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Unformatted text preview: Department of Mechanical Engineering, Yeditepe University ME 342 MACHINE ELEMENTS 1a Namik Ciblak
MIDTERM EXAM 11, May 6*, Spring 2008 NAME: 5 0 L U T1 0 N 5 Time: 150 Minutes Open books and notes. Four questions.
Q. l. (10%) Answer the following questions as true (T) or false (F). C_______IRCLES ONLY. Desi _n is as art as it is science.
DET 15 the most conservative static desi theo The vonMises stress is not really a stress.
IV. In static design of ductile materials, the designer may choose not to
include stress concentration effects since they occur in highly localized
regions and are relieved due to strainhardenin_ henomenon.
If the stress levels are below the yield strength of the material, fatigue
does not happen. Fatigue fracture somewhat resembles the static brittle fracture. la] VI. The endurance limit modiﬁcation factors are used because the stress—
life ex neriments are not reliable. IX. The surface ﬁnish modiﬁcation factor is used to account for the
differences in crack creation in different manufacturing processes.
All materials exhibit certain endurance limits. Q.2. (30%) The stresses at a point in a machine part made of Gray Cast Iron ASTM grade 40 are
found, in MPa, to be ox 2 200,0), : ~500,0z = —250,ryz = 200,233, 2 400,er = 300
Determine the static factor of safety. Assume brittle material and use ModifiedMohr theory. Let
0A =01 and 0'3 :03. Q.3. (30%) The sign post in the ﬁgure is subjected to a uniformly distributed load due to a steady
wind with a velocity of 36 kmfh in '
—z direction as shown. As you
learned in fluid mechanics, one
can apply the Bernoulli’s equation
between a point upstream and a
point on the sign post, where
stagnation is assumed, to get the
following relation Ty'fézrr AP=PS Pa =épu2 where ,0 =12 kg/m3 is the air
density. Determine the ratio — corresponding to a static factor of
safety of 3 according to the
distortion energy theory. Material: A181 1030 HR Q.4. (30%+10%) The ﬁgure shows a traditional swing that you could see in many playgrounds. The post
(the column) of the main structure is to be designed. Assume that the swing executes a circular motion from — 9m“ to + 6m,1x . Note that the maximum velocity is attained at 6 = 0 and velocity is zero at maximum swing angle. We are interested in determining the maximum moment at the bottom of the post due to tension
2". For this, one should ﬁrst complete the dynamic analyses. Using the free—body diagram of the swing seat as shown in the figure at right, the following equilibrium
equations in curvilinear coordinates can be written: v2 2F, =—Wsin6=ma, and SF" =T—Wcose9=man =m? 172 Therefore the tension at any angle is given by: T = W 00319 + m T . Now, if one writes the energy equation, assuming no dissipation, between the lowest position and any angle, the following is obtained:
v2 = 2gL[cos€  cos 6mm ].
Therefore,
T = W 0056 + 2mg[cos 6 — cos Qmax] = W[3 cos 6 — Zcosﬂmax] a) (BONUS: 10%, Please try to solve this as last) Show that the maximum and minimum
moments at the base of the post occur at following angles. am, if 9m 3 30° 6 . . z i
critical COS—1 [.16. [005 6m“ + m1] if 6m” 2 300 b) (5%) Assuming that the result in part (a) is correct. Determine the angle and value of maximum
moment if m = 100 kg, L = 2.5 m, amax = 60°. Take g =9.81 III/52. c) (20%) The post is to be made of A181 1030 CD round tubing, with an outer diameter of 60min
and thickness of Smm. Estimate the life in cycles corresponding to a reliability of 99.99%. (1) (5%) Assume that the swing is used 10% of the time in a day and the period of one swing
(oscillation) is about 3 seconds. Now, state the life in days. YEDITEPE UNIVERSITY FACULTY OF ENGINEERING AND ARCHITECTURE
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 Spring '08
 NamıkÇıblak

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