# WIND FORCES.

## Presentation on theme: "WIND FORCES."— Presentation transcript:

WIND FORCES

Wind Load is an ‘Area Load’ (measured in PSF) which loads the surface area of a structure.

SEISMIC FORCES

Seismic Load is generated by the inertia of the mass of the structure : 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 VBASE wx hx S(w h) VBASE = (Cs)(W) ( VBASE ) Fx =

Where are we going with all of this?

categories of external loading: DL, LL, W, E, S, H (fluid pressure)
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

( + ) SM1 = 0 0= -200 lb(10 ft) + RY2(15 ft) RY2(15 ft) = 2000 lb-ft
RY2 = 133 lb ( ) SFY = 0 RY1 + RY lb = 0 RY lb lb = 0 RY1 = 67 lb ( ) SFX = 0 RX1 = 0 RX1 10 ft 5 ft RY2 RY1 200 lb 0 lb 10 ft 5 ft 67 lb 133 lb

w = 880lb/ft RX1 RY1 RY2 24 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 RY2 24 ft

resultant force - equivalent total load that is a result of a distributed line load resultant force = area loading diagram resultant force = w( L) = 880 lb/ft(24ft) = 21,120 lb RX1 12 ft 12 ft RY1 RY2 24 ft

resultant force - equivalent total load that is a result of a distributed line load
resultant force = area loading diagram resultant force = w( L) = 880 lb/ft(24ft) = 21,120 lb RX1 12 ft 12 ft RY1 RY2 24 ft ( ) SM1 = 0

( + ) 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) = 880 lb/ft(24ft) = 21,120 lb RX1 12 ft 12 ft RY1 RY2 24 ft ( ) SM1 = 0 –21,120 lb(12 ft) + RY2(24 ft) = 0

( + ) 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) = 880 lb/ft(24ft) = 21,120 lb RX1 12 ft 12 ft RY1 RY2 24 ft ( ) SM1 = 0 –21,120 lb(12 ft) + RY2(24 ft) = 0 RY2(24 ft) = 253,440 lb-ft

( + ) 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) = 880 lb/ft(24ft) = 21,120 lb RX1 12 ft 12 ft RY1 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

( + ) 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) = 880 lb/ft(24ft) = 21,120 lb RX1 12 ft 12 ft RY1 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

( + ) 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) = 880 lb/ft(24ft) = 21,120 lb RX1 12 ft 12 ft RY1 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

( + ) 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) = 880 lb/ft(24ft) = 21,120 lb RX1 12 ft 12 ft RY1 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 RY1 + 10,560 lb – 21,120 lb = 0

( + ) 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) = 880 lb/ft(24ft) = 21,120 lb RX1 12 ft 12 ft RY1 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 RY1 + 10,560 lb – 21,120 lb = 0 RY1 = 10,560 lb

( + ) 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) = 880 lb/ft(24ft) = 21,120 lb RX1 12 ft 12 ft RY1 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 RY1 + 10,560 lb – 21,120 lb = 0 RY1 = 10,560 lb ( ) SFX = 0 RX1 = 0

( + ) 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) = 880 lb/ft(24ft) = 21,120 lb RX1 12 ft 12 ft RY1 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 RY1 + 10,560 lb – 21,120 lb = 0 RY1 = 10,560 lb ( ) SFX = 0 RX1 = 0 w = 880 lb/ft 0 lb 24 ft 10,560 lb 10,560 lb

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 P V M

INTERNAL FORCES Axial (P) Shear (V) Moment (M)

P + + M V +

- P - M V -

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 +20k -10k -10k -20k -10k +20k 20k compression

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

P V Area of Loading Diagram +10.56k -0.88k/ft * 24ft = -21.12k
w = lb/ft 0 lb 10,560 lb 24 ft 10,560 lb P Area of Loading Diagram -0.88k/ft * 24ft = k 10.56k k = k +10.56k -880 plf = slope k V k -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

P V M Area of Loading Diagram +10.56k -0.88k/ft * 24ft = -21.12k
w = lb/ft 0 lb 10,560 lb 24 ft 10,560 lb P Area of Loading Diagram -0.88k/ft * 24ft = k 10.56k k = k +10.56k -880 plf = slope k V k zero slope -10.56k 63.36k’ k-ft Slope initial = k k-ft pos. slope neg. slope Area of Shear Diagram (10.56k )(12ft ) 0.5 = k-ft M (-10.56k)(12ft)(0.5) = k-ft

Wind Load spans to each level
W2 = 30 PSF SPAN 10 ft 1/2 + 1/2 LOAD SPAN 10 ft W1 = 20 PSF 1/2 LOAD

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 is generated by the inertia of the mass of the structure : 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 question : FX VBASE wx hx S(w h) VBASE = (Cs)(W) ( VBASE ) Fx =

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) (hx) S (w h) Fx =

Load Flow to Lateral Resisting System :
Distribution based on Relative Rigidity Assume Relative Rigidity : Single Bay MF : Rel Rigidity = 1 2 - Bay MF : Rel Rigidity = 2 3 - Bay MF : Rel Rigidity = 3

Distribution based on Relative Rigidity :
SR = = 4 Px = ( Rx / SR ) (Ptotal) PMF1 = 1/4 Ptotal