Structural Analysis Prof

FE Exam Review for Structural Analysis Prof. V. Saouma Oct. 2013

 Structural Analysis is part of the afternoon exam.  In the afternoon, you are to answer 60 questions, and Structural Analysis is about 10% of the test content (or about 6 questions).  Each question is worth 2 points.  You are expected to know: 1. Structural analysis of statically determinate beams, trusses and frames. 2. Deflection analysis of statically determinate beams, trusses and frames. 3. Stability analysis of beams, trusses and frames. 4. Column analysis (e.g. buckling, boundary conditions). 5. Loads and load paths (e.g. dead, live, moving). 6. Elementary statically indeterminate structures.  The only page in the “Supplied-Reference Handbook” related to Structural Analysis (shown in the next page).  Make sure that you know how to make best use of it, as it contains: 1. Reminder of what do we mean by “Moving Loads”. 2. Beam-Stiffness and moment carryover: to use for the analysis of statically indeterminate beams (unlikely that you get a SI frame). 3. Equations for the calculations of the deflections of trusses and beams using the virtual work method. Careful it is the virtual force/moment time the actual displacement (FL/AE for trusses, and M/EI for beams). 4. Member fixed end actions for uniform and concentrated load.

 I strongly recommend that you also memorize: for the maximum deflection of

a uniformly loaded, simply supported beam.  Careful with the SI units, GPa is 109 Pa or 109 N/m2 Many problems use the SI system.  In most cases, you will be dealing with round numbers, which greatly simplify your calculations.  Do not be tricked in believing that all triangles are 3-4-5.

26 26 (i vil Disdpline 回 Specific Review for the FE/EIT Ex目 m

From the PVC ， the low point is located at Solution Divide 七he beam in 七o two shapes as shown L 00FL的町一つm +パヨ4 A m 十AA m 2一せ一一一一一一一せ吐 GA AA せ m 吐斗 A

Determine Determine the tangent offset ぅ仏 at the low point.

2 y=iG 2 - G 1)X 2L - (0.016 一 (-0.02)) (444 m)2 (2)(800 (2)(800 m)

ニ 4.4 4 m eleVlow eleVlow point = 742.12 m 十 4.44 m = 746.56 m (747 m) Al = (0.08 m)(0.15 m) 2 The answer is B. = 0.012 m

A 2 = (0.05 m)(0.15 m) 2 ェ 0.0075 m STRUCTURAL STRUCTURAL ANA 町SIS The dis 七ance from 七he top of section 2 to the centroid is Problem Problem 38 Uc=Z 生Yc ，i The y-coordinate (mea 師su 町1 2乞:A ん4 tむro ぱid for the T -shaped beam is ((川「一「ーr--- I X 2 十( (0.0075 m ) (平十 0 叫))

J08m 比ニ 0.08 m

The answer is A. 0.23 0.23 m Problem Problem 39 The force in member HE from the 七russ shown is mos 七 nearly nearly

y G

(A) (A) 0.08 m (B) (B) 0.16 m (C) (C) 0却 m (D) (D) 0.38 m

12.0 12.0 m

(A) (A) 1110 N compression (B) (B) 1110 N tension (C) (C) 2490 N compression (D) (D) 2490 N 七ension

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Solution Solution Problem 40 8um moments about A to find the vertical reaction at E. The magnitude of the vertical reaction force at support port A is most nearly :Z= MA = 0 (A) (A) 3.3 kN RE ，匂 (B) 6.7 kN dEA (0) 10 kN

R H:" = i2224 N)( 伊6 m) (D) 16 kN

円'げU ν1ロ2m =1112N Solution

8ince 8ince support D is a roller support う the horizontal re-

The free-body diagram abou 七point E is ac 七ion force う RAx ，is 0 kN. To find the vertical reaction at at support A ，RAy ，a free-body diagram is drawn of the 、....._. EH entire 七russ and moments are summed about suppor 七D.

DE DE !RE ，ν RAy(15 m) - (20 kN)(5 m) =0 kN

RA" = i20 kN)(5 m)

吋 15 m = 6.67 kN (6.7 kN) 8umming forces in the horizontal direction gives the horizontal horizontal reaction at E as 0 N. The answer is B.

The sum of forces in 位1e vertical direction is Problem Problem 41 FEH ，υ-RE ，り= 0 lbf The magnitude of the compressive force in member AB = 1112 N compression is most nearly

The horizontal component of the force in member EH is (A) 4.2 kN (B) (B) 6.7 凶 FEH ，h = (2)(1112 N) (0) 8.5 kN = 2224 N (D) 11 kN

The resultan 七force in member EH is Solution Fr om 80 1. 40 ， the reaction at support A is 6.67 kN. FEH=V( Fi凹，出(Fi叫 2 Next ぅ a free 同 body diagr 乱m is drawn for support A with member forc 田 and 七heir force components. Using 七he (1112 N)2 十(山 N)2 =ゾ Pythagorean theorem ， the relative magnitudes of each

ニ 2487 N compre 回 on (2490 N ∞mpr 出品on) force and each force う s horizontal and vertical compo- m 臨 can be found. (In this case ，6.403 ，5， and 4 are The answer is C. the relative magni 七udes of the member AB force and

the the forces う s horizontal and ver むical componen 七s，respec- Problems Problems 40-42 are based on the following information tively.) and illustra 七ion.

A plane truss is loaded as shown.

F AEx

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For For equilibrium ，all forces on a free body mus 七sum to The posi 七ive sign in the calculated member AE force o kN. Summation of vertical forces giv< 回 means that the assumed direction of the force on 七he

free-body free-body diagram う which indicates tension ， is in the FABy + RAy = 0 kN same direction as the calculated force.

This This can be rearranged to give A free-body diagram of joint E will show that the vertical tical member BE is unable to sustain any horizontal FABy = -RAy = -6.67 kN force. Therefore ， the force in member ~F is the same as as the force in member A E. Recall Recall that joints in 七russes are frictionless ， so no bending ing moments exist. F EF =FAE = 8.33 kN (8.3 kN) The force and its components are proportional to the geometric geometric leng もhs of the triangle sides. The answer is D. 一 (6 .4 03 m ¥一 .l' AB = I 一一了一一一一 I .l' AB 'U ¥ 4 m I ν Problem 43 (6 .4 03 m ¥ If the truss members are made of steel and the cross- ) (-6.67 kN) =(τ'~--- sectional area of each member is 1000 mm 2 ， the magnitude tude of the vertical de fl. ection 抗 joint E is most nearly = -10.68 kN (-11 kN) (A) (A) 0.70 mm The answer is negative. This means that the calculated (B) 1. 1 mm force force is in the opposi 七e direc 七ion to 七he assumed force di- (C) 1. 6 mm rection rection on the free-body diagram. 8ince the member AB (D) 2.8 mm

force force w 回 assumed to apply tension on the free 伊 body diagram ， the negative answer means that member AB is Solution Solution m compreSS lO n. Use the principle of vi 巾 al work (unit load method). The answer is D. The actual forces in each member can be determined by applying the equations of equilibrium to each truss Problem Problem 42 joint and are shown 回 follows. (80me round-off error The magnitude of the force in member EF is mos 七ne 乱rly exists in these calculated numbers.)

(A) (A) 4.2 kN τ'r uss member lengths in meters are as follows. (B) (B) 5.3 kN

Solution Solution A T< 41 々ず'" 14 仏 Fr om 80 1. 41 ，F ABy is -6.67 日~. The horizontal com- / I 、士一 '::"'D ponent ponent of the member AB force is 5 E 5 F /ell-¥/1111¥¥1111/¥ F vhum 凡ト AB 一一一一一一EUρhu Z 一A 一 UFO 住民mm The actual forces in kilonewtons are shown.

