INFORMATIO
strengthsadiq.pdf

2 2 2 2 1 2 1 y point a 41 10 40 2 10 40 8 4 3 4 3 3

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2 2 2 2 1 2 1 Y Point A 41 . 116 ] ) 10 30 ( ) 10 40 [( 2 10 40 8 4 3 4 3 3 J Tr A MPa 859 . 101 ] ) 10 30 ( ) 10 40 [( 4 10 40 5 . 3 4 3 4 3 3 I My A MPa σ x =-101.859 MPa , σ y =0 , xy =116.41 MPa c= 2 y x = 9295 . 50 2 0 859 . 101 MPa R= 2 2 2 2 ) 41 . 116 ( ) 2 0 859 . 101 ( ) 2 ( xy y x =127.063 MPa A(-101.859,116.41) Draw Mohr’s circle R C 1 =-50.9295+127.063=76.1335 MPa R C 2 =-50.9295-127.063 =-177.9925 MPa 2 2 2 2 1 2 1 Y (76.1335) 2 -(76.1335)( -177.9925)+( -177.9925) 2 ≤(250) 2 51100≤62500 since 51100<62500 so these loadings will not cause failure. MPa 41 . 116 MPa 859 . 101 ? A C 1 2
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6 Example:- The solid shaft shown below has a radius of 0.5 in. and is made of steel having yield stress Y 36 ksi. Determine if the loadings cause the shaft to fail according to the maximum shear stress theory and the maximum distortion energy theory. ksi A P 1 . 19 ) 5 . 0 ( 15 2 ksi J Tr A 55 . 16 ) 5 . 0 ( 2 5 . 0 25 . 3 4 1 . 19 x ksi , 0 y , 55 . 16 ksi 2 , 1 2 y x ± 2 2 ) 2 ( xy y x = 2 0 1 . 19 ± 2 2 ) 55 . 16 ( ) 2 0 1 . 19 ( =-9.55±19.11 ksi 56 . 9 1 ksi 66 . 28 2 Maximum shear stress theory Y 2 1 36 66 . 28 56 . 9 38.2>36 So the failure will occur according to this theory. maximum distortion energy theory 2 2 2 2 1 2 1 Y (9.56) 2 -(9.56)(-28.66)+(-28.66) 2 =(36) 2 1186.515<1296 the failure will not occur according to this theory.
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7 Example:- The solid cast iron shaft shown below is subjected to a torque of T=400 lb.ft. Determine the smallest radius so that it does not fail according to the maximum normal stress theory ult =20 ksi. 3 4 8 . 3055 ) ( 2 12 400 r r r J Tr 0 x , 0 y , 3 8 . 3055 r psi 2 , 1 2 y x ± 2 2 ) 2 ( xy y x 2 , 1 2 0 0 ± 2 3 2 ) 8 . 3055 ( ) 2 0 0 ( r 3 1 8 . 3055 r , 3 2 8 . 3055 r ult 1 3 8 . 3055 r =20000 r=0.535 in.
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8 Columns Columns are long slender members subjected to an axial compressive force. The force may be large enough to cause the member to deflect laterally or sides way, this deflection is called buckling. Critical Load The maximum axial load that a column support when it is on the verge of buckling is called the critical load (P cr ). Any additional loading will cause the column to buckle and therefore deflect laterally. Ideal Column with Pin Supports The column to beconsideredis an ideal column, meaning one that is perfectly straight before loading, is made of homogeneous material, and upon which the load is appliedthrough the centroid of the cross section. It is further assumed that the material behaves in a linear-elastic manner and that the column buckles or bends in a single plane.
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9 In order to determine the critical load and the buckled shape of the column we will apply the following equation: M dx v d EI 2 2 0 sec tion M M+Pv=0 M=-Pv Pv dx v d EI 2 2 ) 1 ...( .......... .......... 0 ) ( 2 2 v EI P dx v d The general solution of equation (1) is: ) 2 ..( .......... .......... ) cos( ) sin( 2 1 x EI P C x EI P C v C 1 and C 2 are determined from the boundary conditions : v=0 at x=0 C 2 =0 v=0 at x=L 0 ) sin( 1 L EI P C C 1 ≠0 therefore 0 ) sin( L EI P n L EI P .... , ......... 3 , 2 , 1 2 2 2 n L EI n P
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