Direct Stress + Bending Stress: Problem No.14

 

DESIGN OF MACHINE ELEMETS II

10ME62

ri. Direct stress + bending stress  F   = +  M  c   A b ii  

 Aer i   50 =

 + (5000x)(43.3)

5000

7853.98 

∴.. 

 

..

7853.98X6.7X50

 

  Maximum offset distance. 

x. 

Problem no.14 

An Open ‘S’ made from 25mm diameter rod as shown in the figure determine the maximum tensile, compressive and shear stress 

Solution:  (I) 

Consider the section P-Q 

MENT OF MECHANICAL ENGG, ATMECE, MYS

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DESIGN OF MACHINE ELEMETS II

10ME62

Draw the critical section at P-Q as shown in the figure. 

Radius of centroidal

axis  r c  =100mm 

Inner radius r i =100 

= 87.5mm 

Outer radius r o = 100+ 

= 112.5mm 

Radius of neutral axis 

r n = 

= 

= 99.6mm  Distance of neutral axis from centroidal axis e =r c - r n  =100 - 99.6 = 0.4mm Distance of neutral axis to inner fiber ci = r n –   –  r   r i  = 99.6 –  87.5  87.5 =12.1 mm 

DEPARTMENT OF MECHANICAL ENGG, ATMECE, MYS

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DESIGN OF MACHINE ELEMETS II

10ME62

Distance of neutral axis to outer fiber co = r o -r n 

Area of cross-section 

A = 

=112.5 –  99.6  99.6 = 12.9 mm  π π 2 2  d =  x25  = 490.87 mm   4 

4 

Distance from centroidal axis  I = r c = 100mm  Bending  moment about centroidal axis M b = F.l = 100 x 100   = 100000Nmm 

Combined stress at the outer fiber (i.e., at Q) =Direct stress +bending stress  σro= 

F 

M bCo 

A 

Aeo 

100000 X12.9 

1000 

= 

 –  

490.87  

490. 87 X 0.4 X 112.5 

= - 56.36 N/mm2 (compressive)  Combined stress at inner fibre (i.e., at p) 

σri= Direct stress + bending stress  =

F 

 A 

M c 

= 1000  +  100000 X 12.1 

 b Aer   i 

490.87  

490.87 X 0.4 X 87.5 

i 

= 72.466 N/MM2 (tensile)  (ii) Consider the section R -S Redraw the critical section at R – S as shown in fig. 

r c = 75mm  25 

=62.5 mm r i = 75 - 2  DEPARTMENT OF MECHANICAL ENGG, ATMECE, MYS

 

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DESIGN OF MACHINE ELEMETS II

r o = 75 + 

10ME62

25

 = 87.5 mm 

2 

A = π  d    =  π 

2 

2 

2

= 

4 X 25 

4 

490.87 mm 

2 

 ƒ rr o  +

( 

r n =

  4 4

r i  

2 

= ( 487 .5 +4 

 )

462.5

 

=74.4755 mm 

e = r c - r n = 75 -74.4755 =0.5254 mm  –  62.5 =11.9755 mm  ci = r n - r i =74.4755  –  –  74.4755 = 13.0245 mm  co = r o - r n = 87.5  – 

l = r c = 75 mm  M b = Fl = 1000 X 75 = 75000 Nmm  Combined stress at the outer fibre (at R) = Direct stress + Bending stress  75000 X13.0245  1000  F  M b co  σro = 

 – 

A 

 Aer    

o

= - 41.324 N/mm

= 490.87  

- 

490.87 X 0.5245 X 62.5 

2 

(compressive) 

DEPARTMENT OF MECHANICAL ENGG, ATMECE, MYS

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DESIGN OF MACHINE ELEMETS II

10ME62

Combined stress at the inner fiber (at S) = Direct stress + Bending stress   σri = 

F 

M bco 

A 

Aer o 

75000 X 11.9755 

1000  490.87  

490.87 X 0.5245 X 62.5 

2 = 55.816 N/mm  (tensile) 

 

2 at

∴ Maximum tensile stress = 72.466 N/m m  

P

 

2 

Maximum compressive stress = 56.36 N/mm  at Q  Maximum shear stress τmax=0.5 σmax= 0.5 X 72.466  2 

= 36.233 N/mm

at P 

Stresses in Closed Ring  Consider a thin circular ring subjected to symmetrical load F as shown in the figure. 

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