U 、 111I/Lιnhv可 L MH 一一 t 且 an 笹口m 20 お N 一δ

For For equilibrium at support A ， the sum of the horizontal forces forces must be equal to 0 kN. The force in member AE is

FAE = FAEx D = -FABx = -(-8.33 kN) 6.67 13.33 = 8.33 kN

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Application Application of a vertical unit load at joi 凶 E results Problems 44 and 45 are based on the following illustra 司 in in the virtual member forces in kilonewtons as follows. tion. (Some round off eロor exis 匂 in these calculated numbers. bers. )

町5m ↑ 5m 手 5m J D

Problem Problem 44 0.67 0.67 0.33 The magnitude of the ver 七ical reac 七ion force at sup- It It is recommended that a table be used to keep all vari- port. A is most nearly ables ables organized. Use negative values to represent com- (A) (A) 1. 3kN press lO n. (B) 5.0 kN F p FQ L FQFpL (C) 13 kN 2 member (kN) (kN) (m) (kN 岨) (D) 20 kN ABCAEFBCBBCDEFDEFF 10.67 10.67 -1. 07 6.4 73.1 -16.67 -16.67 -0 .4 2 5.0 35.0 Solution -2 1. 34 0.53 6.4 72.4 Since suppor 七C is a roller suppor 七ぅ there is no horizon- 十8.33 +0.83 5.0 34.6 tal reaction force at that point. The vertical reaction

十8.33 +0.83 5.0 34.6 force at support A う RA 約 can be found by converting +16.67 十0.4 2 5.0 35.0 the uniformly distributed load ，ω， into a resultant point

十0.00 +1. 0。 4.0 0.0 load. /III-凶一¥、、 -6.67 -6.67 +0.33 4.0 8.8 W 一一一一ωLi二到vhu EB vhum 、¥mt/』 +10.67 -0.53 6.4 -36.2 ノ引制 I: I: = 239.7 11 11 The modulus of elasticity of steel is E = 2.1 X 10 Pa. This resul 七ant load is located a七the centroid of the uni- Since Since the area and modulus of elasticity are the same formly distributed load.

for for all truss members ，七 heir product ，AE う is common to all all members and can be taken outside of the summation for for simpli 五cation. Thereforβ ， the vertical deflection at point point E can be found by Hr A ZRlM¥l/εvEA ?QL山一 E bz一AE 仁い一E いP

2 (1000 (1000 mm ) (吋 :m) 5m 5m

×川1 Pa) (品) The support A vertical reaction can then be found by summing moments about support C. This results in a x (1000 誓) vertical reaction at support A of

ェ1. 14 rnrn (1. 1 mr 吋 RAy(10 m) - (15 kN)(5 m) ISince ISince the unit load in the virtual force system was down- +(25 kN)(2.5 m) = 0k N- m .ward and the answer is positive in sign ， the actual de- 丑ection is also downward. RA" ニ (15 kN)(5 m) - (25 kN)(2.5 m)

吋 10 m The answer is B. = 1. 25 kN (1. 3 kN)

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Since Since the answer is positive in sign ， the direc 七ion of the STRUCTURAL DESIGN calculated calculated reaction is the same as that of the assumed Problems Problems 46 and 47 are based on the following informa- reactioni reactioni that is ， the direction of the reaction is upw 乱rd. tion tion and illustration. The answer is A. The cross section of a reinforced concrete beam with tension tension reinforcement is shown. Assume that the beam Problem Problem 45 is underreinforced. The magnitude of the maximum vertical shear in the 2 beam is most nearly 五= 3000 lbfjin 2 (A) (A) 1. 3 kN ん= 40 ， 000 lbfjin (B) (B) 14 kN As = 3 in 2 [three no. 9 bars] (C) (C) 25 kN (D) (D) 39 kN

Solution Solution One way 七o determine the answer to this problem is to construct construct a shear diagram. The change in shear is the area area under the applied loading. Although a momen 土

diagram diagram is not required for this problem う it follows that the the change in momen 七is the area under the shear diagram ， so a moment diagram is usually included. 20in 23in

/ llF15kN LkN/m I 1c~ ~ ~ ~ ~ ~

V(kN) V(kN) Problem 46 In In accordance with American Concrete Institute (ACI) 凶rength design ， the allowable moment capaci もy of 七he beam is most nearly (A) (A) 160 ft-kips (B) (B) 180 ιkips M(kN'm) M(kN'm) 0 (C) (C) 200 ft-kips (D) (D) 210 ft-kips

-62.5 -62.5 Solution ん一3-{2i ρ'一一一一ニお一如 Mω一1iqG一一一 As can be seen ， the maximum value of vertical shear is m n 25 25 kN at support C. ーー

The answer is C. In an actual design and analysis situation ，乱 check should should always' be made to see th 乱， t the actual reinforcing ing steel ratio falls between the allowable maximum and allowable allowable minimum steel ratios ， even though this check is is not required to solve this specific problem.

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The minimum required length of spiral transition be- Do these characteristics satisfy 七heir corresponding su- tween tween the curve and road is most nearly perpave requirements?

(A) (A) 28 ft (A) Yes ，all the p釘 ameters are within 組 acc 叩七回 (B) (B) 36 此 able range. (0) (0) 44 氏 (B) No ， the VMA is excessive

(D) (D) 72 ft (0) No ， the dust-to 田 asphalt ratio is too high

(D) (D) No ，G mm at N m 日 is too high. 36. 36. The design requirements for a section of highway with with a 1. 5% grade are as follows. 39. A road leading to as 七one quarry is traveled by 40 40 trucks ，wi 七 h each truck making an average of 10 trips design design speed = 80 kmjh per day. When fully loaded ， each truck consis 七s of a coefficient coefficient of friction = 0.35 front single axle transmitting a force of 10 ， 000 lbf and driver driver reaction time = 2.0 s two rear tandem axles う each axle transmitting a force driver driver eye height = 1. 2m of 20 ， 000 lbf. The load equivalency factor for the front

object object (to be 師 oided) height = 0.2 m single axle is 0.0877. The load equivalency f配 tor for each each rear tandem axle is 0.1206. The downh i1l design braking distance for this highway is is most nearly The 18 ， 000 lbf equivalent single axle load (ESAL) for the the truck tra 伍c on this road for 5 yr is most nearly (A) (A) 45 m (B) (B) 75 m (A) 0.33 ESAL (0) (0) 100 m (B) 130 ESAL (D) (D) 120 m (0) 48 ， 000 ESAL (D) (D) 240 ， 000 ESAL

37. 37. A crest on a section of highway consists of a vertical curve curve with a 1500 m radius and a posi 七ive %1 grade fol- Problems 40 and 41 町 e based on the following i1l us 七ra- lowed lowed by a negative 3% grade. The design requirements tion. are are as follows. design design speed = 80 kmjh ↓ド 15kN i「↓ 51NK driver driver eye height = 1. 2m object object (to be avoided) height = 0.2 m 日七 opping sight distance デ300 m FJT:; 矛;;「 The minimum required length of vertical curve needed to to sa 七isfy the design stopping sight dis 七ance is most nearly nearly 40. 40. The magnitude of 七he maximum bending moment (A) (A) 680 m in in the beam is most nearly (B) (B) 700 m (0) (0) 760 m (A) 6.3 kN- m (D) (D) 840 m (B) 14 凶 m (0) (0) 25 kN.m (D) (D) 63 kN- m 38. 38. A superpave design mixture for a highway with 7 ESALs < 10 h回 a nominal maximum aggregate size of of 19 mm. The mixture has been tested and has the 4 1. If the beam is made entirely of steel and the whole following following characteristics: beam has a moment of inertia about the axis of bending 8 of of 2.0 X 10 mm 4， the magnitude of the vertical de fl_ ection tion at point D is most nearly air air voids = 4.0% VMA ニ 13.2% (A) 0.20 mm VFA = 70% (B) 2.3 mm dust-to-asphalt dust-to-asphalt ratio = 0.97 (0) 23 mm

at at N = 8 gyrations う G mm ニ 87.1% (D) 50 mm

at at N = 174 gyrations ぅ G mm = 97.5%

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Problems Problems 42 and 43 are based on the following informa- 44. In the x-direction as shown ， the m a.x imum influ- tion tion and illustra 七ion. ence line ordinate for tensile force in member BF is most nearly nearly A truck is facing in its intended direction of travel along the the beam as shown. (A) 0.18 kN/kN (B) (B) 0.36 kN/kN (C) (C) 0.53 kN/kN (D) (D) 0.71 kN/kN

45. 45. In the x-direction as shown ， the m a.x imum influence ence line ordinate for compressive force in member BF is is most nearly

(A) (A) -0.71 悶 /kN トx (B) -0.53 kN/凶 (C) (C) -0 却 kN/kN 下予 lfm 手ょ :iD (D) -0.18 kN/kN

Problems 46 and 47 are based on the following information tion and illustra 七ion.

