P1.2 w = 1.35g k +1.5q k = = 4.35kN/m 2 M = wl 2 /8 = /8 = 34.8kN.m V = wl /2 = /2 = 17.4kN

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Arnold Woods
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1 Chapter Solution P. w = =.5k/m (or.5/) US load =.5 g k +.5 q k =.5k/m = / =.5 / =.k.m (d) V = / =.5 / =.k P. w =.5g k +.5q k = =.5k/m = / =.5 / =.k.m V = / =.5 / = 7.k 5 5( ) 000 0, P. The weight of the r.. dek impoed on eah beam = = 5k/m Total UD = x5 =.k/m SS defletion w =.5 (5+.) +.5 x 5 x.0 =.k/m. = V =. 5.k. m. 7.k Chapter Solution P. US load, w =.5 g k +.5 q k = 5k/m US oment, d = W / = 5k.m The minimum plati etion modulu i Wpl f el (d) (e) W pl V = / = 07.5k The minimum hear area i m

2 Vd v f 75 P. Sine thi i a Cla o-etion pl, f W pl k.m Sine thi i a hot rolled o-etion, hear bukling an be afel ignored and the hear trength i given b the ield value of hear tre multiplied b the hear area V pl, vf = The elati itial hear tre i 5.5k t / d 77. The ield hear tre i f / 5./ Sine >>, failure b hear bukling will not our (d) Sine the applied hear fore equal the full deign hear apait, the applied hear tre equal the ield hear tre, i.e. x f Von-ie ield iteria how x z Thi beome x z xz f x f f Therefore, x = 0. Thi mean that the hear area annot reit bending tree, beaue it i full utilied reiting the hear tree. The moment apait i therefore determined for the o-etion minu the hear area. The plati etion modulu of a retangular blok of width, t, and depth D i W pl t D Therefore, the plati etion modulu of the beam hear area (hown in rror! Referene oure not found.a) i W pl nd the moment apait of the hear area Thu, the plati moment apait i pl, k.m k. m

3 (e) Sine the applied hear tre equal ¾ of the ield hear tre, i.e. x f Von-ie ield iteria beome x Therefore, x = /. Thi mean that the bending trength of the hear area ha fallen from 75/ to / due to the applied hear tre. The redution in moment apait of the hear area i (75) 750.5k.m Thu, the plati moment apait i pl, k.m P.. pl, = f W pl. = 75xx0 x0 - = 0k.m ffetive length = 5000 (not detabiliing) x xx0 0770x.55x0 / 5.k. m (d) b, k.m Chapter Solution P. a) =.0 = 0,000 (length between web member) pl, f k,z z k b, pl, k 0. FoS =. 0 >.0 therefore thi i OK 00 b) = 0.5 = 5,000 (length between web member),z z x k b, pl, k

4 . FoS =. 5 >>.0 therefore thi i ineffiient 00 ) = 0.5 = 5,000 (ditane between web member), π π k b, pl, FoS =. >>.0 therefore ver ineffiient k P. a) w =.5g.5q =.5 ( )+.5 =.k/m ul k k. 0.k.m el, f W el k.m. YS b) w l = g k q k = (5 0.) (0.+ )= 7.k/m wul.0 ) V.5k V pl, k.5kYS d) ffetive width = 500/ = 50 pl, 50 = π π =7.75k k b, pl,rd k.5k pa P. a) rea

5 pl, f leg k For loal bukling, the effetive length, =.0 = 00. Therefore, the elati itial bukling fore and atual bukling fore of eah leg are: π π b, pl,rd ,loal k 5k Sine there are leg member, the fore required to aue loal bukling i: loal b, rd 5 05k b) Firt alulate the eond moment of area of the entire prop: The effetive length of the global bukling model i 0,000. π , global 0 k 0000 (d) FoS.0 00 oment apait 77k 5. 0.k.m nd the applied moment due to the aidental idewa fore i: d P 00 00k.m The amplifiation of moment fator i: nd finall uing the interation equation, the degree of utiliation i: FoS = =.. Chapter Solution P. a) oment at the ¼ pan point qk 5

6 m xial fore at the ¼ pan point gk f f qk f f / b) Cro-etional area pl, f rea 5000 k 7 ) ln S m.0.5m π π k d) b, 0k pl, el, f k.m α /. 7/ / / b, el Sine thi i le than.0, the arh i unlikel to exhibit anti-etri bukling under US load. P. a) oment at the ¼ pan point

7 7 q k 7.5k.m 5 5 xial fore at the ¼ pan point k k f f q f f g 5.5k / b) Cro-etional area 70 ) 50 (500 7k rea f pl, ) ) 5 (50 5m ln 7 5 S 5.5m 5.0 k π π d) k 7 pl, b, 5k.m f el,. 5.5/ / α el / b, / FoS

