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# APPLIED ARCH STRUCTURES ARCH 631

Texas A&M

GPA 3.82

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This 10 page Class Notes was uploaded by Eleanora Ernser III on Wednesday October 21, 2015. The Class Notes belongs to ARCH 631 at Texas A&M University taught by Staff in Fall. Since its upload, it has received 28 views. For similar materials see /class/225806/arch-631-texas-a-m-university in Architecture at Texas A&M University.

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Date Created: 10/21/15

ARCH 631 Note Set 52 F2008abn Examples Trusses and Columns Example 1 Example Problem 4l Method of joints asymmetrical roof truss shown in Figure 44 supports two verucal moi loads Delermine the support reactions at n hma than u 39 39 J lnink 39 quot her forces Summarize the results of all memth mots on a 3000b 1 FED mm B E A At J K 2 solve for support forces 3000m 5 Ax 3 120010 15 fbv ZMA3000 10quot1200quot 20quotE30quot0 152 E A 223 10 D F 20 25 1800 420039 4000 A 0 AV 2400 180m 2400 3 look for special cases 1200 lb C so CE BC and CD 420039 a I A F FsoDFAFandBFD 4 choose a joint with 2 or less unknowns E or A will work C won39t 2F 1800 EC 1 jo 6 last joint needs only one equation AB 7 EC A ZFy 2400 Alen45 0 m 2236 l 4 AB 39 240039 73394 ED EC 71800 BC 2400m1 sin45 180039 2400 20 20 ZF 4400 473394560545 0 F 50 4025 quot 0 ED 4025D 3600M X Z H 2236 2236 2236 need 30 AB AF or DF which leaves joints B D amp A F won t work 5 choose ajoint with 2 or less unknowns B D orAwilI work F won t 1200quot D 8D 5 4200 BDsm45 O BD SIMS 1539 350m 3 ZFX 460039 DF 1697 Cos45 0 3 DP 240039quot AF C A I 551800 ID Ay2400 lb ARCH 631 Note Set 52 FZOOSabn 4k Example 2 B1 c 2 3 Example Problem 43 Method of Sections 4quot 12 A 64foot parallel chord truss Figure 430 supports hori A E zontal and vertical loads as shown Using the method of H G F sections determine the member forces BC HO and GD 4k 3k 4 18 16 1639 1 look for sections 2 F B D 4k 5 c n E H 39 G F AX I 4k Ian I i6quot 539 539 1 1539 E 3 solve for support forces ZFX AX4K 0 AX 4 25 Ay 4 4k 3k E0 ZMA 4k 12 4 16 4k 32 3k 48 E64 r 0 384Mt k k E 64 andsub A 4 draw section 4k 3 B e 1M B 4kltr H I 4k HG 16 5k IE 1 5 look for intersection for summing moments B or G 6 write equilibrium equations kvii 2MB HG12ft 5k16 4k 12 0 HG121 21067k ZMG 4 16 5quot 32 4 12 BC12quot 0 144H C 12 ZFy 5 4 BG0 BG 467quot 7 repeat with other section 2 D L 51 12 GF 7 F 6k 3k GD 0 GD 5 Z y 20 16 6k 12 20 ARCH 631 NuteSel s 2 anuaabn Example 3 0 Q Erased Column quotquot COLUMNS MEMBER quotMBER D 2 llquot Mmmus a Elashmly E 1x 10E Izlm d3m 1 39 Cmsmng Stress Fr 2m IMn 2nmm LY L IZ BRACED COLUMNS m hudllng mm mm m necked Bracing Level V mm A L P 39r p DquLVa oa COLUMNS MEMBER YIMSER Modulus 0f Ejasilgvly a Iann39mmr ulzn14un Crushmg S ress I F 39 c2 an b n OUT OF PLANE BUCKLING PmU Lenglh Ovemll Physmal Lengm Ll Momemot lnema It Lquot 7 39 2 39 12 39As W EL Fm T wus w39nm um a 23m Cri calBuckhng Slress 2 m L r E Member Buckles ARCH 631 NuteSel s 2 anuaabn Exam le3 conunued can a caanlumn mn COLUMNS MEMBER mm b2m Mmu usol asumly 1 an I E39 6Ko39lnm Crushingslress 12n14un L FEW Mquot IN PLANE BUCKLING L12 Lengm vemllPhys malLengm Momenlollnema IV I i I P Bracing Level 1 H y L Lynn2 r mm a Munbu sucklps 3 o a Brand Column COLUMNS MEMEER MEER quot 2 quot Modulus a1 Easmy x 1n Mn I Crusmng Stress Lammm F2Aoa W1 cmuLaumuN cnmmsumm mum nsPLAME umEcmN mnewmrsumcnw P 423 lbs Py 6086 lbs Pm LV L J2 L 3mm PM 7 ARCH 631 NuteSel s 2 anuaabn Exam le3 contmued 00 Brand Column m COLUMNS MEMeEa IIMEER w MmumsmElasmty 7 I x s1smwm aJm Cmshmg Sness Laman 1v Fame M RR SINCEF lt P THE coLUMN ACTUALLY aucKLEs IN THE OUTOF PLANE mREcnoN Bracing Lave La Critical Buckling I k L Load for Column 39 39 3 lbs h o 9 Bracsd Column m COLUMNS MEMBER muss P quot2 Mndnlus DVEasucMy Fm I z1sx1n lbm 39 crushmg snags L nznunn 139 FADD hlln NOTE THAT IF THE MID HEIGHT BRACING WERE quotREMOVEDquot THEN THE COLUMN WOULD BUCKLE ATA LOWER LOAD IN THE OTHER DIRECTION Swwe P0 L PM we cmumn buckles as Shawn