42. 42. For a truck traveling in the direction shown ，七 he The cross sections of two short ，concentrically loaded m a.x imum ver もicallive load shear at suppor 七C is most reinforced concrete columns are shown. nearly nearly (A) (A) 27 kN だ= 4000 Ibf/in 2 (B) (B) 76 kN fy = 60 ，000 Ibf/in 2 (C) (C) 100 kN 18 in 18in (D) (D) 110 凶

43. 43. For a truck traveling in the direction shown う the m a.x imum live load bending moment 抗 support C is most ne 紅 ly (A) (A) 80 kN- m longitundinal (B) (B) 90 kN.m reinforcement (C) (C) 140 kN.m round round spiral column square tied column (D) (D) 360 kN.m (cross section) (cross section)

(Prob 圃 46) (Prob.47) Problems 44 and 45 are based Dn the following information tion and illustration. 46. 46. For the short round spiral column ， the applied a.x ial A plane truss span is shown. The roadway behaves as dead load is 150 kips ，and the applied a.x iallive load is simply simply supported beam spans between the supporting 350 kips. Assuming that the longitudinal reinforcing lower lower chord truss joints. By convention ，positive forces bars are all the same size ， the minimum required size of are are tensile forces う and negative forces are compressive each longitudinal reinforcing bar is forces. forces. (A) (A) no.7 (B) (B) no.8 IaTill--47 (C) no. 9

A m (D) no. 10 「F 『 47. 47. For the short square tied column ， the applied a.x ial dead load is 150 kips ，and the applied a.xi al live load 5m 5m 5m is 250 kips. Assuming tha 七七 he longitudinal reinforcing

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'7. 3'7. There are two equa 七ions to chec k. For 5 yr ， the tota1 ESAL is

First ， where s七opping sigh 七distance ，8 ，is 1ess than the curve curve 1ength う L: 札 r ニ州 2 L=- A8 = 240 う097 ESAL (240 ， 000 ESAL) (100%) (100%) (VWi +伊豆 ;y

(1% 一 (-3%))(300 m)2 The answer is D.

一 (100%) (何日早川南可 f 40. 40. Construct shear and mome 凶 diagrams. The change = 756 .4 m in shear is the area under the applied 1oading ，and the change change in mome 凶 is the area under 七he shear diagram. Second ， where stopping sigh 七 distance is greater than the the curve 1eng 七h: (200%) (200%) (VHi +♂'2/ L=2S A N| 一:-1 (200%) (200%) (Vl五五 +VO 万五 :)2 トベー一 = (2)(300 m) 一 AL 一一円円h 川一-qJA-νIEq4 1%-(-3%) 』且晶 mill-- = 48 1. 0 m

whuIVAM Since 也e previous two equations show that the stop- Ay 川 ping ping sight distance is 1ess than the curve 1ength ， the minimum required vertica1 curve 1ength is L = 756 .4 m (760 m) The answer is C. V{kN)

38. 38. This mixture wou1d be designated as a 19 mm su 伺 perpave perpave mixture. The limits for such a mixture are air air voids = 4.0% minimum VMλ= 13% M{kN.m) 0 VFA = 65-75% dusιto-aspha1t dusιto-aspha1t ratio = 0.6- 1. 2 at at N init = 8 gyratior おう maximum G mm = 89% -62.5 at at N max = 174 gyrations ，maximum G mm = 98%

All All parameters in this mixture are within superpave The 1argest magnitude of bending 江lO ment 1S specifica 七ions. -62.5 kN.m (63 kN- m) at support C.

The answer is A. The answer is D.

39. 39. The tota1 ESAL per truck for each trip is 4 1. Use the princip1e of virtua1 work to 五nd 七he vertica1 ESAL 七r叫= (1 sing1e 砿 1e)(0.0877) de 丑ection at point D. + (2 tandem ax1es)(0.1206) = 0.3289 ESALjtruck-trip ムD= ~(J m( 自白) The tota1 daily ESAL for 40 trucks ， each making 10 10 trips a day ，is The moment functions in the directions indicated by the the 1oca1 x-coordinate for each beam segment under the E 叫=山 cks) (10 寄) 乱ctua110ading are shown on the following moment dia 同 gram. gram. X (0 印刷) truckωtrip

= 13 1. 56 ESALjday

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The modulus of elasticity of steel is E = 2.1 X 10 11 Pa.

Fr om the principle of virtual work ， the vertical de fl. ection tion at poin 七D is

5 ムD = 1 (- EI

+J +J EI

X X イー云 2

M(kN.m} 0 呼判:+一山 1: I M= 6.25- EI 十半 1: 十平 1: 十年 1: The moment functions in 七he directions indicated by the local local x-coordinate for each beam segment under the vir- - 26.042 kN 2.m 3 - 78.125 kN 2.m 3 tual tual unit loading at point D are shown on the following momen 七diagram. 十 390.625 日~2.m3 + 286 .4 58 kN 2.m 3 2 十 390.625 kN .m 3

11 11 (2.1 (2.1 X 10 Pa) (品)

/紅 l立1¥ X Il lJlJ U -一一ー l ¥ m / = 22.9 mm (23 mm)

Since Since the unit load in the virtual force system was downward and the answer is positive in sign ， the actual de- fl. ection is also downward.

The answer is C.

A table summarizes the moment functions as follows.

reg lO n regron reg lO n

A-B B 同 C C 自 D function function 0

M 1. 25x 6.25 - 13.75x -2.5x 2 m -0.5x -2.5 - 0.5x -x EI EI EI EI

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42. 42. Construct and use an in 丑uence line for ver もical shear at at support C. The in 乱uence line is constructed by plot-

ting ting the change in response on a free 四 body diagram of a section of beam at suppo 吋 C as a unit load travels across across the structure. 74 74 5m 5m 。 t t load load position 1.5 1.5 for maximum 。 bending moment influence influence line for V c (kN/kN) t t The load position shown gives the maximum response load load position 71.17 kN 17.79 kN in the beam at suppor 七C for the speci 五ed direc 七ion of for for maximum trave The maximum bending moment can be found vertical vertical shear 1_1 l. by superposition 出

The load position shown results in 出 e maximum re 司時 max = 1(-5 守)川 sponse sponse in the beam at suppor 七 C for 出e speci 五ed direction rection of trave l. The maximum vertical shear can be = 355.8 kN- m found found by superposition 回 This This problem asked for the magnitude of maximum bending bending moment at support C for a truck traveling in 日 x= (1. 5 ~~) (7 1. 17 凶) the direction sho ωn. If this problem had asked for the magnitude magnitude of maximum bending moment at support C

for for th 巴 given truck αxle configur αtion う the axle load po- sitions sitions would have to be switched around and both axle This This problem asked for the magnitude of maximum ver- loads placed on the infiuence line with the heavier axle tical tical shear at support C for a truck tn αveling in the load load on the larger in 宜uence line ordinate and the lighter direction direction shown. If this problem had asked for the mag- axle load on the smaller in 丑uence line ordinate. This nitude nitude of maximum vertical shear at support C for the would result in a larger numerical answe r. given given truck αxle configur α， tion ， the axle load posi 七ions would would have to be switched around and both axle loads The answer is D. placed placed on the infiuence line with the heavier axle load on the larger infiuence line ordinate and the lighter axle load load on the smaller in 丑uence line ordinate. This would 44. Construct an infiuence line for the force in mem- result result in a larger numerical answe r. ber BF. The in 丑uence line is constructed by plotting the the change in response in member BF as a unit load The answer is D. travels across the structure.