8 Chapter 5 Solution P5. id-pan moment = 0x0 0x7 50k. m id-pan hear fore = 0k Support moment = 0x7 0k. m Shear at HS of firt upport = -0x7 = -0k Support reation = x0/ = 00k Shear at RHS of firt upport = = 00k 0 = applied / 0.0 Calulate the bukling oeffiient: k σ σ 0.5 k b/a π t ( ν ) b / 0.5 0/ pa applied 500 (00 0) / 0.0 k / 0 0 b (d) k 5. a 0 k / pa / 00 0 τ σ 0 5/

9 Sine i taken a the lowet value of and, the deign hear tre i /. applied / () pa The maximum applied ompreion tre in the web i applied d / 0.0 d F (e) k σ σ applied P / b/a 0.5 (00 5)/ / / Fail P5. The area of the tiffened plate i t The ditane to the entroid of the tiffened plate from the top of the main plate i (0 75) Uing the parallel axi theorem, the eond moment of area of the tiffened plate i t 0000 (7.5) 050(0 757.) t = Bukling of the whole plate between the diaphragm: The elati itial bukling tre i σ π t t π nd the deign tre i / σ 5..5/ nd / 00

10 (d) k= / / 50 (e) 5 / Sine a = b, the hear bukling oeffiient i k t b a t t b b k 5. a k The elati itial hear tre i τ k t π t ( υ ) b (f) The ield hear tre i 5.0 /./ The deign hear tre i the leer of and, therefore 5.0 /. Thi how that hear tree will aue ielding before hear bukling. d d (g) τ d τ 75/ d Chapter Solution P. eer of paing (.5m) or x/ = x/=.0m P. a) ffetive width i the beam paing or x0000/ = 500, whihever i le US load under the wet weight of the onete w.5(0.5..5) k/m 0

11 00 5k.m d 50 0 / 500 Sine the maximum tre (±/ ) i le than (75 / ), beam remain elati. beff Taking moment of area about the top of the lab x x =.0 from the top of the lab The eond moment of area of the ompoite etion i.7 0 omp.70( 0) (0 0) 0 The ontrution load of 0.75k/m i not ating on the beam in SS. The load on the beam i w k/m d.50 0./ 500 Total load in SS w.5 5.0k/m k.m Stre at the top of the lab = omp / Bottom of the lab = omp (-0) 0.5/ (-0) Top of the teel etion =../ (00) Bottom of the teel etion =. 77.5/ d) Defletion = P. Working load under the wet weight of the onete w k/m

12 . 0.k.m d / Defletion b eff = x000/ = 000 Width of top flange of the tranformed etion = Taking moment of area about the top of the lab:.0 beff x (50 000) x =. from the top of the lab The eond moment of area of the ompoite etion (uing the parallel axi theorem) i omp UD =.5x = k/m and point load = 0k 0 5k.m Stre at the top of the lab = omp ( )../ Bottom of the lab = omp (50.)./ (50.) Top of the teel etion = 5 -/ (50 00.) Bottom of the teel etion = 5 5/ 050 (d) Full defletion under working load = wet weight defletion + defletion under impoed load = P Working load under the wet weight of the onete w.05.k/m 0.

13 . k.m d 0 0 7/ b eff = 000/ = 50 Width of top flange = beff Taking moment of area about the top of the lab: 7. 0 x = x ( ) 7. 0 omp 7. 0(.7 0) UD = = k/m (0 0.7) 700 k.m Stre at the top of the lab = omp ( ).7./ Bottom of the lab = omp (0.7).5/ Top of the teel etion = omp 0 (0.7) / 700 Bottom of the teel etion = 0 (0 0.7) 7 / 700 (d) Full defletion under working load = wet weight defletion + defletion under impoed load = (e) b eff = 000/ = 50 < beam paing (000) therefore OK T k 00 0 X z = d teel X d lab x 0 0.5

14 k.m P.5 Weight of the onete and teel w =.5 [ ] = 7.k/m 7.0 0k.m d / 500 ffetive width of the ompoite beam, b eff = x0000/ = 500 b eff = 500x7/0000 =.7 Taking moment of area about the top of the lab: x(70 700) x = from the top of the lab The eond moment of area of the ompoite etion (uing the parallel axi theorem): omp ( 0) 500 US impoed load =.5.5 =.5k/m US Point load =.5 0 = 5k/m 700(0 0) 0 P k.m Stre at the top of the lab = omp 7.50 ( ) 5./ Bottom of the lab = omp 7.50 (0).7/ Top of the teel etion =.50 (0) 5 5/ 0.50 (0 0) Bottom of the teel etion = 5 7/ d) US deign hear fore =.5.5 5k t the end of the beam, Shear flow, q = V' (00.5) / 0 Connetion Strength 700 Spaing Shear Flow

