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ME 10W E. Evans Stresses in a Plane Some parts eperience normal stresses in two directions. hese tpes of problems are called Plane Stress or Biaial Stress Cross Section thru Bod z angent and normal to the surface Force ΔF τ z Force ΔF τ z Force ΔF z z Convenient coordinate sstem Make two more cuts in the original bod, each perpendicular to the others. ME 10W 1

General State of Stress Cubic volume element Characterized b three components on each face Changes with a different coord. sstem Rotating the element in an direction changes the state of stress. General State of Stress & Strain z τ τ z τ z τ z τ z τ Positive orientations shown ε ε ε z ν νz + E E E ν νz + E E E ν ν z + E E E ε strain ν Poisson s Ratio Finding Maimum Stress here are three maimum stresses to be found: he maimum normal stress, P/A ± Mc/I he orientation of maimum stress Mohr s Circle ime dependent d maimum stress endurance strength Purpose of Visual tool used to determine the stresses that eist at a given point in relation to the angle of orientation of the stress element. here are 4 possible variations in Mohr s Circle depending on the positive directions are defined. ME 10W

Sample Problem Some Part & orientation A particular point on the part - ksi 6 ksi τ 3 ksi - ksi -ais 6 ksi (6 ksi, ) 6 3 τ 3 Center of (- ksi, -) -ais - ksi ( avg, τ ma ) - ksi ( avg, τ ma ) ( ksi, 5 ksi) 6 ksi τ 6 ksi τ R + (- ksi, -3ksi) avg ksi avg ksi ( avg, τ min ) 1 1 avg + R 7 ksi avg R ksi R ( 3 ki) ksi) + ( 4 ki) ksi) 5 ksi R τma 4 ksi ( avg, τ min ) ( ksi, -5 ksi) 1 ME 10W 3

- ksi ( avg, τ ma ) ( ksi, 5 ksi) Principle Stress - ( avg, τ ma ) ( ksi, 5 ksi) 6 ksi τ 1 θ an 4 ksi θ 36. 869 θ 18. 435 4 ksi θ ( avg, τ min ) ( ksi, -5 ksi) 1 Principle Stress Element θ 18.435 1 7 ksi Rotation on element is half of the rotation from the circle in same direction from -ais 4 ksi θ ( avg, τ min ) ( ksi, -5 ksi) 1 Shear Stress ( avg, τ ma ) ( ksi, 5 ksi) Relationship Between Elements avg ksi avg ksi avg ksi Maimum Shear Stress Element φ 6.565 τ ma 5 ksi φ 90 θ φ 90 36. 869 φ 53. 130 φ 6. 565 φ 4 ksi θ ( avg, τ min ) ( ksi, -5 ksi) 1 - ksi φ 6.565 6 ksi θ 18.435 τ 3k ksi θ + φ 18.435 + 6.565 45 τ ma 5 ksi avg ksi - 1 7 ksi ME 10W 4

What s the stress at angle of 15 CCW from the -ais? Some Part U & V new aes @ 15 from -ais A particular point on the part V? ksi? ksi U 15 τ? ksi Rotation on 15 on part and element is 30 on ( V, τ V ) ( avg, τ ma ) ( U, τ U ) 30 avg ksi ( avg, τ min ) 1 Rotation on U avg + R*cos(66.869 ) U 3.96 ksi V avg R*cos(66.869 ) V 0.036 ksi τ UV R*sin(66.869 ) τ UV 4.60 ksi ( V, τ V ) R 66.869 ( avg, τ ma ) ( U, τ U ) avg ksi ( avg, τ min ) 1 What s the stress at angle of 15 CCW from the -ais? Some Part A particular point on the part V V.036 ksi U 3.96 ksi U 15 τ 4.60 ksi ME 10W 5

Special Case Both Principle Stresses Have the Same Sign Y Questions? Z X p D t Clindrical Pressure Vessel p D 4 t τ τ ma τ z z 0 z z 0 since it is perpendicular to the free face of the element. 3 1 his isn t the whole stor however for X-Y Planes 1 and z 3 and 1 τ ma 1 3 for X-Z Planes ME 10W 6

Pure Uniaial ension z τ ma τ z 0 P/A τ ma 3 1 0 1 z 0 since it is perpendicular to the free face of the element. 1 > > 3 Ductile Materials end to Fail in SHEAR Note when S, S s S / Pure Uniaial Compression Pure orsion 0 P/A τ ma 1 0 τ c J -τ τma τ 1 τ CHALK 1 Brittle materials tend to fail in ENSION. ME 10W 7

Uniaial ension & orsional Shear Stresses Rotating shaft with aial load. Basis for design of shafts. P/A τ R ma (, τ ) Equivalent orque Special case of bod subjected to BENDING and ORSION. Point on top of bar is in shear and in tension. M/S τ c/j R + τ /-R (0, τ ) τ ma / 1 /+R Fied F τ /Z p M Bending Moment S Section Modulus orque Zp Polar Section Modulus τ τ R ma ma M S M + Z p Equivalent orque + τ + Z p Z e p /-R (0, τ ) τ R ma (, τ ) / M + e 1 /+R Strain Gage 1-8 Bolt Eample orque on nut is 10000 in-lbs We want to appl a strain gage to the bolt to measure the maimum strain. Determine the proper orientation and predict the strain to be measured. For round bars, Z p S, e Equivalent orque Strain gages onl measure normal strain, NO shear strain! ME 10W 8

Aial Force in Bolt he aial force in the bolt can be found using Eqn. 18-3: K*D*P (Pg. 719) where torque in in-lbs D nominal outside diameter of threads, in. P Clamping load, lbs K constant depending on the lubrication present K.15 for average commercial applications with an lubrication present at all. K.0 for well cleaned and dried threads. Values of K are approimate & should be verified ME 10W 9