43. 43. Construct and use an infiuence line for bending moment 抗日uppo 凶 C. The in 丑uence line is constructed by

plotting plotting the change in r田 ponse on a 丘ee 四 body diagram of of a section of beam at support C as a unit load travels across 七he structure.

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_J.，" _J.，" Pr 目di(e Ex am 1 Solutiolls S9

I七is required that ゆPn 2: Pu. For axial compression

with with spiral reinforcement ，ゆ= 0.70. 8etting ゆPn = P，叫 and solving for the area of longitudinal reinforcing steel gIVes gIVes

ゆ E;fd 務 lllJzr lllJzr J i iλ哩~I (附 /kJ 山 ps)(1000 2)

When moving the uni 七load across a truss structure ，this load load must be dis むibuted to the two joints. It is stated 485) (4000 主) (年) 七hat the roadway acts as simple beam spans between truss truss join 七s. Therefore ，when a load is placed between 。主一 (0.85) (4000 ~~;) truss truss join 七s， the two joints adjoining the beam span act as as simple beam supports and the magni 加 des of the loads loads applied to 出e two truss joints are the same as for As As = 6.69 in 2 ，required for the given applied axial com- the the calculated reaction forces of 七his beam. pressive pressive loads ，is greater than As = 2.54 in 2 based on The maximum ordinate for tensile force in member BF the minimum allowed reinforcement ratio ， Pg = 0.0 1. is is 0.53 kN /kN The minimum required area of reinforcement is As 二 2 6 司 69 in . The answer is C. The column has six longitudinal reinforcing bars. The 45. 45. Fr om the in fl. uence line in 80 1. 44 ， the maximum required area of each longi 七udinal reinforcing bar is ordinate ordinate for compressive force in member BF is -0.53 kN/kN A=~一-一日主主 nbars nbars 6 bars The answer Is B. = 1. 11 in 2 /bar 46. 46. Determine the amount of reinforcing s七eel required by the minimum required reinforcement ratio ，Pg ， of A bar area of 1. 11 in 2 is satisfied by a no. 10 bar ， 0.0 1. The minimum area of reinforcing s七eel required is which has a nominal area of 1. 27 in 2.

In In an actual design/ analysis situation ，乱 check should also also be made to see that the actual longitudinal rein- = (0 川(年) forcement ratio does not exceed the maximum allowable ratio ratio of 0.08.

The answer is D. Determine Determine the required amount of reinforcing steel based on the factored axialload ， Pu. 47. Determine the amount of reinforcing steel required by the minimum required reinforcement ratio ，内， of Pu ニ1. 2Pdead 1. 6P !i ve + 0.0 1. The minimum area of reinforcing steel required is ヱ(1. 2)(150 kips) + (1. 6)(350 kips) = 740 kips As = pgAg = (0.01)(18 in)2 The nominal axial compressive load capacity is given by = 3.24 in 2

凡 = 0.85 P. 。 Determine the required amoun 七 of reinforcing steel based

= (0.85)(0.85f~Aconcrete +ん As) on the factored axi ，al load う Pu. = (0 部 )(0.85 五(Ag - As) + んAs) Pu = 1. 2P de 吋+ 1. 6P !i ve = (1. 2) (150 kips) + (1. 6) (250 kips) = 580 kips

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The 七ra 伍c flow relationship is given by q = kv ， in which Problems 41 and 42 are based on the following informa- q is the tra 伍c volume in veh/hr. The maximum tra 伍c tion and illustration. volume volume for 七his road is most nearly The beam shown is loaded with two 1000 N point loads. (A) (A) 760 veh/hr The separation is maintained 抗 2m ，but 出e loads may (B) (B) 880 veh/hr be moved to any location on the be 乱m. (C) (C) 900 veh/hr (D) (D) 960 veh/hr 門…什「一助|h

38. 38. The s七opping sight dis 七ance is 430 ft for a design speed speed of 50 mph on a section of highway. The grades for for this highway section are -1% followed by 3%. The required required length of vertical curve needed to satisfy 七he AASHTO stopping sight distance for this design speed 蕩Z， is is most nearly (A) (A) 270 氏 10 m 10 m (B) (B) 380 氏 (C) (C) 410 ft (D) (D) 450 ft 4 1. The maximum value for shear at support A is most nearly nearly

39. 39. A one-lane rural road has a 10 0 curve extending for (A) 2000 N 230 230 m along its centerline. The road is 5m wide with (B) 2800 N 3m wide shoulders. The design speed for this road is (C) 3000 N 75 75 km/h. (D) 3800 N

The superelevation needed so that side friction is not needed needed is most nearly 42. The maximum value for moment at suppor 七A is most nearly (A) (A) 0.00050 (B) (B) 0.034 (A) 1. 8 kN.m (C) (C) 1. 1 (B) 8.0 kN- m (D) (D) 1. 9 (C) 10 kN.m (D) (D) 18 kN- m

40. 40. The worn surface course of a high-volume pave 四 43. A triangular pin-connected truss carries a load of ment is being replaced with a design requiring a total 4448 N as shown. Each member has the same modulus structural structural number of 6.6. The engineer has decided to of elasticity and cross-sectional area. replace replace 6 in of the surface with recycled-in-place asphalt concrete concrete having a surface course strength coe 血cient of :%:l M 0.4 2 ，leaving in place 3 in of sound original pavement having having a strength coefficient of 0.3. Under the original nal pavement are a 10 in ceme 凶 -treated base having a strength coe 血cient of 0.20 ぅ and 阻 8 in sandy gravel subbase. subbase. What is the minimum strength coefficient for 3.0 3.0 m the the subbase?

(A) (A) 0.05 (B) (B) 0.10 (C) (C) 0.15 P (D) (D) 0.20 N

4448 N

4.6 4.6 m

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6 The truss member properties are E = 200 X 10 kPa Problems 47-49 are based on the following information 2 and A ニ 2580.6 mm . The vertical deflection at point and illustration. P is most nearly The span length and cross section of a reinforced con- (A) (A) 0.25 mm crete beam are shown. The beam is underreinforced.

(B) (B) 0.4 8 mm The concrete and reinforcing steel properties ぽ e 五= (C) (C) 0.51 mm 2 2 2 3000 3000 lbf/in うん = 40 ， 000 lbf/in ，and As = 3 in • (D) (D) 0.75 mm

6ft 6ft Problems Problems 44-46 are based on the following illustration.

l~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ L~ A 易易 18 18 i日目 3m 3m beam cross cross section section 44. 44. If the reaction at support A is 18.75 N ， the reaction 10 ft at at each of 出e outer supports is most nearly (A) (A) 5.6 N 47. 47. Neglec 七ing beam self-weight and based only on the (B) (B) 7.2 N allowable moment capacity of the beam as determined (C) (C) 11 N using using American Concrete Institute (ACI) strength de 白 (D) (D) 14 N sign sign specifications う the maximum allowable live load is most nearly 45. 45. The maximum value of vertical shear at any point (A) 23 ， 000 lbf along along the beam is mos 七nearly (B) 29 ，000 脳 (A) (A) 4.7 N (C) 35 ， 000 lbf (B) (B) 9.4 N (D) 50 ， 000 lbf (C) (C) 14 N (D) (D) 18 N 48. The beam supports a concentra もed live load of 50 ， 000 lbf. Neglec 七 beam self-weight. The minimum amoun 七 of shear reinforcement required for a center- 46. 46. The maximum value of moment at any point along to-center stirrup spacing of 12 in under ACI strength the the beam is mos 七nearly design specifications is most nearly (A) (A) 1. 2 N.m (A) 0.18 in 2 (B) (B) 3.1 N.m (B) 0.36 in 2 (C) (C) 4.6 N.m (C) 0.67 in 2 (D) (D) 5.7 N.m (D) 0.78 in 2

49. 49. The balanced reinforcing steel ratio for this beam in in accordance with ACI specifications is most nearly (A) (A) 0.037 (B) (B) 0.043 (C) (C) 0.051 (D) (D) 0.058

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The density is 33.125 veh/mi for maximum tra 缶c vol 司 For a side- 台iction factor of f =0 ， the superelevation is ume. Substitution in 七o the tra 白c flow relationship gives given by the the maximum tra 血c volume.