italwadof Lszm ARCH 631 Note Set 24 Z F2008abn Masonry Examgle 1 Determine the maximum lateral force H by wmd as per MSJC 8quot CMU will rm3ooo psi 32quot 2 35 each end of WI Case A neglect all reinforcement Case B consider vertical reint neglect horizontal reinl Case C consider vertical and horizontal reinf Case D design horizontal reinforcement for max shear Case A neglect all reinforcement 76 x80 I 8 moment arm 51 78139 l lexure 96 XH 39I39M 5Fr 419 9 0 Hm174 rbs10zkrps shear 813 114 psi gt 5 fm 822 psi equation for running bond and solid grout 2 d equation for allowed shear stress 120 si is maximum V Fbl 109psi763x80443qus p 3 wall area Case B consider only vertical reinforcement allowed increase of 33for Flexure neglecting l a 0 909 from table combination load with wind MI Afjd 2 x 0 79 in2 133x 24 Im 09x 72 0quot 3268 kiu HM 340 kip H J allowe ress or grade 60 Steel aver d or 2 bars lumping z us39s allowed increase of 33for combination load with wind 6 M 548psr farWgtI F 10Jf39m lt35 psi FVI33135 psi466 psi Vm bdF 763quot 7239 466psi 1000 256 log governs actual width of 8quot nominal CMU block ARCH 631 Note Set 24 z FZOOSabn Case C consider all reinforcement Flexure same as me E allowed Increase of 33for Shear combination load wth wind m VAFd 020 11232 quot24 ksi x L137Z 4A4 3239 s gnverns M I FL 39 S fandgt v 5 f 75 psi Fy 1543000 32 psi gt75 psi Fy 133 75 psi 100 psi 44kx10001bk m 100 h Si 76J7Z F p a Case D design horizontal reinforcement for maximum shear strength m 51111214100 psi763 972l000 49111 34kips okay gaverns IS A 4 de 494ki39psI33x 24ksir72aazzsin pain i v r ARCH 631 Note Set 242 F2008abn Example 2 A 12 in nominal solid brick column 16 ft high is built with brick M mortar and Grade 40 reinforcement There are 4 4 bars with 2 ties at 8 in on center The column must carry an aXial load of 63 kips Check ifthe column design is adequate f m 5300 psi SOLUTION Find the allowable axial load Pa which depends on hr r 1 lbh2bhy 115inx028933in sohr16ftx12inft33in58lt99 h 2 P 025 39A 065A F 1 f Wyn As 4 020 in 08 in2 An 115 inX 115 in 08 in2 1315 in2 F5 20 ksi 2 P 02553ksi1315m2 06508in220ksi 1 m 1527psi 14033m Find the bending stress fb fl MS M Pe where e 01115 in 12 in fb 63k1000 bk12 in115x11526in3 2982 psi s P Fa Fb S 1 or equivalently P S l b Fl warm 5300psi3 1767 psi 63k 2982psz39 1527k 1767 psi 2058 lt1 OK ARCH 631 Note Set 242 F2008abn Example 3 Determine the maximum transverse wind load w per UBC 3 12 2wythe hollowclay tile with Type S Portland cement lime mortar P 4500 psi special inspection provided 10 kipft l x1 cores for a 1foot width of wall A 144 inl S btz6 288 in3 r 0289t 347 considering solld section through mortar joint 347728 b 72 77281nr 1i 1i 744 3464m Case A with wind Weak section has been assumed to be through mortar bed joint This assumes that unit strength will be at least twice that of the mortar ratio of mortar area to clay area 1 at midheighl of wall M Q 2 u M 10 kip x J39quot w 5 x123L 2 8 ft M 338w 5390 I allowed increase of 33 for Where W ka quotd M k39P 39 quot combination load with wind tension criterion Iquot x 133 40 psi x 133 532 psi m kip w 00532 ksi w 601 psf 144 m 288 m Table 2232 Note assume F for solid units since mortar bed is full with respect to tension normal to bed joint ARCH 631 Note Set 242 F2008abn moments from moments from eccentric load P transverse load w Pe k h2 1 h quoth S Jr h2 for large l and small w critical location is at top of wall M Fe for small P and large w critical location is near midbeighl M PeZ wh18 6 39 39 casewngnglogzrzggg 39 allowed increase of 33for quot combination load with wind lt 133 Fa FD M 338 x 00601 ksf 150 353 kip in P 10 M 353 f X T47 0069 ks f F E 0123 ks FD 03357 0334500p5 750005 11 15x12 h z 518 F 02 39 1 216 r 347 0 5f 140 0 fl 970pquot 2 0254500ps 7 97005 740 347m Psl 0071 0082 0153 lt 133 0k 970 1500 CaSe B without wind at top of wall M Pe 30 kip in P M renston cntenon F 40 ps1 A S 10 Iquot 30 39 p 39 quot 5 0040 ksi 9 144in2 zssin 0 0694 ksi 00104 ksi 00348 ksi lt 0040 ksi 0k

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