15 Chapter 7 Solution P7. Ultimate limit tate UD, w = =.k/m US moment,..5 5k. m US hear fore, F = / =..5/ =k ffetive depth, d = / = 57 u = 0.7f k bd = = 55. k.m Sine < u, deign the beam a a ingl reinfored beam f bd k o k k / x0.0/. z d 0.5 o 0.7f k 50 z (d) oment at a ditane a from the end of the beam, a = f a = 50% of the mid-pan moment = 7.5k wa V.a.a a Root = 7.,.5 The 50% mid-pan moment i loated.5 m from the end of the beam. (e) The point of urtailment =.5 m anhorage length nhorage length = rebar diameter = 5 = 5 Therefore, the point of urtailment for 5 bar = 50 5 = 5 P7. d = /= 0 d ' 5 0/ = 7 ' 57 f T m k 7x500 x0.5 k C f m k 57x500 x0.5 5.k 5

16 C 0.5f b k 0.x T = C + C 0.5x0 x x 5.5x0.5 x = 7.5, limiting value for x = 0.5d = 5.5 therefore PSS C 0.5x x7.5x0.5 5.k Taking moment about the tenion teel: ' = C (d d ) C (d 0.x) 5.5(0 ) d 55.k. m 57 V d 0k.m wa a = Vd.a 5.( ) k.m f a = 50% of the mid-pan moment: wa midpan moment V.a 0 5 a 55. 0a 0 a 0a The root of thi quadrati are, a =.05 and 5.75 The 50% mid-pan moment i loated.05m from the end of the beam. The point of urtailment =.05m anhorage length nhorage length = 7 x rebar diameter = 0x = 0 Therefore, the point of urtailment = 05 0 = 5 from the entre of the upport. P7. a) w =.5x5x0.75x x0 =.k/m P =.5x0 +.5x5 = k d = V d P P.x x.k. m.x.5k b) d = / = 7 ) u = 0.7f k bd = 0.7x5x75x7 x0 - = 7. k.m Sine < u, deign the beam a a ingl reinfored beam

17 f bd.x0 5x75x7 k o k k / x0./.. z d 0.5 o 0.7f k.x0 z 0.7x500x. d) rea of a pair of rebar = v = 0 V 0.7f k d w w 0.7x500x7x0 0.7f kd.7 V.5x0 w 0 e) V 0.7f kd 0.7x500x7x x0 k 00 P7. Uing imilar triangle.5 x.55. The ditane from the end of the beam i 55. Point : Cruhing From q. (7.5) C nd from q. (7.) and (7.7) / 0 7.k 7.k nd from rror! Referene oure not found.) 75k Coordinate: (0, 75) Point : Balane From q. rror! Referene oure not found.7.5), the neutral axi depth x 0.d From q. rror! Referene oure not found.) C 0.57f k b0.x k The train in the ompreion teel i greater than the ield train (0.00) when alulated uing q. (7.5), therefore from q. rror! Referene oure not found.(7.) 7.k From q. rror! Referene oure not found.) C k From q. (7.), rror! Referene oure not found.) and rror! Referene oure not found.) z h / d' 0.0/ m 7

18 z z (d h / ) (0.70.0/ ) 0.7m h / 0.x 0./ m From q. rror! Referene oure not found.) C z z z ( 7.) ( 0.7) 50k.m Point : Pure moment From q. rror! Referene oure not found.) (d d') ( 7.) ( ) k.m Chapter Solution P. Z T = Z B = bd / = / = 0 = = 0 0 w = =.k/m =.. / =.k.m P Pe 000 T Z Z 00 T B T P Pe B..7.. / Z Z B / = -. P. dl = 5 0.5h / = 0.h k.m l = 0.h + ( )/ = (0.h + ) k.m Z Z T B 0.(0.h ) 0.0.h 0.h 0.7.h (0.h ) 0.0.h 0.h 0.7.h For a retangular etion, Z B = Z T = bh / = 0.5h / = 0.0h m Therefore: h.h. = 0 h.h. = 0 Whih olve to h = 0.5m and 0.5m, therefore the minimum depth = 0.5m

19 w dl = = 7.5k/m dl = 7.5 / =.75k.m w l = 7.5+ =.5k/m l =.5 / = 5.75k.m = = 0.m Z = bh = e P ZT min.t = 0.0m Z P P 5 dl T 5 P P 0k.5 e P ZB max.t Z P dl B 5 P P k 7.5 e P ZT max.l Z l T P P P P k. e P ZB min.l Z P l B P P k.7 k P k P. d = / = 5.7 k.m = bh / = / = 7.x0