= (53 ::) (33.125 :~) ((75 引(1 吋(品)) (9.81 (9.81 ~) (1318 m) 一 (0.8 品) (33.1 叫

The answer is B. The answer is B.

40. 40. The structural number is the sum of products of 38. 38. Since there is a negative grade preceding a positive the layer depths (thicknesses) and strength coe 血cients. grade う this is a sag vertical curve. Using the sag vertical cal curve equations from the civil engineering section of SN=αrecycleDrecycle 十 αoriginal surface D original surface the the NCEES Handbook ， the algebraic difference between +αbaseDbase 十 αsubbaseDsubbase grades grades is 6.6 6.6 = (0.42)(6 in) + (0.3)(3 in) 十 (0.2) (10 in) A = 1-1% 3%1 十αsubbase(8 in) =4% αsubbase αsubbase = 0.15 Where the stopping sight distance ，8 ，is less 七han 七he vertical vertical curve length ，L う The answer is C.

L=-- 竺L 4 1. Draw an in 丑uence line for shear at suppor 七A with 400 十 3.58 loads positioned as shown for maximum shear value. (4)(430 (4)(430 ft )2 )2 For example う for shear at support A う using basic beam 400 十 (3.5)(430 此) statics ， = 388.2 ft

Where the stopping sight distance is greater than the curve curve leng 七h，

400 400 + 3.58 L 二 28 一一 A一一

400 400 + (3.5)(430 氏) 20 m = (2)(430 此)-4 2.0 = 383.8 ft influence influence line for shear at support A

Fr om the 七wo values for length of curve う it can be seen that that the s七opping sight distance is greater than 七he ト --1 curve curve length. Therefore ， the required vertical length of of curve is L = 383.8 抗 (380 此) The answer is B. loads loads 一一一 placed for maximum value shear :51~20at support A

39. 39. The radius of curvature is The maximum value for shear at support A is

VA = (1000 N)( 1. 8) + (1000 N)(2.0) R=~= 川寺) = 3800 N

The answer is D.

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42. 42. An infiuence line for bending moment at support A By inspection ， the x-component of force in member MP can can be drawn with loads positioned as shown for max- is the same magnitude and opposite direction as the imum mome 凶 value. As in 五nding 七he infiuence line reaction at point M. for for shear う a unit load is moved across the beam and the 広明一向一〉一一十一イ、一一2M一 Mpi RUv dlu RI variation variation in bending moment at point A is graphed. M 二一白川41 -m eU ‘，一円U

伝山、必一勺一十一弘で一 G 400N

剖「|

mAす姦 M 一-6820 N

10 10 m 10 m

8142 N influence influence line for moment at support A 3.0 m ON

P 一一 6820 N N 6820 N

loads loads placed for maximum value at support A 4.6 4.6 m

The maximum value for moment at support A (dis- regard regard the sign ぅ since the maximum value is wanted) Applying a unit load at point P produces the following lS lS virtual forces.

MA = (1000 N)(8.0 m) + (1000 N)(10.0 m) M ....1.5N ....1.5N 二 18000 N.m (18 kN .m)

The answer is D. 1.8 1.8 N

3 開 Om I 0N 43. 43. The principle of virtual work can be used to find the the defiection a七poin 七P. P Choose the positive directions as upward and to the 1.5 N N 1.5 N right. right. Choose positive moments as clockwise.

LMN=ON

= R M ，， (3.0 m) 十 (4448 N) (4.6 m) 4.6 m RM" = 6820 N [七 o the 1eft] エMM=ON The following table summarizes the actual and virtual forces. forces. = -R N " (3.0 m) + (4448 N)(4.6 m) RN" = 6820 N [to the right] FQ Fp L FQFpL member (N) (virtual force) (m) (N- m) LFy LFy =0 N MP 8142 1. 8 5.4 9 80459 = RM y - 4448 N NP 6820 1. 5 4.6 47058 RMν= 4448 N [upw 乱rd] MN 0 0 3.0 0 total total = 127517

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Calcu1ate Calcu1ate the vertica1 de 丑ection 抗 point P. The change in momen もis given by the area under 七he shear shear diagram up to that point.

ムp = LFQ5L= 玄FQ22

=占 LFQFpL (t) = (t) (5.6 叩 2 m)

川却別ル一川2川い川Oω O川山川1叩川O併6 印刷叫叫)べ中(や 1∞ c ル 3 m =凡=1. 12 m 十(訪問問問 m)

2 x (2 附 mm ) (抗日

x (127517 N- m) (1000 空)

The answer is A.

44. 44. Fr om the 1aws of equi1ibrium う each reaction at an outer outer suppor 七is ???

x= 1.12 m 1.88 m 1司 88 m ニ (t) ((5 :)州 -18 同

The answer is A.

45. 45. The maximum shear can be determined from a 3 膚 1 3.1 shear shear diagram. (A1though no 七necessary for this prob- 1em ，a moment diagram has a1so been constructed ，and M(N'm) this this is used for solving the next prob1em.)

The change in shear is the area under the 10ad diagram up to that point. Up to the center of the be 叩 1， point A ， 5.7 5.7

ド 5.6 N- (5 :) (3 m) Fr om the shear diagram ， the maximum va1ue for shear occurs occurs 剖 support A.

At poin 七A ， Vmax = 9.4 N

V = -9 .4 N + 18.75 N The answer is B. = 9.35 N (9 .4 N)

46. 46. Fr om the moment diagram in 80 1. 45 ， the va1u 巴for The moment is equa1 to the area under the shear dia- maximum moment occurs at support A. gram. gram.

M max = [- 5.7 N.m[ = 5.7 N- m

The answer is D.

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哩50 Appendix Aftemoon Sa 円lple Ex amination

42. 42. 1n using ch1 0rine for disinfection a. . a. concen 住ation of HOC1 is pH-dependent b. b. [HOCl] equals [OCll at pH 7 c. c. e旺'e ctiveness increases with time of contact d. d. total elimination of bacteria is possible

43. 43. A 61 ・cm-diameter circular concrete sewer is set on a slope of 0.00 l. The flow flow when it is halιfull is ajof 伽叫制l b. b. 0.08 m""/s c. c. 0.20 m 3/s d. d. 0.28 m 3/s

44. 44. A rigid bar BCD is supported by a hinged connection at support B and by a cable AC as shown. The reaction force at support B is

D B

ド 2.0m 、、レ O.5m .. ..J a. a. 2.50 kN b. b. 14.20 kN c. c. 9.70 kN d. d. 14.42 kN

45. 45. A simply supported beam AB is subjected to a uniformly distributed load load and a concentrated load as shown. The midspan de f1 ection is

20 kN

ド 3.0m

E= 200 GPa 1= 3x10 6 mm 4

a. a. 16mm b. b. 38 mm c. c. 22 mm d. d. 8mm .. ..

Sample Ex am 15.

46. 46. A steel column (W21 x 93) has the bottom end supported rigidly so as r--r・一一一一r- to prevent both rotation and translation. The top of the column is free to to translate and rotate in 出e plane of the diagram but pinned (rotation allowed ，translation prevented) in the perpendicular direction. The length length of the column is 32 ft. The critical buckling load as given by Euler' Euler' s theory is a. a. 13 .4 kips b. b. 36.7 kips c. c. 366 kips d. d. 1002 kips

47. 47. Th e three-hinged arch ABC carries a uniformly distributed load from a horizontal horizontal deck as shown. 百le value of the horizontal thrust at the supports is

[}=[] [}=[] ー

15m 10m ド当6 オ

a. a. 282 kN b. b. 197 kN c. c. 218 kN d. 252 kN

48. 48. Given the truss shown ， the vertical deflection at B is

E = 29 ，000 ksi for both members A = 2 in 2 for AB A = 3 in 2 for BC 10 抗

20 kips ド 8 ft 当|

a. a. 1. 2 in b. b. 0.05 in c. c. 0.2 in d. d. 0.1 in 152 Ap pendix Aftemoon Sam ple Ex amination

49. 49. Th e maximum bending moment for the beam shown is

3.5m 1.5 m ド ~〈

a. a. 12.60 kN m b. b. 8.82 kN m c. c. 9.4 1 kN m d. d. 10.25 kN m 50. 50. A client requires a new fac i1i ty quickly to replace one destroyed by a hurricane. hurricane. He is not certain exactly what it should include. The best type type of contract to be used in 出 is situation is a. a. unit price b. b. lump sum c. c. cost plus fixed fee d. d. surety 51. 51. What is the height (白) of a conical spoil pile ， if 100 ft bank cu yd of common earth with an angle of repose of 32 0 and a 12% swell is deposited? deposited? a. a. 9.6 b. b. 19.2 c. c. 10 .4 d. d. 33.3 52. 52. Determine the volume of fi1l for a whole station (100 ft) of cut with the the end areas shown below.

