Transcript Document

WIND FORCES
wind load
Wind Load is an ‘Area Load’
(measured in PSF) which loads the
surface area of a structure.
SEISMIC FORCES
seismic load
Seismic Load is generated
by the inertia of the mass of
the structure : VBASE
VBASE = (Cs)(W)
( VBASE )
Redistributed (based on
relative height and weight) to
each level as a ‘Point Load’
at the center of mass of
each floor level: FX
Fx = VBASE wx hx
S(w h)
Where are we going with all of this?
global stability & load flow (Project 1) tension, compression, continuity
equilibrium: forces act on rigid bodies,
and they don’t noticeably move
boundary conditions: fixed, pin, roller
idealize member supports & connections
external forces: are applied to beams & columns
as concentrated & uniform loads
categories of external loading: DL, LL, W, E, S, H (fluid pressure)
internal forces: axial, shear, bending/flexure
internal stresses: tension, compression, shear, bending stress,
stability, slenderness, and allowable compression stress
member sizing for flexure
member sizing for combined flexure and axial stress (Proj. 2)
Trusses (Proj. 3)
EXTERNAL FORCES
200 lb
( + ) SM1 = 0
0= -200 lb(10 ft) + RY2(15 ft)
RY2(15 ft) = 2000 lb-ft
RX1
RY1
10 ft
5 ft
RY2
RY2 = 133 lb
(
+) SFY = 0
RY1 + RY2 - 200 lb = 0
RY1 + 133 lb - 200 lb = 0
RY1 = 67 lb
200 lb
(
+) SFX = 0
RX1 = 0
0 lb
67 lb
10 ft
5 ft
133 lb
w = 880lb/ft
RX1
RY1
24 ft
RY2
resultant force - equivalent total load that is a result of a distributed line load
resultant force = area loading diagram
resultant force = w( L)
RX1
RY1
24 ft
RY2
resultant force - equivalent total load that is a result of a distributed line load
resultant force = area loading diagram
resultant force = w( L)
RX1
RY1
= 880 lb/ft(24ft) = 21,120 lb
12 ft
12 ft
24 ft
RY2
resultant force - equivalent total load that is a result of a distributed line load
resultant force = area loading diagram
resultant force = w( L)
RX1
RY1
= 880 lb/ft(24ft) = 21,120 lb
12 ft
12 ft
24 ft
RY2
( + ) SM1 = 0
resultant force - equivalent total load that is a result of a distributed line load
resultant force = area loading diagram
resultant force = w( L)
RX1
RY1
= 880 lb/ft(24ft) = 21,120 lb
12 ft
12 ft
24 ft
RY2
( + ) SM1 = 0
–21,120 lb(12 ft) + RY2(24 ft) = 0
resultant force - equivalent total load that is a result of a distributed line load
resultant force = area loading diagram
resultant force = w( L)
RX1
RY1
= 880 lb/ft(24ft) = 21,120 lb
12 ft
12 ft
24 ft
RY2
( + ) SM1 = 0
–21,120 lb(12 ft) + RY2(24 ft) = 0
RY2(24 ft) = 253,440 lb-ft
resultant force - equivalent total load that is a result of a distributed line load
resultant force = area loading diagram
resultant force = w( L)
RX1
RY1
= 880 lb/ft(24ft) = 21,120 lb
12 ft
12 ft
24 ft
RY2
( + ) SM1 = 0
–21,120 lb(12 ft) + RY2(24 ft) = 0
RY2(24 ft) = 253,440 lb-ft
RY2 = 10,560 lb
resultant force - equivalent total load that is a result of a distributed line load
resultant force = area loading diagram
resultant force = w( L)
RX1
RY1
= 880 lb/ft(24ft) = 21,120 lb
12 ft
12 ft
24 ft
RY2
( + ) SM1 = 0
–21,120 lb(12 ft) + RY2(24 ft) = 0
RY2(24 ft) = 253,440 lb-ft
RY2 = 10,560 lb
(
+) SFY = 0
resultant force - equivalent total load that is a result of a distributed line load
resultant force = area loading diagram
resultant force = w( L)
RX1
RY1
= 880 lb/ft(24ft) = 21,120 lb
12 ft
12 ft
24 ft
RY2
( + ) SM1 = 0
–21,120 lb(12 ft) + RY2(24 ft) = 0
RY2(24 ft) = 253,440 lb-ft
RY2 = 10,560 lb
(
+) SFY = 0
RY1 + RY2 – 21,120 lb = 0
resultant force - equivalent total load that is a result of a distributed line load
resultant force = area loading diagram
resultant force = w( L)
RX1
RY1
= 880 lb/ft(24ft) = 21,120 lb
12 ft
12 ft
24 ft
RY2
( + ) SM1 = 0
–21,120 lb(12 ft) + RY2(24 ft) = 0
RY2(24 ft) = 253,440 lb-ft
RY2 = 10,560 lb
(
+) SFY = 0
RY1 + RY2 – 21,120 lb = 0
RY1 + 10,560 lb – 21,120 lb = 0
resultant force - equivalent total load that is a result of a distributed line load
resultant force = area loading diagram
resultant force = w( L)
RX1
RY1
= 880 lb/ft(24ft) = 21,120 lb
12 ft
12 ft
24 ft
RY2
( + ) SM1 = 0
–21,120 lb(12 ft) + RY2(24 ft) = 0
RY2(24 ft) = 253,440 lb-ft
RY2 = 10,560 lb
(
+) SFY = 0
RY1 + RY2 – 21,120 lb = 0
RY1 + 10,560 lb – 21,120 lb = 0
RY1 = 10,560 lb
resultant force - equivalent total load that is a result of a distributed line load
resultant force = area loading diagram
resultant force = w( L)
RX1
RY1
= 880 lb/ft(24ft) = 21,120 lb
12 ft
12 ft
24 ft
RY2
( + ) SM1 = 0
–21,120 lb(12 ft) + RY2(24 ft) = 0
RY2(24 ft) = 253,440 lb-ft
RY2 = 10,560 lb
(
+) SFY = 0
RY1 + RY2 – 21,120 lb = 0
RY1 + 10,560 lb – 21,120 lb = 0
RY1 = 10,560 lb
(
+) SFX = 0
RX1 = 0
resultant force - equivalent total load that is a result of a distributed line load
resultant force = area loading diagram
resultant force = w( L)
RX1
RY1
= 880 lb/ft(24ft) = 21,120 lb
12 ft
12 ft
RY2
24 ft
( + ) SM1 = 0
–21,120 lb(12 ft) + RY2(24 ft) = 0
RY2(24 ft) = 253,440 lb-ft
RY2 = 10,560 lb
(
+) SFY = 0
RY1 + RY2 – 21,120 lb = 0
w = 880 lb/ft
RY1 + 10,560 lb – 21,120 lb = 0
RY1 = 10,560 lb
0 lb
10,560 lb
24 ft
10,560 lb
(
+) SFX = 0
RX1 = 0
SIGN CONVENTIONS
(often confusing, can be frustrating)
External – for solving reactions
(Applied Loading & Support Reactions)
+ X pos. to right
- X to left neg.
