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Unformatted text preview: ENFD 375 — Basic Strength of Materials NAME: K 59% i5? VG” BASIC STRENGTH OF MATERIALS
ENFD 375 — WINTER 2008 A, n”, {MIDTERM #1 DR. MILLER
60“” WEDNESDAY, JANUARY 30, 2008; 45 PM NO BOOKS OR NOTES NOTE: There are multiple versions of this exam, all using the same ﬁgure. The numerical
values are different. The various versions are randomly distributed to the class. Directions: 1) Put your name on each page of the exam. Exams without names receive “0”. Put
your name on each page in case the pages are accidently separated. 2) Do the problems on the test sheet. If you need additional sheets, put “continued” at
the end of the problem sheet and staple the additional sheet to the end of the exam. Additional sheets MUST have your name and problem number. ONE PROBLEM
per additional page. 3) Show ALL calculations. No calculation — no credit. 4) All answers must have proper value, units and sense. Sign (+ or ) is not enough
unless the sign is properly deﬁned in the problem.
5) Be neat. Messy, unreadable or indecipherable calculations receive a grade of “0”. P
OAXIAL = :4: = E 8
5
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Tavg = Z OBEARING : A BEARING V = SHEAR FORCE FS = Oultimate Oapplied ENFD 375 — Basic Strength of Materials NAME: ALL PROBLEMS REFER TO THE FOLLOWING DIAGRAM: See Detail for
Bearing Plate Steel
Column
Beaﬁng
Plate
See
H——~l
40 in. Expanded
Concrete Top View Detail on
Column 12 ft Probs 4
,_ \/—and 5.
o O )
Footer 1 ft 6 in R
O O
l‘—‘—.—'">l
7 In Bearing Plate Detail There is an upward force (T), P, at the top of the steel column. There is a downward (1) force,
Q, 11 feet from the top (or 10 feet from the bottom) of the steel column. There ‘is a downward
force (T), S, at the top of the concrete column and a lateral force (—>), R, at the bearing plate. The drawing is not to scale to enhance detail. Note that the values at PI Q, R and S change in each problem. ASSUME: The steel column has E = 30,000 ksi The concrete column and footer have E = 4,000 ksi
The steel has a yield strength of 36 ksi. 1 kip = 1k = 1000 pounds ENFD 375  Basic Strength of Materials NAME: 3 L ué K531 Problem #1: Assume P = 100 kips, R=Q=S=O. a) Draw a Free Body Diagram and axial force diagram of the steel column, only. (5 points)
b) If the area of the steel column is 5 inz, what is the stress in the column? (5 points)
c) How much does the steel column (only) change length? (5 points) x
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has an area of 5 inz. ' a) What is the change in length of the steel column (only)? (10 points) b) Draw a Free Body Diagram showing the force at the bottom of the footer. What is the
value of this force? (5 points) c) The soil has a bearing capacity of 0.050 ksi. What is the safety factor against bearing
failure of the soil? (10 points) STEEL GOLuMaJ 'L k
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column has an area of 5 inz.
a) What is the stress in the concrete column? (10 points)
b) Assuming the soil under footer is incompressible, what is the total change in length of the
system (in other words, how much does the top of the steel column move?) IGNORE
THE DEF ORMATION OF THE BEARING PLATE AND THE FOOTER. (10 points)
c) In this calculation, you ignored the deformations of the footer and the bearing plate
because these deformations are orders of magnitude smaller than the deformation of the steel and concrete columns. Why are the deformations so small? (5 points) l0
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/ ' l ENFD 375 — Basic Strength of Materials Side View T°P View Bolt, Washer and NAME: Problems 4 and 5 refer to the
diagram on the right, which is
expanded detail of ONE anchor
' bolt. The anchor bolts and the
bolt holes are V2 inch diameter.
Remember, R is applied to the entire bearing plate, not a single
bolt! Nut Not Shown for Clarity. Problem4—IfP=Q=S=0andR=40kips, a) Draw a free body diagram to show why the bolt is in single shear (5 points).
b) What is the average shear stress in the bolts? (10 points) 00 "i 12
V': $IN51LE’ SH‘GWK u.— 50,01 "s’ ENFD 375 — Basic Strength of Materials NAME: 3 4 U5" ’4 (‘7 Z Problem 5: Assume the SHEAR FORCE in ONE anchor bolt is 8 kips and the plate is % inch
thick: a) What is the bearing stress on the bolt hole? (10 points); b) Assume the plate fails in shear along the dotted lines. What is the shear stress on
these planes? (10 points) 3 0/ :— ”an—if“ : Zl'glcs’
a 804%“: (Ugwﬂoﬂiv) bl A»! = 75 "”(IISMNZ Paws} = 2,25,; 4“ r: 2:“ = 3.5!“! . L
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 Spring '08
 Miller

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