Al=120ft'/ ¥ Al=120ft'/ ¥ ι= 150 が/ ¥A'= 100 ft ，/

a. a. 820 yd 3 b.456.8ydj c. c. 506.2 yd J d. 569 .4 yd J 53. 53. ABC Construction Company is the low bidder at $500 ，000 on a building projec t. The company has provided the typical bid bond. After award of of the contract ， the company decides to withdraw its bid. The second lowest lowest bid was $520 ，000. How much is the sur~ty . company required to reimburse reimburse the client for the failure of ABC to sign? a. a. $500 ，000 b. b. $50 ，000 c. c. $20 ，000 d. d. $5000 Solutions 唱63

41. 41. d. All of the statements are co 町 ec t.

42. 42. a.

43. 43. b. Qf= 0.2 m 3/s

Q/Qf= 0.4 内 Q = 0.08 m'>/s

44. 44. d. Th e free body diagram for BCD is shown. Note 白at a common mistake couldbe 脳出国ng that cable AC forms a3 -4- 5 凶angle when it d<∞ sn' t.百 le ve 同cal and horizontal components of T 訂 e O.66T and 0.75 T， respectively.

T 10

ヱM B =0.66Tx2-10x2.5=0 司 T= ω4

ヱぇ=久一 0.75 x 18.94 = 0 弓 Bx = 14.20

ヱπ=久+0 “x 18.94 -10 = 0 => By = -2.50

RB= 石弓 4

45. 45. b. EI = r 2 x ぽ型 )X(3X 1O-6 m 4) = 6∞kN m 2 ¥ m- J Deflection Deflection at rnidspan (x = 2 m) due to point load is given by

Pb j三句今l 20xO.5 _ _， j l o= 一一一 [-x +(L ーか )x] =_ -~:. ~~:~ [-2 + (4 - OS) x 2] = 0.016 6LEr " 6x4xω 。 Deflection Deflection at rnidspan (x = 2 m) due to distributed load is given by

δ 5wL 4 5x4x4 4 =一一=一一一一 =0.022 384EI 384x600 Total Total midspan deflection = 0.038 m = 38 mm

46. 46. c. Buckling about the strong (x) 鉱 is:

K = 2.0; L = 32 ft; rx = 8.70 in; KUr = 88 .3 Buckling Buckling about the weak (y) axis:

K = 0.7; L = 32 ft; ry = 1. 84 in; KUr = 146.1 唱64 Appendix Aftemoon Sample Ex amination

Rπ 2EA π2 x29 ，000x273 内〆" . =ナーーす= ヲ =jOO 阻 p: (子}

47. 47. b. Th ere are four extemal reactions-A_ A. c.. and C-and two intemal x' "Y ~x' hinge hinge forces-B x and By 一旬 be deterrnined. Recognize that Ax and Cx are are equal and opposite. Solving for either Ax or C x is sufficien t. Taking Taking moments about A (entire structure):

ヱM A =500x 山 +4ι -25C y =0

Taking Taking moments about B (right half of structure):

ヱMB.right = 200x5-6 に一以 =0

Solving Solving these equations ，we get Cx = ー197 .4 kips. 48. 48. d. Step 1: U sing method of joints at joint B:

凡B =+ 25.6 k (T)

F BC = ー16.0 k (c)

Step Step 2: Calculate the member forces due to virtualload (note: in 出is case ，出e 毘 alload and the virtualload look similar ， so we can use scaling):

んB = + 1. 28 (T) fBC = -0.80 (c)

Step Step 3:

=す r FfL 1=~5 似1. 28 x (12.81 x 12)+-16x -O .8x(8xI2 L= 0.10 血 """"'l """"'l AE ) 2x29.000 3x29.000

49. 49. b. Using the fixed end moment load cases in 白e FE Supplied-R そference Handbook with the following data:

P = 12 kN ;α= 3.5 m; b = 1. 5 m; L =5 m

...... Solutions 唱65

2 2 Pab = 12 --_._.__--_.- x 3.5 x 1. 5 =3.78 kN rn AB =ーで一72 L~ 2 2 =ーで一=Pa b --_._.__ 12x3 .5 -._.- x 1. 5 ==8.82 kN rn BA r2 L~

Taking Taking rnornents about A ，

ヱM A =+3.78 一以3.5-8.82+ 玖 =0 叫 =9 .4 08 釦 dAy =2.592

The shear diagrarn and bending rnornent diagrarns 紅 e:

2.592

9.408 9.408

+5.292 kN m

-3.78 kN m -8 .82 kN m

50. 50. c. Cost p1us fixed fee. Choices a and b require a detailed design; choice d is is a bonding cornpany.

51. 51. c. Loose cu ft = 100 cu yd x 1. 12 x 27 cu ft/ cu yd = 3024 cu ft Base diarneter (B) == (7.64 x 3024/ tan 32 0 )1/3 = 33.3 ft 0 Height Height == (33 .3 /2) (tan 32 ) == 10 .4 ft

52. 52. d. 百1i s problern can be solved using the prisrnoidal rnethod. Vo1urne == (1 cu cu yd / 27 cu ft) 1∞ ft [120 sq ft + (4 x 150 sq ft) + 100 sq ft] /6= 506.2 506.2 cu yd

53. 53. c. The client can only recover the 10ss incurred by the default ， which is 出e di 百erence between the low bid and the second 10west bid: $20 ，000.

54. 54. b. Total cost = $50 ，000 + $5000 == $55 ，000 Unit Unit cost == $55 ，000/200 == $275.00/ft Note 出at the unit price rnust include include the contractor's profi t/ rnarkup.

=-.: Determinacy and Stability Determinacy and StabilityI

Trusses are statically determinate when all the bar forces can be determined from the equations of statics alone. Otherwise the truss is statically indeterminate. A truss may be statically/externally determinate or indeterminate with respect to the reactions (more than 3 or 6 reactions in 2D or 3D problems respectively). A truss may be internally determinate or indeterminate. If we refer to j as the number of joints, R the number of reactions and m the number of members, then we would have a total of m + R unknowns and 2j (or 3j) equations of statics (2D or 3D at each joint). If we do not have enough equations of statics then the problem is indeterminate, if we have too many equations then the truss is unstable. 2D 3D Static Indeterminacy External R > 3 R > 6 Internal m + R > 2j m + R > 3j Unstable m + R < 2j m + R < 3j

Victor E. Saouma; CVEN 3525; Univ. of Colorado Reactions 5/23 Method of Joints Method of JointsII

This method should be used when all member forces must be determined. In truss analysis, there is no sign convention. A member is assumed to be under tension (or compression). If after analysis, the force is found to be negative, then this would imply that the wrong assumption was made, and that the member should have been under compression (or tension). On a free body diagram, the internal forces are represented by arrow acting on the joints and not as end forces on the element itself. That is for tension, the arrow is pointing away from the joint, and for compression toward the joint.

