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STRUCTURE
- REFERS TO A (^) SYSTEM OF (^) CONNECTED PARTS USED (^) TO
SUPPORT
ALOAD.
IMPORTANT EX'S^ :^ BUILDINGS^ , BRIDGES, O (^) TOWERS (^) (CIIL (^) ENGINEERING)
SHIP AND AIRCRAFT FRAMES, TANIS
, PRESSURE VESSELS, MECHANICAL SYSTEMS , AND^ ELECTRICAL SUPPORTING STRUCTURES=IMPORTANT WHEN DESIGNING A (^) STRUCTURE (^) MAKE SURE (^) IT SERVE A^ SPECIFIED FUNCTION FOR PUBLIC^ USE.
- (^) ENGR (^). MUST TAKE ACCOUNT SAFETY , AESTHETICS & (^) SERVICEABILITY , WHILE^ TAKING^ INTO^ CONSIDE- RATION ECONOMIC^ AND^ ENVIRONMENTAL^ CON- STRAINTS.
STRUCTURAL ELEMENTS-
TIE RODS-STRUCTURAL MEMBERS SUBJECTED
or TO A^ TENSILE^ FORCE (^). RACING (^) STRUTS)
BEAMS-USUALLY STRAIGHT HORIZONTAL (^) MEM. USED PRIMARILY TO^ CARRY VERTICAL
LOADS.
=> BEAM CROSS^ SECTIONS^ MAY ALSO^ BE^ "BUILT^ UP"^ BY^ ADDING^ PLATES
tO THEIR TOP AND^ BOTTOM. LOAD ·
simply ,^ fixed^ ,^ continuous^
(^) common (^) buildings Or saschool. · cantilevered =^ ex
: Overhang
CABLES-USUALLY FLEXIBLE^ AND CARRY^ THEIR^ LOADS^ IN^ TENSION.
CABLE-TENSION
ARCHES-COMPRESSION
FRAMES-OFTEN USED^ IN^ BUILDINGS^ AND^ ARE^ COMPOSED^ OF^ BEAMS^ &
COLUMNS THAT (^) ARE EITHER (^) PIN OR (^) FIXED CONN
INDETERMINATE
- "UNKNOWN VALUE "
- DILI MASOLVE SURFACE STRUCTURES^
- made from material having a small thickness^ compared
to its^ dimensions
STABILITY r(3m (^) UNSTABLE PERPENDICULAR r3 m STABLE (^) IF REACTIONS/FORCES ARE^ NON-CONCURRENT, ↑
UNSTABLE IF^ MEMBER REACTIONS ARE CONCURRENT
/PARALLEL OR^ SOME^ COMPONENTS^ FORM^ A^ CO-
LLAPSIBLE MECHANISM.
EX :
BX (^08) BX Stable A -- V=^3 M=^2
E 2( Ay UNSTABLE BUT
THROUGH
INSPECTION ITS STABLE
· UNSTABLE
STRUCTURAL (^) THEORY : ANALYSIS OF^ STRUCTURES 2 DETERMINE^ THE REACTIONS (^) ON THE BEAM (^). 151 M 10KYm ↓ (^) + (^) SOL'Ni ↑ 51/ --^ REG'D : Ax, (^) Ay , Ma *x -^7 5km^ E =^ O; (^2) Fy1 =^0 ; EM = (^0) ;
A /
vvvvvvvv v (^) - (^3) UNIFOR EEX- = (^0) ; MA
AyT 12 In^ Ax =^ 0 kN
[FyT=^0 ; 3
DETERMINE THE^ REACTIONS ON^ THE BEAM . ASSUME
SUPPORT &^ A^ IS^ PIN^ &^ B^ IS^ ROLLER(SMOOTH SURFACE, 500 1b/ft -B vvvvvvvv 4 ft A
