HW 2 - 172 CHAPTER 2 Axially Loaded Members Solution 2.12-2...

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172 CHAPTER 2 Axially Loaded Members Solution 2.12-2 Bar between rigid supports !~A~¥~d;1~~~~CE~~;;;;~d~23 B ~---·I, r-~ ·1· d = 20mm d = 25mm U y = 250MPa l 2 DETERMINE THE PLASTIC LOAD P p : At the plastic load, all parts of the bar are stressed to the yield stress. Pointe: ~ .~j;";; = uyA l F AC P = F + F AC CB P p = uyA l + uyA Z = uy(A) + A z ) _ SUBSTITUTE NUMERICAL VALUES: P p = (250 MPa) (~)cd~ + di) = (250 MPa) (~) [(20 mm)2 + (25 mm)z] =201 k:N - Problem 2.12-3 A horizontal rigid bar AB supporting a load P is hung from five symmetrically placed wires, each of cross-sectional area A (see figure). The wires are fastened to a curved surface of radius R. (a) Determine the plastic load P if the material of the wires is p elastoplastic with yield stress uy- (b) How is P p changed if bar AB is flexible instead of rigid? (c) How is P p changed if the radius R is increased? A Solution 2.12-3 Rigid bar supported by five wires F F F F F A A F =uyA (b) BAR AB IS FLEXIBLE At the plastic load, each wire is stressed to the yield stress, so the plastic load is not changed. - (c) RADIUS R IS INCREASED Again, the forces in the wires are not changed, so the plastic load is not changed. - (a) PLASTIC LOAD P p At the plastic load, each wire is stressed to the yield stress. :. P p = 5uyA _
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CHAPTER 3 Torsion Problem 3.3-4 An aluminum bar of solid circular cross section is twisted by torques T acting at the ends (see figure). The dimensions d and shear modulus of elasticity are as follows: L = 1.2 m, :T~~.;;q;~q;;;::;:;.;:q;:;:;.;;;;;;;; T d = 30mm, and G = 28 GPa. ... (a) Detennine the torsional stiffness of the bar. I I (b) If the angle of twist of the bar is 4°, what is the maximum <-. ------ L --------. shear stress? What is the maximum shear strain (in radians)? Solution 3.3-4 Aluminum bar in torsion From Eq. (3-11): T = Tr = Td = (GIpcP)(~) max I p 2Ip L 2Ip L = 1.2 m d = 30 mm (28 GPa)(30 mm)(0.069813 rad) G = 28 GPa 2(1.2 m) (a) TORSIONAL STIFFNESS = 24.43 MPa GIp G'TTd 4 (28 GPa) ('TT)(30 mm)4 T max = 24.4 MPa +-- kT=T= 32L 32(1.2 m) MAXIMUM SHEAR STRAIN k T = 1860 N . m +-- Hooke's Law: (b) MAXIMUM SHEAR STRESS T max 24.43 MPa cP = 4° = (4°)( 'TT /180)rad = 0.069813 rad 'Ymax = G = 28 GPa From Eq. (3-15): 'Ymax = 873 X 10- 6 rad +-- TL cP = GI p Problem 3.3-5 A high-strength steel drill rod used for boring a hole in the earth has a diameter of 0.5 in. (see figure).The allowable shear stress in the steel is 40 ksi and the shear modulus of elasticity is II ,600 ksi. What is the minimum required length of the rod so that one end of the rod can be twisted 30° with respect to the other end without exceeding the allowable stress? ~ [c:z::::::z::::::z:::::¢rZiiz:z.Oz.·5z . 1z'.nz· :Z::::::Z::::::Z::::::Z::::::Z::::::Z::::::Z:::::::=DO --.;:. 1--'
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This note was uploaded on 11/14/2011 for the course ECIV 310 taught by Professor Huckelbridge during the Spring '11 term at Case Western.

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HW 2 - 172 CHAPTER 2 Axially Loaded Members Solution 2.12-2...

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