CivE 270.Final Exam B

# CivE 270.Final Exam B - CIVE 270 Final Exam Spring 2001...

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Unformatted text preview: CIVE 270 Final Exam Spring 2001 Name 1. A compressed air tank has a 600 mm outside diameter and an 8 mm wall thickness. The pressure inside the tank is 150 kPa. A 50 kN force P is applied to a wire wrapped around the tank as shown. Determine the shear and normal stresses at locations a and b. Sketch the elements with the stresses you have calculated. Note: a is located at mid—height and b is located at the top ﬁbre. CIVE 270 Final Exam Spring 2001 Name 2. Several planks are glued together to form the box beam shown in cross-section in the ﬁgure. Knowing that the beam is subjected 20 to a vertical shear force of 3 kN, determine the average shear 20 stress in the glued joints at A and at B. 30 30 20 CI VE 270 Final Exam Spring 2001 Name 3. CIVE 251 (Survey school) students are commonly known to carry their levelling rod when it is fully extended and in a horizontal position as shown in the ﬁgure below. The rod is 5000 mm long when fully extended and its total weight (accounting for dynamic effects while carrying the rod) is 90 N. The cross-section of the rod is shown at the right of the ﬁgure. If the rod is supported on the student’s shoulder at 3000 mm from one end as shown in the ﬁgure, calculate: (a) the maximum normal stress in the levelling rod when it is carried with the x—axis in the vertical direction. (b) the maximum normal stress in the levelling rod when it is carried with the y-axis in the vertical direction. (0) the factor of safety in each case if the ultimate stress of the material is 65 MPa. Note: The weight of the rod can be assumed uniformly distributed over its length. Levelling rod WI 15 Cross-section CIVE 270 Final Exam Spring 2001 Name 4. The cantilever beam shown has a length L and rests on a ﬂexible support at L/2 from the ﬁxed end. The ﬂexible support has a stiffness k and prior to applying any load to the beam, it is unstressed. The moment of inertia of the beam cross-section is I and the modulus of elasticity of the beam material is E. Neglect the self-weight of the beam. A concentrated downward force, F, is applied at the free end, as shown. (a) Find the reaction at the ﬂexible support as a function of k, F, E, I, and L. (b) Find the reaction at the same location if the spring is replaced by a roller. CIVE 270 Final Exam Spring 2001 Name 5. The statically determinate pin—ended truss shown is made up of members that have the cross- section given at the right of the ﬁgure. The modulus of elasticity of the material is 72,000 MPa and the yield stress in tension and compression is 270 MPa. A force, F, is applied at point A, as shown. (a) Which member will yield or buckle ﬁrst as F is increased? (Clearly indicate whether it yields or buckles.) (b) At what load, F, will this failure occur? Hint: Members AC and BE are critical in tension (they have the same force) and member ED is critical in compression. ' Cross section CI VE 270 Final Exam Spring 2001 Name 6. The complete state of stress on an element taken from a structure is shown in 'the diagram on the left. Using Mohr’s Circle, determine the state of stress at the same point in the structure on the element shown in the diagram on the right. Draw a neat sketch of the rotated element clearly showing the stresses you have calculated. “5:?” ‘— 3.951 We 333244004’ 5-849 = /,30‘-/x/osmm‘f x M 6731000206 0)? ) 0’5: if? = éS’Zf/o 00‘ I 34.8\ MP“ 1' = 3/; = 652 x/o‘wﬁ (c It Qxxo mg - 6.719 Ma ﬂ [email protected]%9>-<Z—(§yﬂ% = 1.402, X/oé’ ME A“ 2.4314470 = iolMﬁa *‘L "Mink 7; 2—H fame] mpg J ~92.738+%,8| ” : 39.5%. 1,548 "Wax ‘LS‘WMPA (00 Q9) (a? joint E S'OHVE 3: [E] Cnv €170 \ Q: @onx§o) = 60,000 “M3 ( o! 3 o! q I : '0 ‘15)) — [1264’ Iigsﬁ (Bngoxzyj‘ﬂ é : H.88X/O mvﬂ _ 1g _ Goooléo'boo _ ’2: ' TE WILB‘Bx/oszxw) ' 0'379 MP“ 6) “c: n (5:0) 0 O H (‘2 Ca Co) 3 I: ’ p : £0?” = o.O\8 H/MM M =Q),OI8X3000)L15‘00) = Buooo (4mm q;= 5% : Ci—X—J‘gi, 75 = . \.7 Mk 0 [\$4353 —_ gggao W” 6M:’AA=C-:L>—’?g%x—3@ = 24.4qu a 1:5,. = = /.053 F5“ gr Cu/ E 1’70 % 4 NJ I: ECIOOq- 800 = 2.898x/O A - rr(so)“‘- new)" = Z8 27 W? Swat R : FM 1 [a F Fe!» : F CC) Teﬂn’k (1,1 Pkg :. \: CC) ﬁ‘au : 6) A 2 (2749mm) F“, * F (T) = 743.3 Ml. juhxh S : Fm : FM = 2F Cc) F - MM My?“ —> O’=%73 = S’OO.Zx/o (“F 270' m2 macaw) F: W = 533.8 erl A” ,cowxqartsste, medal-Xe,“ are, Zm '. MWLAI-r mI-LCRS TV 151’ = ﬂaw z.” @000)" 6&3 occurs a): 257.4 Ml 30mg sch,“ = ‘zfzéig—O ‘— 3].! M9c~< ZZOM’C‘ 296,“ (OK) Truss Q53: ‘93 BUMVB. F=3Bl¥4 (5) Maya?” 2mg EUJJW. CC) GCQWS OJC F: 257,‘/ LF= “m. % C(I¢,o) R=I€1+é"=8.48\$m 75° IZo°-4/S° = SBA 73° '5 7%: 'Z = 8.20 PAPA og—Ié aos7f°= (R (5;: ; M14 0:5, ; )Q—(18.Zo—/®= I380 MEL ‘ /’ \y/v X\|\ Q'ZOMPa [8,210MPd \ \$809“? M52 ...
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## This note was uploaded on 10/04/2009 for the course CIVE 270 taught by Professor Eddie during the Spring '09 term at University of Alberta.

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CivE 270.Final Exam B - CIVE 270 Final Exam Spring 2001...

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