C C

c c -ve -ve

A B A B t +ve

Victor E. Saouma; CVEN 3525; Univ. of Colorado Reactions 9/23 Example; Method of Joints Example; Method of JointsI

8 12

G 10' H F

32'

E BCD A 20 k 40 k 40 k

24' 24' 24' 24'

√ 1 R = 3, m = 13, 2j = 16, and m + R = 2j 2 We compute the reactions

E + ΣMz = 0; ⇒ (20 + 12)(3)(24) + (40 + 8)(2)(24) + (40)(24) − RAy (4)(24) = 0

¡ ⇒ RAy = 58 k 6 ? + ΣFy = 0; ⇒ 20 + 12 + 40 + 8 + 40 − 58 − REy = 0

⇒ REy = 62 k 6

Victor E. Saouma; CVEN 3525; Univ. of Colorado Reactions 10/23 Example; Method of Joints Example; Method of JointsII

3 Consider each joint separately: Node A: Clearly AH is under compression, and AB under tension.

FAH

A FAB

58 k + 6 ΣFy = 0; ⇒ −FAHy + 58 = 0 F = l (F ) AH ly AHy p 2 2 ly = 32; l = 32 + 24 = 40 40 ⇒ FAH = 32 (58) = 72.5 k Compression - + ΣFx = 0; ⇒ −FAHx + FAB = 0

F = Lx (F ) = 24 (58) = 43.5 k Tension AB Ly AHy 32

Node B:

Victor E. Saouma; CVEN 3525; Univ. of Colorado Reactions 11/23 Example; Method of Joints Example; Method of Joints III

FBH

B 43.5 k FBC

20 k - + ΣFx = 0; ⇒ FBC = 43.5 k Tension + 6 ΣFy = 0; ⇒ FBH = 20 k Tension

Node H: 12 k

F H HG FHGy FAHx F FHCy HGx FAHy FHCx 72 k 20 k FHC

Victor E. Saouma; CVEN 3525; Univ. of Colorado Reactions 12/23 Example; Method of Joints Example; Method of JointsIV

+- Σ = ; ⇒ − + = Fx 0 FAHx FHCx FHGx 0 24 24 43.5 − q (FHC ) + q (FHG) = 0 242+322 242+102 + Σ = ; ⇒ + − + − = 6 Fy 0 FAHy FHCy 12 FHGy 20 0 32 10 58 + q (FHC ) − 12 + q (FHG) − 20 = 0 242+322 242+102

This can be most conveniently written as

0.6 −0.921 F 43.5 HC = −0.8 −0.385 FHG 26.0

Solving we obtain FHC = −7.5 and FHG = −52, thus we made an erroneous assumption in the free body diagram of node H, and the ﬁnal answer is

FHC = 7.5 k Tension

FHG = 52 k Compression

Node E:

Victor E. Saouma; CVEN 3525; Univ. of Colorado Reactions 13/23 Example; Method of Joints Example; Method of JointsV

FEF

FED E

62 k q 242+322 ΣFy = 0; ⇒ FEFy = 62 ⇒ FEF = 32 (62) = 77.5 k C 24 24 ΣFx = 0; ⇒ FED = FEFx ⇒ FED = 32 (FEFy ) = 32 (62) = 46.5 k T

The results of this analysis are summarized below

Victor E. Saouma; CVEN 3525; Univ. of Colorado Reactions 14/23 Example; Method of Joints Example; Method of JointsVI

8 12

52 52 43.5 46.5

58 .5 62 7 32 2 7

.5 4 0 7 .5 .5 2 20 7

43.5 43.5 46.5 46.5 58 20 40 40 62

4 We could check our calculations by verifying equilibrium of forces at a node not previously used, such as D

Victor E. Saouma; CVEN 3525; Univ. of Colorado Reactions 15/23 Examples Beams Example Beam; 1

11 k 10 k 3 4 2 k/ft

BCD A E 4ft 6ft 4ft 4ft

Reactions are determined from the equilibrium equations (+ )ΣFx = 0; ⇒ −Ax + 6 = 0 ⇒ Ax = 6 k

(+)ΣMA = 0; ⇒ (11)(4) + (8)(10) + (4)(2)(14 + 2) − Ey (18) = 0

¡ ⇒ REy = 14 k (+ 6)ΣFy = 0; ⇒ Ay − 11 − 8 − (4)(2) + 14 = 0 ⇒ Ay = 13 k

Victor E. Saouma; CVEN 3525; Univ. of Colorado Internal Forces 11/31 Examples Beams Example Beam; 2

11 k 10 k 3 4 2 k/ft

BCD A E 4ft 6ft 4ft 4ft

Shear are determined next. 1 At A the shear is equal to the reaction and is positive. 2 At B the shear drops (negative load) by 11 k to 2 k. 3 At C it drops again by 8 k to −6 k. 4 It stays constant up to D and then it decreases (constant negative slope since the load is uniform and negative) by 2 k per linear foot up to −14 k. 5 As a check, −14 k is also the reaction previously determined at F. Moment is determined last: 1 The moment at A is zero (hinge support). 2 The change in moment between A and B is equal to the area under the corresponding shear diagram, or ∆MB−A = (13)(4) = 52. 3 etc...

Victor E. Saouma; CVEN 3525; Univ. of Colorado Internal Forces 12/31 Examples Beams Example Beam; 3

10 k A

11 k 8 k

E 13 k 2 k C B 2 k -6 k

13 k 2 k A=(13)(4)=52 A=(6)(2)=12

A=(-6)(4)=-24 A=-4(6+14)/2=-40 -6 k

Slope= dV/dx=w=-2 -14 k

x=+2 dM/d dM/d x=-6 52+12=64 0+52=52 64-24=40 A B CD E

Victor E. Saouma; CVEN 3525; Univ. of Colorado Internal Forces 13/31 arches Three Hinged Arch; Point Loads Three Hinged Arch; Point Loads I

B 30 k 20 k 30 k 20 k HB HB B V V 20' 20' 33.75' B B

80 k 80 k HC C C 30' VC 26.25' HA A A

80' 60' VA

Four unknowns, three equations of equilibrium, one equation of condition statically determinate. ⇒

+ ΣMC = 0; (R )(140) + (80)(3.75) (30)(80) (20)(40) + R (26.25) = 0 ¡ z Ay − − Ax 140RAy + 26.25RAx = 2.900 - ⇒ + ΣFx = 0; 80 R R = 0 − Ax − Cx ! (22) + 6 ΣFy = 0; RAy + RCy 30 20 = 0 ! − − B + ¡ ΣM = 0; (Rax )(60) (80)(30) (30)(20) + (R )(80) = 0 z − − Ay 80R + 60R = 3, 000 ⇒ Ay Ax

Victor E. Saouma; CVEN 3525; Univ. of Colorado Cables & Arches 19/24 arches Three Hinged Arch; Point Loads Three Hinged Arch; Point Loads II

Solving those four equations simultaneously we have:

15.1 140 26.25 0 0 RAy 2, 900 RAy k 0 1 01 R 80 R  29.8 k     Ax  =    Ax  = (23) 1 0 10 R 50 ⇒ R    Cy     Cy   34.9 k   80 60 0 0   R   3, 000   R      Cx Cx 50.2 k                    

We can check our results by considering the summation with respect to B from the right:

B + ΣM = 0; (20)(20) (50.2)(33.75) + (34.9)(60) = 0√ (24) ¡ z − −

Victor E. Saouma; CVEN 3525; Univ. of Colorado Cables & Arches 20/24 Examples Beam Example: BeamI

Determine the deﬂection at point C. E = 29, 000 ksi, I = 100 in4.

2 k/ft 1 k A B A C C B C 0.5 k 1.5 k 15 k 45 k 20' 10' x x 15x-x 2 Real Moment Virtual Moment

-x 2 -0.5x -x

Element x = 0 M δM AB A 15x − x2 −0.5x BC C −x2 −x

Victor E. Saouma; CVEN 3525; Univ. of Colorado Virtual Work 10/23 Examples Beam Example: BeamII

Applying the principle of virtual work, we obtain

Z L M(x) ∆C δP = δM(x) dx | {z } 0 EIz ∗ | {z } δW ∗ δU Z 20 (15x − x2) Z 10 −x2 (1)∆C = (−0.5x) dx + (−x) dx 0 EI 0 EI 2, 500 = EI (2, 500) k − ft3(1, 728) in3/ ft3 ∆C = (29, 000) ksi(100) in4 = 1.49 in

Victor E. Saouma; CVEN 3525; Univ. of Colorado Virtual Work 11/23 Examples Truss; Simple Truss; SimpleI

Determine the deﬂection at node 2 for the truss.

