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ENFD%20375%20-%20Midterm%201%20-%20Blue%20Key

ENFD%20375%20-%20Midterm%201%20-%20Blue%20Key - ENFD 375...

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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; 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 O-AXIAL = :4: = E 8 5 8 = — L V P Tavg = Z O-BEARING : A BEARING V = SHEAR FORCE FS = O-ultimate O-applied 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 6~) «moo loo—“o ,4 1; T:/oo"’{/ u“ ‘s’ A H wLTiIOo" F89 $212! ptM'MM by 0' - f. , 533:: 20’“r TErJS‘ON ST ‘ A 5.5: lot»K (Z‘FWOZQIFT : .1981.” 4‘ = / Fr f‘ a) a} it” ‘ ) O '0 9 A5 giv~1(3oooct‘5t 1 FY COLUMA} szamauj .165 ’4 :: ,ol‘l ENFD 375 — Basic Strength of Materials NAME: 5 w e; #191 Problem #2: Assume P = 0 kips, Q= 10 kips, S = 40 kips and R = 0 kips and the steel column 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 6‘) 4‘ifir:0= ‘10 *4 4:” 1‘ Hr( - P’- - OK(:|FTXI2‘7M) W':*, '0‘ 6“ Z A: 5‘ 1(30' 000‘“) $7.1(3wmfin) l, ”Fr P ; : r003 ~5H0£nry$ Amok 2 .0007}:r SWIZ’F‘4S 1:) gm ’1‘ 27‘1’0: '10L‘H°L+C' '0‘ l C: 50"? 90" Tam)"- .5’0M Mu ‘ c p / - gmeuvk on- 0’” C) (D’Bcfletufi ‘ 73:3 ('10. ) '02'6 , ENFD 375 — Basic Strength of Materials NAME: 8 L L26 ' MY Problem #3: Assume P = 10 kips, Q = 50 kips, S = 80 kips, R = O kips, Wt and the steel 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 0‘) E rliF-Ké,:0=l0 -gv—50+C “Fr Cs! I'LD'é ‘f ’OFI’ 10‘ T [0'4 ,0 IL C 30l< a“, K I g .4 Sb p¢H m *«Llo 00’”; (10K l/ l L C 89% I?) ’lIZU" ILOK .. I 88153! COMP 0“) OZONC, 8.¢( IJ) " PL l3> C5: 2 354—: r (Jamar-T) r< FY) ”0100 ) "3% y: Lilli—fl”! — 5m ('5 000‘”) 6%“ (WOW f 5.3-1' (bpoodk‘l) .. Fr —_- 000733 r~,oawc7 C— 00%“ 9 _ LL: .5 {s vab/ €m“"‘ ”:5 C) d" ’46 L : Sm/lLL. A 13 m l 'r " .4” <7 A'LL fiwD I} ’ ‘ CV ['13 Vé,u,‘/ .J’V“ Fol. nrc Pull’tq rack, L 1: t‘l\l 1 AM! («a 50 J‘ ‘v§ SmA Lab‘ / ' 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 z.’2..§u~ ...
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