Question

2. (55 points) Compute the horizontal deflection at point D, the vertical deflection at C and rotational displacement at C, Given: I = 120 in', E=29,000 kips/ in? 2EI 4kips 96kips. ft

Answer 1

Solusion 12 2ET 4kips 2 EI F1 12 MA Ax 96 kips-tt SuPPost Peachion Calenlotion Con sideing 2 Talina moment abont + (4x12)+ 16O -HA 0, MA= 144 kips-ft () Ax -4-0 o, Ax- 4kips will be generJed et A ventical torce i aing on No Venticol echion Smee there are Note St sncture the 1fnd the seehinal moment in main stucture as the abore in At inst we Seetons ase +the the dingranm Shoun Seution -0 Considening lower posn ton of hat seeron Co4x 12) Fog potion AB

M2 Concidoineg 2 Ma -144 + 4 = O Mm 144-4x 4hies A Seetion - Comeidering left par tion ofit FoA 144 K-ft Ponton BC ( o<x12) Considaoni rg S 4x12) u4+ Mox-0 12 4kips 144 uip-ft Seetion Considering loer pontion of it For Pos tion CD ( os x12) Considein -Mz+96-0 (12-n) 09, M2-96 96 hipsaft Now we have to fmd the horizome defleetion a vintual we will Con stnuc at pont D. To do thout Stnuctune where onla an umit hegn 12ental load is arplied at D. Then the sectiona moments we will fnd

Apraina f-o; 12 124 A D +Tuip ar section (D- Consi dea m lowen postion AB o Consideorina 2 12) 1. uip Section - Comsidesning left pon tim BcCoxs12 considesina 2 OS212 B 12 691,mr 12- seetion ()-) Considen lowen pontion. [eD (0x< )2 Comsidening 2 (12-7) mx +(12-2)x1-o 1wip o9, mu12-X

Now Can find the hosiont defteeron at D we this (L) (A), Mrmdx E1 In our case 12 2 Mam 12 Mardx 2 Et EI EI 12 12 (Ah 12 9C12-n) 96 X12 ET 12 12 I 12 o, (Ah Su-)d 52-96)x 56 dz 1 tt ET EI 12 31 O(A)T1 12 576 E1 3 /2 EI 522712 Et 6o912 13821 10368 2 304 CO12 + EL E1 EI EL .(Ph 21888 (1233 uips. in3 ups. im2 kips.ft hips in m 2 21888 29000x120 (Ah= - 10.868 mch (Positve sign mdiofes that the dinectron of detleetion k along the evoplred unit lo0nd diection

the ventical defteepron at tue we have to find Now we have to comstnuct amd to do thot am ventica we will md Stnncture at c onla. Then agm unit load ihe seefronal moments. 1uip ara-1-0 12 2' Tai ng moment A; 2 abont A D 09, mA 12 ips-fe) Seetion -0 Considering Considening 2 Tower portin [AG Cos12 my 12= O s12 Section - 12 uip ft 1uip Comsidening lelt pos bon Bc (orsa (1xx) + 12 +mx= 6 K-12 o9, m- /2 12 Lip-ft A 4u P

Sectin 3) 3 loner poston L cD (ox12) Considina Comsidesing 2 C12-1n) L D Now C1)x (A)Y dz EI 12 12 Mnmx 2EI EI 2 2 u(iwu-un) -12) L 96x F12) +96x0 ET 12 12 -1728+48r) du 8576) d 12 728 Ft 2 12 oy, (A)v 20736 + 1 34 56 + E T 3456 6912 Ef E T CA 3 26736x12 ips. t3 wips, i 20736 EI 29000x 120 kips. in 2 (A -10-29 6 inch Cnagatve sign indi cotes that diection of defleabron is the ofPosite of applpi unit lond ie. t upwand

Cwe wil arpla As fon Jotuton at moment ot C onl then agoum find the en unit We will seetional moments Considesig 2 B 1 uip-fe A; 12' 12' (3 YA seetion C Comsi des ing lorer pention. ous 12AB Considenin ? -1 -0 A 1uip-st Section Consi desing te st pon tisn 6xS12; B Considesnim 2 091, 12 1 1 uip-1t

Sechon Comsidening lower pontiom Comsi deing 2& (2-7) 0aue 0 L D S0, C1)r (0,)-Mmod Et 12 2 o or, (a)0)- 12 Mn me Mmmo daf d t ET 2EI E I 12 12 12 091 96x/0 dx ec-Gu4-47)x1 96x1 FI 0.91, e ET 12 12 뉴 Lwtn,-급 [4210 FT FT ET 611 = ?e EI 288 576 + 1구28 2016 2 hips. tt EI 2 2016 X12 ipsinz 29000% 120 isin 2 O. 0834 Jad (positive sion imdicades thot the soteution will occur 091 c alon Hhe arrlied unit iie. anticlock wise) CAnswer)