+ Y pos. up
- Y down neg
+ Rotation pos. counter-clockwise
- CW rot. neg.
Internal – for P V M diagrams
(Axial, Shear, and Moment inside members)
Axial Tension (elongation) pos. | Axial Compression (shortening) neg.
Shear Force (spin clockwise) pos. | Shear Force (spin CCW) neg.
Bending Moment (smiling) pos. | Bending Moment (frowning) neg.
STRUCTURAL ANALYSIS:
INTERNAL FORCES
PVM
INTERNAL FORCES
Axial (P)
Shear (V)
Moment (M)
P
+
+
+
V
M
P
-
-
-
V
M
RULES FOR CREATING P DIAGRAMS
1. concentrated axial load | reaction = jump in the axial diagram
2. value of distributed axial loading = slope of axial diagram
3. sum of distributed axial loading = change in axial diagram
-10k
-10k
-10k
-20k
0
-10k
-10k
+20k
+20k
-20k
compression
0
+
RULES FOR CREATING V M DIAGRAMS (3/6)
1. a concentrated load | reaction = a jump in the shear diagram
2. the value of loading diagram = the slope of shear diagram
3. the area of loading diagram = the change in shear diagram
w = - 880 lb/ft
0 lb
10,560 lb
0
0
+10.56k
0
Area of Loading Diagram
-880 plf = slope
+10.56 k
V
10,560 lb
-0.88k/ft * 24ft = -21.12k
0
+10.56 k
P
24 ft
-10.56k
10.56k + -21.12k = -10.56k
RULES FOR CREATING V M DIAGRAMS, Cont. (6/6)
4. a concentrated moment = a jump in the moment diagram
5. the value of shear diagram = the slope of moment diagram
6. the area of shear diagram = the change in moment diagram
w = - 880 lb/ft
0 lb
24 ft
10,560 lb
0
0
+10.56k
-0.88k/ft * 24ft = -21.12k
0
M
+63.36 k-ft
63.36k’
zero slope
0
+10.56 k
0
+10.56 k
V
Area of Loading Diagram
-880 plf = slope
10.56k + -21.12k = -10.56k
-10.56k
-63.36 k-ft
P
10,560 lb
0
Slope initial = +10.56k
Area of Shear Diagram
(10.56k )(12ft ) 0.5 = 63.36 k-ft
(-10.56k)(12ft)(0.5) = -63.36 k-ft
Wind Loading
W2 = 30 PSF
W1 = 20 PSF
Wind Load spans to each
level
1/2 LOAD
W2 = 30 PSF
SPAN
10 ft
1/2 +
1/2 LOAD
SPAN
W1 = 20 PSF
1/2 LOAD
10 ft
Total Wind Load to roof
level
wroof= (30 PSF)(5 FT)
= 150 PLF
Total Wind Load to second floor level
wsecond= (30 PSF)(5
FT) + (20 PSF)(5
FT)
= 250 PLF
wroof= 150 PLF
wsecond= 250 PLF
seismic load
Seismic Load is
generated by the
inertia of the mass of
the structure : VBASE
VBASE = (Cs)(W)
( VBASE )
Redistributed (based on
relative height and
weight) to each level as
a ‘Point Load’ at the
center of mass of the
structure or element in
: FX
Fquestion
x =
VBASE wx hx
S(w h)
Total Seismic Loading :
VBASE = 0.3 W
W = Wroof + Wsecond
wroof
wsecond flr
W = wroof + wsecond flr
VBASE
Redistribute Total Seismic
Load to each level based on
relative height and weight
Froof
Fsecond flr
VBASE (wx)
Fx = (hx)
S (w h)
Load Flow to Lateral Resisting System :
Distribution based on Relative Rigidity
Assume Relative
Rigidity :
Single Bay MF : 2 - Bay MF :
Rel Rigidity = 1 Rel Rigidity = 2
3 - Bay MF :
Rel Rigidity = 3
Distribution based on
Relative Rigidity :
SR = 1+1+1+1 = 4
Px = ( Rx / SR ) (Ptotal)
PMF1 = 1/4 Ptotal