A=5.0 in 2 each; 60 k 120 k E=10x10 3 ksi 60 120

7 -45.0 -0.50 -0.54

4 5

8 6

8 -

. 6

5 -0.56

6 2 .

6 6 -117.3 -0.56 5

-1.124 2 .

3 4 5 -83.8 . 2 0 12' 1 -1 1.574 0 .

0 0 - 1 3 37.5 52.5 0.25 0.25 0.45 0.63 1 A 2 0.5 12' 12' 75.0 105.0 0.5 1.0

(e) (e) (e) P(e)L δP P , L, A, E, δP AE Member kips kips ft in2 ksi 1 +0.25 +37.5 12 5.0 10 × 103 +22.5 × 10−4 2 +0.25 +52.5 12 5.0 10 × 103 +31.5 × 10−4 3 -0.56 -83.8 13.42 5.0 10 × 103 +125.9 × 10−4 4 +0.56 +16.8 13.42 5.0 10 × 103 +25.3 × 10−4 5 +0.56 -16.8 13.42 5.0 10 × 103 −25.3 × 10−4 6 -0.56 -117.3 13.42 5.0 10 × 103 +176.6 × 10−4 7 -0.50 -45.0 12 5.0 10 × 103 +54.0 × 10−4 +410.5 × 10−4

Victor E. Saouma; CVEN 3525; Univ. of Colorado Virtual Work 18/23 Examples Truss; Simple Truss; SimpleII

The deﬂection is thus given by

7 X (e) PL δP∆ = δP AE 1 ∆ = (410.5 × 10−4)(12 in/ ft) = 0.493 in

Victor E. Saouma; CVEN 3525; Univ. of Colorado Virtual Work 19/23 Examples Truss with initial camber Truss with initial camberI

It is desired to provide 3 in. of camber at the center of the truss shown below

2 3

1 36'

1 k

6 @ 27' by fabricating the endposts and top chord members additionally long. How much should the length of each endpost and each panel of the top chord be increased? Assume that each endpost and each section of top chord is increased 0.1 in.

(e) (e) Member δPint ∆L δPint ∆L 1 +0.625 +0.1 +0.0625 2 +0.750 +0.1 +0.0750 3 +1.125 +0.1 +0.1125 +0.2500

Victor E. Saouma; CVEN 3525; Univ. of Colorado Virtual Work 22/23 Examples Truss with initial camber Truss with initial camberII

Thus, (2)(0.250) = 0.50 in Since the structure is linear and elastic, the required increase of length for each section will be 3.0 (0.1) = 0.60 in 0.50 If we use the practical value of 0.625 in., the theoretical camber will be (6.25)(0.50) = 3.125 in 0.1

Victor E. Saouma; CVEN 3525; Univ. of Colorado Virtual Work 23/23 Solve for the vertical displacement at C of the following structure.

Aluminum rod, A =1 in27 ,E =10 psi D

Aluminum Beam, A =10 in247 , I = 171 in ,E =10 psi 3 60" 2k 4

A B C

80" 160"

Step 1. Draw free body diagram and calculate all forces due to real load and virtual load . D D x (CCW +ve)ΣM A = 0

Dx (60)−= 2(240) 0

Dy Dk= 8 2k x +

→ΣFx =0 A B C −+=DAxx0 Ax Akx = 8 80" 160" +↑ΣFy =0 Ay −+ADyy−=2 0(1)

V. Saouma; CVEN‐3525 1 D y 3 TD= (2) From Eq.(2) 5 y D Dk= 6 4 y 8k 3 T = Dkx = 8 FromEq .(1) 4 5 T T =10k A y = 4k

Likewise, we can get the reaction due to unit virtual load at point C by similar manner.

8k 4k

6k 2k 3k δP =1

A B C A B C 8k 4k 80" 160" 80" 160” 160" 4k 2k

Free Body Diagram due to real load Free Body Diagram due to virtual load

V. Saouma; CVEN‐3525 2 Step 2. Determine forces in cable and beam for each elements due to real load and virtual load

(i) Cable. Tension is constant along the cable length.

Element x=0 P δP DB D 10 5

(ii) Beam. δP =1 6k 2k 3k A 8k B C A 4k B C 8k 4k 80" 160" 80" 160" 4k 2k

Free Body Diagram due to real load Free Body Diagram due to virtual load

Element x=0 P δP M δM AB A ‐8 ‐4 ‐4x ‐2x BC B 0 0 2x‐320 x‐160

V. Saouma; CVEN‐3525 3 From (i) and (ii) PM Δ=δδPPdxMdx δ + C ∫∫ LLAE EI δ w* δu* 100 (5)(10)80 (−− 4)( 8) Δ=(1) dx + dx C ∫∫ 00AEcb AE 80 (−− 2xx )( 4 )160 ( x − 160)(2 x − 320) ++∫∫ dx 00EIbbEI 100 (5)(10) 80 (−− 4)( 8) =+dx dx ∫∫27 27 00(1in )(10 psi ) (10 in )(10 psi ) 80 (2− x)(−−− 4xxx )160 ( 160)(2 320) + + dx ∫∫7474 00(10psiin )(171 ) (10 piis )(171n )

Δ=+++C 0.0005 0.0000256 0.0007984 0.001597 =+0.0005256 0 .0023954 Axial Contribution Flexure Contribuion t

Δ=C 0.0029in ( ↓) V. Saouma; CVEN‐3525 4 For the truss shown below: a. Determine the horizontal deflection at C, Assume AE constant.

600 lb C

1500 lb B D

A E 8'

V. Saouma; CVEN‐3525 10 (a). Determine the horizontal deflection at C. Assume AE constant.

600 lb C δ P = 1 C

4 2 4 8' 2 8'

1500 lb B 3 D B 3 D

6 8' 1 6 8' 1 5 5 A E A E 8' 8'

Real load Virtual load

Member δP(lb) P(lb) L(in) AE δP.(PL/AE) 1 2 1,200 96 AE 230,400/AE 2 2 1,200 96 AE 230,400/AE 3 0 ‐1,500 48 AE 0 4 ‐2.24 ‐1,341.64 107.33 AE 322,560/AE 5 0 1,677.05 107.33 AE 0 6 ‐2.24 ‐3,018.69 107.33 AE 725,759.6/AE

1,509,119lb− in 6 ∑= 1.51×− 10 lb in AE Δ=Ch ()→ AE V. Saouma; CVEN‐3525 11 (b). Remove the loads and determine the horizontal displacement of C if members AB and BC experience a temperature increase ΔТ=200°F, Take A = 2 in2, E = 29,000 ksi, and α=10‐6/°F

C Member L(in) α(1/°F) ΔT(°F) ΔL=αΔTL(in) δP δP.ΔL 4 2 1 96 10‐6 200 0.0192 2 0.0384

D Δ=TF200 3 ‐6 B D 2 96 10 200 0.0192 2 0.0384 6 3 48 10‐6 0 0 0 0 1 5 4 107.33 10‐6 0 0 ‐2.24 0 A E 5 107.33 10‐6 0 0 0 0

6 107.33 10‐6 0 0 ‐2.24 0 ∑ = 0.0768

ΔCinh =→0.0768 ( )

V. Saouma; CVEN‐3525 12 c. Remove the loads and determine the horizontal displacement of C if member CD is fabricated 0.5 in too short.

C Member L(in) ΔL(in) δP δP.ΔL Fabricated 0.5 in 2 4 1 96 0 2 0 too short. 2 96 0 2 0 3 B D 3 48 0 0 0 6 4 107.33 ‐0.5 ‐2.24 1.12 1 5 5 107.33 0 0 0 A E 6 107.33 0 ‐2.24 0 ∑ =1.12

ΔCinh =→1.12 ( )

V. Saouma; CVEN‐3525 13