7ft 3ft
PROBLEM 3 :^ STRUCTURAL THEORY
" DETERMINE THE HORIZONTAL (^) AND VERTICAL (^) COMPONENTS OF (^) REACTION
AT THE PINS A , B, AND C Of THE TWO MEMBER FRAME SHOWN IN
FIG.^ 1-32a. uniformly 8 kN^
3kN/m &
distributed ↑Xv (^) load zm
M
=- 2m - i L (^1). 5 m ↓ em- (. )
FBD :^ (Free^ Body
Diagram) Simplify^ : (^) = - (8kN)(2m) =^
- 16kNm = - -kn(2m) =-^6 knm
= - In kN/m-GkN/m +^ Bx^ (1. (^5) m) = - (^22) + Bx(1. 5) BX (1.^ 5)^ =^22 Fy](8) BX =^22 j Fx (^) = (^) (s) (^) ti Soln (^) / #B :^ sol'n
Em
= 0 , ↑ y^ =^3
2 =^0 ,
3kN At-By- (8)^ = 0
(By] (2m)^ +^ (6^ kN)^ (1m)^ =G
Rea =- 3kN-E
=
3 kN
[Ft =^ O,
- (^) By -^ 6kN +^6 =^0 Ex -^ 3kN EEEKN
SAMPLE PROBLEM
: (^) INDETERMINATE BEAMS.
DETERMINE THE^ REACTIONS ATTHE SUPPORTS A , B , AND C^
,
THEN DRAW^ THE^ SHEAR^ AND^ MOMENT^ DIAGRAMS.^ El^ IS^ CONSTANT
G M X
SOLUTION :
RA I
X
RB I
RC
EQUILIBRIUM (^) EQUATIONS : (^2) y =^04 + (^) &Mc = 0 Ra +^ Ri + (^) Rc
- 12 - [3(127] =^0 - Ra(24) - R (^) +(12) + (^) + 2 (18) +^ [3(1)7( Ra (^) + (^) RB + Rc =^48 .... (^) " 22A
Ry
36
M = Rxx +^ Ry(X
12(x
M =
Ry
m(x
2 RB(x-^ 12) Fly'
1273
- 1)" - 2(x - 3) - t(x - 12)4 +^ c^ , x (^) + 2
BOUNDARY CONDITIONS^ : 1)8x =^0 , y
- ex = 12 , y = 0 C
( (1) (^) - 2(b)3 +^ 2(12) 6 24RA +^ C =^36
@X = (^24) y
S 0 =^
Rx(24)
(12)
2 (18)
+(12)
+, (4) 4 94kx
12Rp
c
594 ----- (^4) COMBINE 3 & 4 :^ CELIMINATE
C
, ) COMBINE + ,
, 5
⑨ 24RA^
+ 4 =^36
Rx +^ RB + (^) R =^ #S
96 Ry^ HERBEC
= 594 2RA +RB =
SUBTRACT s (^) from 4
: 72RA^ +
12Rp = (^558) 96 Ra +^ 12Rg
24RA +^ c =^36
RA =^2. (^) 625K RB :^30. 75K Rc =^14. 425K 72 Ra (^) + (^) 12 R = 558----
DOUBLE INTEGRATION^ METHOD
EXAMPLE 2 :^ DETERMINE THE (^) EQUATION OF (^) THE ELASTIC CURVE^.^ El^ IS^ CONSTANT
- Ji & L^ X FBB (^7) * ↑ Mo M (^) & L (^) ↑ EMp^ = Ot By (^) Mp-Mo = 0
MB =^ Mo
Cut &X:
Boundary
Condition: 5 ↑ LT M
& x^0
: O
"Slope
↑B= Mo Shear
Internal
El
= Mox +
El
= M
L = (^0) El = (^) Mo
& x =^0 , V
= (^0) El = MoX +^4 EI^ = Mox+ YtLe &x 2
El X^ =^ Mox+^ C, x^ +^ Ce
02 =^0 (IF ASA^ ANG^ SHEAR TUA ANG^ DISPLACEMENT
El = (^) Mox ElE =^ MoX EIV = Mox ? = EXAMPLE 3 :^ THE^ BEAM IS^ SUBJECTED^ TO^ A^ LOAD P
