Answer to Question #137241 in Mechanical Engineering for ONKAR laha

Question

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
.

Answer 1

Consider the diagram below

Given FB =500 mm; W = 5 kN = 5000 N; FA = 150 mm ; $σ_t=75 MPa=75 N/mm^2, τ=70 N/mm^2; p_b=10 N/mm^2$

W X 500 = P X 150

$P=\frac{5000×500}{150}=16666.67 N$

Reaction at F

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

Design for fulcrum pin

$17400.51=d×l×p_b \implies d= \sqrt{\frac{17400}{10×1.25}}=38 mm$

l= 1.25 d = 1.25 × 38 = 47.5 mm

Check for shear stress

$17400.51 = 2×\frac{\pi}{4}×d^2×\tau=2×\frac{\pi}{4}×38^2×\tau=2268\tau$

$\tau= \frac{17400.51}{2268}=7.7 Mpa$ 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 = $\frac{\frac{1}{12}×45(76^2-44^2)}{76/2}=34913.684$

$σ_b=\frac{2500000}{349136.84}=7.2 MPa$ 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

$5000=d×l×p_b = d_1×1.25d_1×10=12.5d_1^2$

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

Check for shear stress

$5000 = 2×\frac{\pi}{4}×d^2×\tau=2×\frac{\pi}{4}×20^2×\tau=628.4\tau$

$\tau= \frac{5000}{628.4}=7.96 Mpa$ 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,

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

$M =\frac{5}{24}×5000×25=26041.7 Nmm$

Section modulus, z = $\frac{\pi}{32}×d_1^3=\frac{\pi}{32}×20^3=786 mm^3$

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

Design of lever

Bending moment,

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

Section modulus, z = $\frac{1}{6}×tb^2=\frac{1}{6}×t(3t)^2=1.5t^3$

$σ_b=\frac{M}{Z}=\frac{2250000}{1.5t^3}$

$t^3=\frac{1.5×10^6}{75}\implies t= 27.1 = 28mm$

$b=3t=3×28=84 mm$