Question

Q3. A rectangular cross sectioned cantilever beam is 300 mm long and has a half bridge gauge at a distance 200.0 from the free end, whose width is twice the height. If the load applied at the free end is 15 kg, the V, is measured to be 150 micro volts. The gauge factor is given as 2.14 and the applied voltage V, is given a 5V and hence the equation for half bridge is given as E a) Determine the dimensions of the beam if the young's modulus is given as 200 GPa. (Marks - 4) pg. 3 b) Determine the maximum load (in kg) the beam can withstand if the factor of safety of the beam is 2 using Von Mises criterion. (Yield stress = 200 MPa) (Marks - Total marks 8

Answer 1

Detailed answer of above asked question is solved as below.

- Freebody diagram P-1589.81 = 147.15 N 15kg = 147.15 N 3 А loomm 200mm b= 2h cross-section In given conlilever beam bending stresses are induced. so thad Bending moment at A MA= 144.15x200 - 29430 Nmm Normal bending stress at A (6) -position of neutral axis 7 = hl2 - Moment of inestia (IX) zhh3 h4 Ixa bh3 - Normal bending stress use flexural formula MA 294304h/2 = 6 = MAJ 29930x3 Ix h416 63 (Given) = 88290 h3 strain at A 5=2.14 Yo= 150 MN = 150x10 by V8-54 2 x 150x106 S.VS 21485 & = 28.037 X10 6 - we know. by hook's laws E = 200GPO (Given). E = 2 No h = 25.063 mm. 6=2h = 50.1274 mm (Anga) 200x103 = 88290/63 28.03113 x 10 6 h: 25.063 mm Dimensions of beam!

200 a) Given 2 Allowable yon-mises stress is by syt syt - 200MPa NA Nf= 2 by a 100 mipa von mises stress is by' is equated to the allowable tensile stress ball ball = 100 MPa. → moment of inertia of beam cross section Ix=h4 25.0634 6 position of neutral axis 7 = hig= 25.063 12- 5315 mm Maximum moment is at point Bie at support Momax= Px300 = 300 pm Nam - Maximum bending tensile stress induced in the beam. Mmax 65.7629 x 100 mmy 2 max ball 300 Pmax 65.7629x103 12.5315 Pa 1749.2691 N P = 1749.2691 = Praz 178.3149 kg. max (Ansb) max. load the beam can with stand