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Unformatted text preview: ENGR 200 Final Exam Fall 2008 Name 1) Determine the force in members BE and EF of the truss shown below and indicate Whether the members are in tension or compression. (From Statics and
Mechanics of Materials by R. C. Hibbeler,) D 4m ENGR 200 Final Exam Fall 2006 Name 2) The two stone blocks have weights of Wa = 600 lb and Wb=500 1b. Determine the
smallest horizontal force P that must be applied to block A in order to move it. The coefﬁcient of static friction between the blocks is 0.3 and between the ﬂoor and each
block is 0.5 (From Statics and Mechanics of Materials by R. C. Hibbeler,) ENGR 200 Final Exam Fall 2008 Name 3) Express the force as a Cartesian vector, then determine its direction cosines. (From
Statics and Mechanics of Materials by R. C. Hibbeler,) ENGR 200 Final Exam Fall 2008 Name 4) The shaft is made from a solid steel section AB and a tubular portion made of
steel and having a brass core. If it is ﬁxed to a rigid support at A and a torque of
T=50 lb—ft is applied to the shaft at C, determine the angle of twist that occurs at
C. The radius of the brass is 0.5 in. and the overall radius is 1.0 inch. G(steel)=l 1.5);103 ksi and G(brass) is 5.6X103 ksi. (Watch out for units) (From
Statics and Mechanics of Materials by R. C. Hibbeler,) ~ 0.5 in . . T: V.
1m C 50 1" ft Prob. 11—49 ENGR 200 Final Exam Fall 2008 Name 5) The rigid bean is supported by a pin at A and Wires BD and CE. If the load P
on the beam causes the end C to be displaced 10 mm downward, determine the load P. Young’s modulus for the Wires is 200 GPa and each Wire has an area of 5
m2. (From Statics and Mechanics of Materials by R. C. Hibbeler,) ENGR 200 Final Exam Fall 2008 Name 6) Draw the shear force and bending moment diagram for thebeam shown below. 3’65 tit/ﬁg, ENGR 200 Final Exam Fall 2008 Name ——F—§ Stress strain 0:153 or 0': E8
0': E(8— aAT) 1 Stressforce Force
displacement : Strain
‘ displacement Areas The coordinates of the centroid of the area A are fxi/i fydA
_ A  A
x= , y= .
. 1A A The moment of inertia about the x axis 1,, the moment of inertia. ,
about the y axis Iy, and the product of inertia Ixy are Ix=/ysz, Iy=fx2dA, Ix,=/xydA.
A A ‘ A The polar moment of inertia about 0 is Jo=/erA=/(x2+yz)dA$Ix+Iy.
A Av l  1
Area = 5 bh Ix _= — 17113, Lg = —— 12h3 Area=bh
1 3 ' 1 3 122
Ix=§bh, Iy=§hb, Ixy=th Circulararea
1 a 1 3 ' 1
Iz=ﬁbh, Iy'=ﬁhb: Ix’y'=0 Area=1'rR2 La: y’=Z’”R4’ Ix’y'=°
Area=£bh
2V
1 3 1 3 1 22 Ar _1 2 _ _1 4 _
Ix=Ebh1 I’_Zhb’ Ixy=§bh ea——27rR If—IyE'rrR, Ixy—O
1 1 ' 1 ‘ 1 ‘ w 8
Ig=§gbhz . Iy=ghb3, .‘ Ixryl=7_2bzh2‘ Ix.=§'nR4, [i=(E—E;>R4, Igyj—‘O ...
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 Spring '08
 mullen
 Statics

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