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# Stress&strain part 2

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### Transcript of "Stress&strain part 2"

1. 1. Prob. 2.15: A 3-mm thick hollow polystrene cylinder ( = )and a rigid circular plate (only part of which is shown) are used tosupport a 250-mm-long steel rod ( = ) of 6-mmdiameter. If a . load P is applied at B, determine (a)theelongation of rod AB, (b) the deflection of point B, (c) the averagenormal stress in rod AB.
2. 2. Solution: ΔL = For the rod AB: ×.ΔL1 = = 1.414*− m ∗ ∗(0.003)2The elongation of rod AB = 0.1414 mmFor the cylinder:The part of which thickness is 3 mm will be affected bythe stress resulting from the force F=3.2 KNThe area of this part (A) = ((. ) − (. )A = 4.429*− −∗.ΔL2 = = -7.225*− m ∗ ∗.∗−UB = ΔL1 – ΔL2 = 1.414*− - (-7.225*− )UB = 2.14 mmNormal stress for rod AB: σ= = = . (.)
3. 3. Prob. 2.17: Two solid cylindrical rods are joined at B and loaded asshown. Rod AB is made of steel ( = ) and rod BC of brass( = ). Determine (a) the total deformation of the compositerod ABC, (b) the deflection of point B.Solution: −(+) − + (.)UB = ΔL1 = = ∗ ∗(.)UB = -1.485*− m −() − (.)ΔL2 = = = −. ∗ − ∗ ∗(.)Total deflection of rod ABC = ΔL1 + ΔL2= 1.8487*−
4. 4. Prob. 2.19: Both portions of the rod ABC are made of an aluminumfor which = . Knowing that the magnitude of P is 4 KN,determine (a) the value of so that the deflection at A is zero, (b) thecorresponding deflection of B.Solution:We will call the portion BC: part (1)And the portion AB: part (2) ΔL = For part (1):F = Q-P = Q-4000 − − (.)ΔL1 = ∗ ∗(.)
5. 5. For part (2): (.)ΔL2 = ∗ ∗(.)Since the deflection at A = 0, thenΔL1 = ΔL2 − − (.) (.) = ∗ ∗(.) ∗ ∗(.)(-Q+4000) (0.5) = (4000) (0.4) (9)-0.5Q+2000 = 144000.5Q = - 16400Q = - 32800 NQ = 32.8 KNProb. 2.20: The rod ABC is made of an aluminum for which = . Knowing that = and = , determine thedeflection of (a) point A, (b) point B. ( )Solution:We will call the portion BC: part (1)And the portion AB: part (2) ΔL = For part (1):F = Q-P = -42+6 -36 KN
6. 6. − (.)ΔL1 = = UB = 9.09− ∗ ∗(.) .ΔL2 = = 1.09*− ∗ ∗ . UA = ΔL1 + ΔL2 = 9.09− + 1.09*− =1.81*− Prob. 2.28: The length of the 2-mm-diameter steel wire CD hasbeen adjusted so that with no load applied, a gap of 1.5 mm existsbetween the end B of the rigid beam ACB and a contact point E.Knowing that = , determine where a 20-kg block should beplaced on the beam in order to cause contact between B and E.
7. 7. Solution: