If E 200 kNmm 2 v 03 Tapered Bar Consider a circular bar uniformly tapered from

# If e 200 knmm 2 v 03 tapered bar consider a circular

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If E = 200 kN/mm 2 , v = 0.3. Tapered Bar Consider a circular bar uniformly tapered from diameter d 1 at one end and gradually increasing to diam- eter d 2 at the other end over an axial length L as shown in Fig. 1.18. Take an elementary disc of diameter d x and length dx at a distance of x from end A . Diameter d x of elementary disc = d d d d L d kx x = + - = + 1 2 1 1 Figure 1.17 20 30 500 mm P P
Simple Stresses and Strains 13 where constant k d d L = - 2 1 Cross-sectional area, A d d kx x x = = + π π 4 4 2 1 2 ( ) ( ) Axial stress, σ π x x P A P d kx = = + 4 1 2 ( ) Axial strain, ε σ π x x E P E d kx = = + 4 1 2 ( ) Change in length over dx , δ π d Pdx E d kx x = + 4 1 2 ( ) Total change in length, δ π L Pdx E d kx O L = + 4 1 2 ( ) = + - - 4 1 1 P E d kx k O L π ( ) = - + - 4 1 1 1 1 P Ek d kL d π = - - 4 1 1 2 1 P Ek d d π as d kL d d d L L d 1 1 2 1 2 + = + - × = Therefore, δ π π L P Ek d d P d d Ekd d = - = - 4 1 1 4 1 2 2 1 1 2 ( ) Putting the value of k d d L = - 2 1 , we get Figure 1.18 Circular tapered bar d 2 d 1 P L B d dx x x A P
14 Chapter 1 δ π L P d d E d d L d d = - × - 4 2 1 1 2 2 1 ( ) ( ) ( ) Total change in length = δ π L PL Ed d = 4 1 2 Example 1.7 A brass bar uniformly tapered from diameter 20 mm at one end to diameter 10 mm at the other end over an axial length 300 mm is subjected to an axial compressive load of 7.5 kN. If E = 100 kN/mm 2 for brass, determine (Fig.1.19): (a) the maximum and minimum axial stresses in bar and (b) the total change in length of the bar. Solution Axial load, P = 7.5 kN. Maximum stress occurs at end A with minimum cross-sectional area: (a) σ π π max ( ) . , . = - = - × × × = - 4 10 4 7 5 1 000 100 95 94 2 P N/mm 2 at end A (b) σ π π min ( ) . , . = - = - × × × = - 4 20 4 7 5 1 000 20 23 87 2 P N/mm 2 at end B δ π L PL Ed d = - 4 1 2 , as the load is compressive = - × × × × × × × = - 4 7 5 1 000 300 100 1 000 10 20 0 143 . , , . π mm Figure 1.19 B P A P 300 mm 10 mm f 20 mm f Exercise 1.7 A 400-mm-long aluminium bar uniformly tapers from a diameter of 25 mm to a diameter of 15 mm. It is subjected to an axial tensile load such that stress at middle section is 60 MPa. (a) What is the load applied? (b) What is the total change in the length of the bar if E = 67,000 N/mm 2 ? [Hint: At middle, the diameter is ( ) 25 15 2 20 + = / mm]
Simple Stresses and Strains 15 Tapered Flat Consider a flat of same thickness t throughout its length, tapering uniformly from a breadth B at one end to a breadth b at the other end over an axial length L . The flat is subjected to an axial force P as shown in Fig. 1.20. Thickness of elementary strip abc is t at a distance of x from end A. Breadth, b b x = x B b L b kx = + - + ( ), where k B b L = - Cross-sectional area of elementary strip, A x = b x t = ( b + kx ) t Stress at elementary strip, σ x x P A P b kx t = = + ( ) Change in length over d x , δ d Pdx Et b kx x = + ( ) Total change in the length of flat, δ L Pdx Et b kx L = + ( ) 0 δ L P Etk B b = - [ ] ln( ) ln( ) = × P Etk B b ln Putting the value of k , δ L PL Et B b B b = - ( ) ln Example 1.8 A steel flat of an axial length of 600 mm and a uniform thickness of 10 mm has a uni- formly tapered width of 40 mm at one end to 20 mm at the other end. If the axial force is P = 18 kN and E = 208 kN/mm 2 , then what is the change in axial length of flat?

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