Answer to Question #137241 in Mechanical Engineering for ONKAR laha

Question #137241
Design a right angled bell crank lever. The horizontal arm is 500 mm long & a load of 5 kN
acts vertically through a pin in the forked end of this arm. The other arm is 150 mm long. The
lever consists of forged steel material and a pin at the fulcrum. Take the following data for
both pin & lever material: Allowable tensile stress = 75 N/mm2
; Allowable bearing pressure
in pins = 10 MPa; Allowable shear stress = 70 N/mm2
.
1
Expert's answer
2020-10-08T15:10:24-0400

Consider the diagram below



Given FB =500 mm; W = 5 kN = 5000 N; FA = 150 mm ; "\u03c3_t=75 MPa=75 N\/mm^2, \u03c4=70 N\/mm^2; p_b=10 N\/mm^2"

W X 500 = P X 150

"P=\\frac{5000\u00d7500}{150}=16666.67 N\n"


Reaction at F

"R_F=\\sqrt{5000^2+16666.67^2}=17400.51 N"


Design for fulcrum pin

"17400.51=d\u00d7l\u00d7p_b \\implies d= \\sqrt{\\frac{17400}{10\u00d71.25}}=38 mm"

l= 1.25 d = 1.25 × 38 = 47.5 mm


Check for shear stress

"17400.51 = 2\u00d7\\frac{\\pi}{4}\u00d7d^2\u00d7\\tau=2\u00d7\\frac{\\pi}{4}\u00d738^2\u00d7\\tau=2268\\tau"

"\\tau= \\frac{17400.51}{2268}=7.7 Mpa\n" The design is safe


Diameter of hole in the lever = d + 2×3 = 38+6 = 44 mm

Diameter of boss at fulcrum = 2×d + 2×38 = 76 mm


Bending moment at the fulcrum, M = W× F_B = 5000×500= 2500000 N-mm


Section modulus, z = "\\frac{\\frac{1}{12}\u00d745(76^2-44^2)}{76\/2}=34913.684"

"\u03c3_b=\\frac{2500000}{349136.84}=7.2 MPa" The design is safe


Design for pin at A.

The effort at A is nit very much different from the reaction at fulcrum

Diameter of pin = 38 mm

Length of of pin = 45 mm

Diameter of boss = 76 mm


Design for pin at B

"5000=d\u00d7l\u00d7p_b = d_1\u00d71.25d_1\u00d710=12.5d_1^2"

"d_1^2=\\frac{5000}{12.5}=400\\implies d_1=20mm"


Check for shear stress

"5000 = 2\u00d7\\frac{\\pi}{4}\u00d7d^2\u00d7\\tau=2\u00d7\\frac{\\pi}{4}\u00d720^2\u00d7\\tau=628.4\\tau"

"\\tau= \\frac{5000}{628.4}=7.96 Mpa\n" The design is safe


"t_1=\\frac{l_1}{2}=\\frac{25}{2}=12.5 mm"


Reduction of wear

Inner diameter of each eye = d + 2×3 = 20+6 = 26 mm

Outer diameter = 2×d + 2×20 = 40 mm


Bending moment at the fulcrum,

"M = \\frac{W}{2}(\\frac{l_1}{2}+\\frac{t_1}{3})-\\frac{W}{2}\u00d7\\frac{l_1}{2}=\\frac{5}{24}Wl_1"

"M =\\frac{5}{24}\u00d75000\u00d725=26041.7 Nmm"


Section modulus, z = "\\frac{\\pi}{32}\u00d7d_1^3=\\frac{\\pi}{32}\u00d720^3=786 mm^3"


"\u03c3_b=\\frac{26041.7}{786}=33.13 MPa" The design is within the safe limits


Design of lever

Bending moment,

"M = 5000 (500-50)=2250000Nmm"


Section modulus, z = "\\frac{1}{6}\u00d7tb^2=\\frac{1}{6}\u00d7t(3t)^2=1.5t^3"

"\u03c3_b=\\frac{M}{Z}=\\frac{2250000}{1.5t^3}"

"t^3=\\frac{1.5\u00d710^6}{75}\\implies t= 27.1 = 28mm"


"b=3t=3\u00d728=84 mm"


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