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Problem 2.5-18 A plastic cylinder is held snugly between a rigid plate and a foundation by two steel bolts (see figure). Determine the compressive stress p in the plastic when the nuts on the steel bolts are tightened by one complete turn. Data for the assembly are as follows: length L 200 mm, pitch of the bolt threads p 1.0 mm, modulus of elasticity for steel Es 200 GPa, modulus of elasticity for the plastic Ep 7.5 GPa, cross-sectional area of one bolt As 36.0 mm2, and cross-sectional area of the plastic cylinder Ap 960 mm2. Solution 2.5-18 Plastic cylinder and two steel bolts FORCE-DISPLACEMENT RELATIONS
s 119 Steel bolt L Probs. 2.5-18 and 2.5-19 Ps L Es As Pp L
p Ep Ap (Eq. 3, Eq. 4) S P S L P Es 200 mm 1.0 mm 200 GPa SOLUTION OF EQUATIONS Substitute (3) and (4) into Eq. (2): Ps L Es As Pp L Ep Ap np (Eq. 5) As Ep Ap n 36.0 mm2 (for one bolt) 7.5 GPa 960 mm2 1 (See Eq. 2-22) Solve simultaneously Eqs. (1) and (5): Pp 2npEs As Ep Ap L(Ep Ap 2Es As ) EQUILIBRIUM EQUATION
Ps Ps STRESS IN THE PLASTIC CYLINDER sp Pp Ap 2 np Es As Ep L(Ep Ap 2Es As ) SUBSTITUTE NUMERICAL VALUES:
Pp N D sp (Eq. 1) Es As Ep 2np N L D Ep Ap 54.0 2Es As 1015 N2/m2 21.6 106 N Ps Pp Pp tensile force in one steel bolt compressive force in plastic cylinder 2Ps 25.0 MPa COMPATIBILITY EQUATION
Ps Pp Ps np P S 2(1)(1.0 mm) N 200 mm D S s p s elongation of steel bolt shortening of plastic cylinder
p np (Eq. 2) ...
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This note was uploaded on 09/20/2009 for the course COE 3001 taught by Professor Armanios during the Spring '08 term at Georgia Tech.
- Spring '08