20 = 700x500 = 50x0.end P /.midpan P Pe de /.average / / 000 P.5 = BD / = 0. 0 Uplift due to tendon fore: 5ααP( δ mid 0.5e end ) αpeend (upward) Gravit load = = 5.5k/m (= 5.5/) (downward) Total defletion = = -. (upward) Uplift due to tendon fore: 5ββP( δ mid 0.5e ) end Gravit load = 5.5+ =.5k/m βpeend (upward) 5,eff (downward) Total defletion = -.+. =. (downward) P. Dead weight of beam = 0. 5 = k/m oment = / = k.m entriit = = 0 0

21 Z T =. 0 /00 =.x0 Z B =. 0 /00 =.7x0 T B P P Pe Z T B Z T Pe Z Z B / w = 5 ( ) = 7.k/m = 7. / = 5.7k.m 0.55 / T 5.7x0 T / Z.x0 B / Z B oment = / = 5k.m = BD / = 0 00 / =. 0 Stre at top of bridge dek = /.0 Stre at top of PSC etion = σ T 50 ( ). 0./.0 Stre at bottom of bridge = B /.0 P.7 f T k ffetive depth, d = 0-0 = 0 C 0.5f b m k 0.x From horizontal equilibrium, C = T.k Tm. 0.5 x. < 0.5d therefore OK 0.5f 0.b k ever arm depth, z = d-0.x = = 70.7 oment apait i = T.z = =.k.m Chek rebar ha ielded: Strain in the teel when the onete uhing train i reahed: 0.005d x x Pretrain in teel =

22 Total train = 0.05 P m = =.m Compreion tre at the entroid of the etion P 0.0. / t the entroid of the etion: b o = = 0.m = / =.m (.0 0.) 0. (.05 0.) 0.7m. bo V ' t t (.0).(.0) k Chapter 0 Solution P0. The da tenile trength of the onete i f t.da f k / 0.0/ ft.da / m width of lab f 500 k Thi equate to 00/ /m top and bottom. Provide 0 diameter 75 entre /m top and bottom in both diretion. The effetive reinforement ratio i p,eff S. 0.φ. p,eff The free hrinkage train due to the hange in temperature i free T 0 (00) 00 The lab i prevented from hrinking b perimeter wall whih urround the lab. uming thee are rigid, the retraint fator, R, i.0, therefore the retrained hrinkage train i r 0.75free R nd the ak width due to earl thermal aking i w S r

23 The uer winter hange in temperature i o C. The free hrinkage train due to the hange in temperature i free 0 0 The retrained hrinkage train i r 0.75free R and (d) w The free hrinkage train due to hrinkage after ear i 0. r 0.75free R Thu, the total ak width i w P0. The total hange in temperature between the top of the lab and the mid-point of the lab = 5/ = 7.5 o C. The indued train at the outer edge of the lab i T 0 The urvature indued i R 50.0 Thi urvature and the aoiate movement ould be indued b equal end moment, therefore the end lope and mid-pan defletion are Sine urvature, Therefore nd R R R.0 5. radian uming that the point of end-rotation our at the top orner of the lab, the upport idewa movement i h Chapter Solution P.

24 Seond moment of area of the etion: = ) / ( k P V paing 000 paing onnetortrength hear q 7./ ) 0.5 ( V q paing 7.5 z.5 7 z =.7m P. m.5k , el k.m ,z ed m 0.5k/ g m w 0.0k.m z k ,z 0k uh.7k. 0 uh, b (d) k ,. 75/5 /,.0 el,z z z el, b,,

25 FoS 0.0. (e) z. 5 / 75/.,z z.7.5 z.0k.m P z.0 P.k.5 P. P V P.5 (.5.5).5 (.5.5) 7k 0k. m timber 000 Width of timber (web) timber width teel = ( (00.) (000) )/ Stre on outide of teel plate 5 / 0 (d) timber (00 5) Stre at top of timber./ teel t the ewed joint: 00/ 5/ hear onnetortrength 500 Shear flow, q paing paing (e) n addition, V / 0 q Rearranging give the maximum permiible paing = 5 7 z. 7 5

26 z =.5m P. The eond moment of area of the timber and teel are t The tiffne () of the timber and teel are t t nd the total tiffne of the ompound etion i.50 The SS load i w l....k/ m nd the mid-pan defletion i The US load i w ul k/ m The mid-pan moment i (d). 5.5k/ m The moment i hared between the timber and teel etion. The proportion eah etion arrie i dependent on the relative tiffne, therefore the moment in the timber etion i t tt k.m.5 nd the moment in the teel i k.m.5 The top and bottom tree in the timber are t. 0 ( )00. /.0 nd the teel tree are.0 ( )0 7.7 / 5. 0

Software Verification

Software Verification Sotware Veriiation EXAMPLE CSA A23.3-04 RC-BM-00 Flexural and Shear Beam Deign PROBLEM DESCRIPTION The purpoe o thi example i to veri lab lexural deign in. The load level i adjuted or the ae orreponding