AT ITS^ END.^ DETERMINE^ THE^ DISPLACEMENT @
C.^ El^ is^ CONSTANT^ ↓ P A , (^000) B C K 2a^ * a^ A -- - P
A
A
EMA =^ ot (^20) = est ↑ ⑪ - ↑ (3a)
- By (2a) =^0 Ay +^24 -^ p =^0 2 By = (^) P(39) (^) = By
Ay (^) +^0
za Ay^ = 1 z
INTEGRATION : Fl = M INTERNAL MOMENT^ @^ REGION^ AB^ :
OLX, 12a^
EMec
= 0ith
Ay = E ( , (^) )M M (^) = 2 REGION RC^ : 2a1X2[3a X 2 ↓
I-^ 2a - B (^) X
h
= O
# -^ exe^ -
3Pa =^ M I 2
DOUBLE INTEGRATION^ : INTEGRATION GOVERNING (^) EQUATION SE^ der-IM dx Ist
#M-
Internal moment
Mx +^ C , e Integration Modulus (^) Moment (^) of Inertia-speed or (^) rotation of of Elasticity^ material Equation for (^) truta of
-^ slope characteristic of material of how^ elastic^ itis
#tuxtre
EXAMPLE 1 :^ Equation for Displacement Each simply supported^ floor (^) joist shown or deflection. is subjected^ to^ a uniform^ design loading of 4 kN/m^
.^ determine^ the^ maximum deflection of^ the joist
4kN/m 401LN (^) Cutting section ↓ (^) ↓I^ ↓ ↓ r- 5m #I is^ constant^ . &
I
I
= zi--^ em--1By^
20 ki Ay EM^ c = (^8) t
= 0 va
At
/
20kN/m
INTEGRATION
: "Ill ,
- #^ A = M ~ A B 12 m nam #^ = 60x
LE12mt =^
12m 10kN/m (^) = E10kN/m X^ Im 1
120kN - 5x
"I
+ &d^
&Y^
! B = 40 kN (^) dy
1 At T A- (^1 ) 4 12
El
=
- taxte A+ 24580kN^ . m] EN = (^) - BOUNDARY^ CONDITIONS = 60kN(8m]^
+ M^ =^0 X =^ R^
- =^0
Ma = 480 kN^.^ M^ X = (^) R y = (^0) 3
E
Fy
= 0 +^1
Elx = (^) - = Ay-60kN = (^0) Ay = 60kN cut (^) X 10kN/m
Elo
=+
M C = 1448 ↑ 60KN EMc =^ of =^ - 60(x) +^ 10x(E) +^ m^ = (^0) EIX =^ Lox- E taxte =- 48x + 5x +^ m^ =^0 M =^ 40 X - 5x2 (^) U 3 1280 El O = (^) 2) -^ E12)"^ + 1440(127^ taz (^8648 4 ) 11280
Ch
= 2792 S
#It = x
& BOUNDAry CONDITIONS (^) : Y =^8
El
= El (^2) 3 #y = -+ X Elo^ = +^4 Cl =^1448 &* E
EN
=+ Y x y = ( ,
Elo (^) = (^) GRE
- ECR) + 1440 (1) the R Ce =^25 , 970 y = E (^) # = +^1440 EM=^ Of = 60kv^ (8m]
+ M
= O - =
cut & (^) X My = 450kN/m^ - & (^) X = 0 ↓ 10kN/M = Oft
u6kN
M (^) EF += At^ As (^) - =^ (^1) ( El n #Mc =^ of^ At =^ - = 40(x) +^ 10x(z)^ +M = 0 EN =
(^1440) X +^ =^ -^ 60x + 5x*+ M =^02590 M =^ u0x - 5x
& =^ X =^0
: -
+ 10X^
