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Unformatted text preview: ENFD 375 — Basic Strength Of Materials NAME: G? @651.) May BASIC STRENGTH OF MATERIALS ‘ L
.56 "i
t? "‘ ENFD 375 — WINTER 2008
MIDTER167 DR. MILLER
WEDNESDAY, JANUARY 30, 2008; 4—5 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 = Z = E8
5
8 = —
L
V P
Tavg : 2 OBEARING : T— BEARING V : SHEAR FORCE F S ___ Oultimate Oapplied . ENFD 375 — Basic Strength of Materials NAME: 6 £55“ 1419/ ALL PROBLEMS REFER TO THE FOLLOWING DIAGRAM: See Detail for
Bearlng Plate Steel
Column
Beaﬁng
Plate
See
I<—————>l
40 In Expanded
Concrete Top View Detail on
Column 12 ft Probs 4
, [and 5.
O O l
Footer 1 ft 5 in R
O O
I+——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. Thereis 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 emblem. 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: Galaw /1€H Problem #1: Assume P = 120 kips, R=Q=S=0. 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 4 inz, what is the stress in the column? (5 points)
0) How much does the steel column (only) change length? (5 points) I20[ ,4 'u’ 1‘ 2F :0: I'ZoLT’
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COLUMN ELDN‘ém 0’ MW @ M® ENFD 375 — Basic Strength of Materials NAME: G metal Maj Problem #2: Assume P = 0 kips, Q= 10 kips, S = 45 kips and R = 0 kips and the steel column
has an area of 4 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 0.05 ksi. What is the safety factor against bearing
failure of the soil? (10 points) l‘
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rs=' = L7; ENFD 375 — Basic Strength of Materials NAME: G lie?” #611 Problem #3: Assume P— — 15 kips, Q= 70 kips, S= 80 kips, R= O kips and the steel column has
an area of 4 1n2.
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 DEFORMATION 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)
I S K 15 " F135” F 5—K. Fl”
15w”) 5122. 2,142.). 3: H,L1'(30400 an) Ll "1(310090th) é‘hvt (9000 55') 2001375” ,00'455’3pr — .oowza”  — .6695 , Cocuvmu Shanna/3 , “1,1930%!” . r . "‘ giltLL.
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1761?. Th: ppﬂm ha P0011“: L" V517 Ado A 1; arm we“: at,“ '1 w13$“C. ENFD 375 — Basic Strength of Materials 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! Top View Bolt, Washer and
Nut Not Shown for
Clarity. Side View Problem4—IfP=Q=S =0andR=120kips,
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) q ENFD 375 — Basic Strength of Materials NAME: 6 WM [L91 Problem 5: Assume the SHEAR FORCE in ONE anchor bolt is 10 kips and the plate is l/z 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) ...
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 Spring '08
 Miller

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