 2 - - 9.2 X10 3p. Imax = x + ²y = 4.68x103p.

А+ 2 - - - Cmax 55 ° 59 х 3р з 9 як хло 3) + (1, 12х1о чр) q175 х 10р + 2. 039 x16 2P - olox to 3р. = 2.оззX10 р. А4 о - 3, 54х10 9 + 35 мяно Зр) 24 (171686 ) - 2 : - 711 3P + 1,332 Уto 3р. 3 602 103 P. Cmax (5) - ) o 832 Xio 3р. comparing all this, we can say the critical section is at б) ауO – Ч. 2 7(o 3p Mpg "Creaчa O - что Хto Эр Мра, oj = он = 9,2 х10 3 Р - Гоо Р= (9102 1031EN emax = 16 60 = Call = 4.68103p. Р= 13, очъKNI Take Minimum Value of p. Vale off, 10.87 KN) Iso Pminimm - (Б) So, Maximum axial shey ?s q+ ), Bending axial. total.