WIND FORCES. wind load Wind Load is an ‘Area Load’ (measured in PSF) which loads the surface...

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Transcript of WIND FORCES. wind load Wind Load is an ‘Area Load’ (measured in PSF) which loads the surface...

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

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

(w h)

( VBASE )

(Cs)(W)VBASE =

Fx =

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

( + ) M1 = 00= -200 lb(10 ft) + RY2(15 ft)

RY2(15 ft) = 2000 lb-ft

RY2 = 133 lb

( +) FY = 0

RY1 + RY2 - 200 lb = 0

RY1 + 133 lb - 200 lb = 0

RY1 = 67 lb

( +) FX = 0

RX1 = 0

RY1

200 lb

RY2

RX1

10 ft 5 ft

67 lb

200 lb

0 lb

10 ft 5 ft 133 lb

= 880lb/ft

RY224 ft RY1

RX1

RY224 ft

resultant force - equivalent total load that is a result of a distributed line load

resultant force = area loading diagram

resultant force = ( L)

RY1

RX1

RY1 RY2

RX1

24 ft

resultant force - equivalent total load that is a result of a distributed line load

resultant force = area loading diagram

resultant force = ( L)

= 880 lb/ft(24ft) = 21,120 lb12 ft 12 ft

( + ) M1 = 0 RY1 RY2

RX1

24 ft

resultant force - equivalent total load that is a result of a distributed line load

resultant force = area loading diagram

resultant force = ( L)

= 880 lb/ft(24ft) = 21,120 lb12 ft 12 ft

( + ) M1 = 0–21,120 lb(12 ft) + RY2(24 ft) = 0

RY1 RY2

RX1

24 ft

resultant force - equivalent total load that is a result of a distributed line load

resultant force = area loading diagram

resultant force = ( L)

= 880 lb/ft(24ft) = 21,120 lb12 ft 12 ft

( + ) M1 = 0–21,120 lb(12 ft) + RY2(24 ft) = 0

RY2(24 ft) = 253,440 lb-ft

RY1 RY2

RX1

24 ft

resultant force - equivalent total load that is a result of a distributed line load

resultant force = area loading diagram

resultant force = ( L)

= 880 lb/ft(24ft) = 21,120 lb12 ft 12 ft

( + ) M1 = 0–21,120 lb(12 ft) + RY2(24 ft) = 0

RY2(24 ft) = 253,440 lb-ft

RY2 = 10,560 lb

RY1 RY2

RX1

24 ft

resultant force - equivalent total load that is a result of a distributed line load

resultant force = area loading diagram

resultant force = ( L)

= 880 lb/ft(24ft) = 21,120 lb12 ft 12 ft

( + ) M1 = 0–21,120 lb(12 ft) + RY2(24 ft) = 0

RY2(24 ft) = 253,440 lb-ft

RY2 = 10,560 lb

( +) FY = 0

RY1 RY2

RX1

24 ft

resultant force - equivalent total load that is a result of a distributed line load

resultant force = area loading diagram

resultant force = ( L)

= 880 lb/ft(24ft) = 21,120 lb12 ft 12 ft

( + ) M1 = 0–21,120 lb(12 ft) + RY2(24 ft) = 0

RY2(24 ft) = 253,440 lb-ft

RY2 = 10,560 lb

( +) FY = 0

RY1 + RY2 – 21,120 lb = 0

RY1 RY2

RX1

24 ft

resultant force - equivalent total load that is a result of a distributed line load

resultant force = area loading diagram

resultant force = ( L)

= 880 lb/ft(24ft) = 21,120 lb12 ft 12 ft

( + ) M1 = 0–21,120 lb(12 ft) + RY2(24 ft) = 0

RY2(24 ft) = 253,440 lb-ft

RY2 = 10,560 lb

( +) FY = 0

RY1 + RY2 – 21,120 lb = 0

RY1 + 10,560 lb – 21,120 lb = 0

RY1 RY2

RX1

24 ft

resultant force - equivalent total load that is a result of a distributed line load

resultant force = area loading diagram

resultant force = ( L)

= 880 lb/ft(24ft) = 21,120 lb12 ft 12 ft

( + ) M1 = 0–21,120 lb(12 ft) + RY2(24 ft) = 0

RY2(24 ft) = 253,440 lb-ft

RY2 = 10,560 lb

( +) FY = 0

RY1 + RY2 – 21,120 lb = 0

RY1 + 10,560 lb – 21,120 lb = 0

RY1 = 10,560 lb

RY1 RY2

RX1

24 ft

resultant force - equivalent total load that is a result of a distributed line load

resultant force = area loading diagram

resultant force = ( L)

= 880 lb/ft(24ft) = 21,120 lb12 ft 12 ft

( + ) M1 = 0–21,120 lb(12 ft) + RY2(24 ft) = 0

RY2(24 ft) = 253,440 lb-ft

RY2 = 10,560 lb

( +) FY = 0

RY1 + RY2 – 21,120 lb = 0

RY1 + 10,560 lb – 21,120 lb = 0

RY1 = 10,560 lb

( +) FX = 0

RX1 = 0

RY1 RY2

RX1

24 ft

resultant force - equivalent total load that is a result of a distributed line load

resultant force = area loading diagram

resultant force = ( L)

= 880 lb/ft(24ft) = 21,120 lb12 ft 12 ft

( + ) M1 = 0–21,120 lb(12 ft) + RY2(24 ft) = 0

RY2(24 ft) = 253,440 lb-ft

RY2 = 10,560 lb

( +) FY = 0

RY1 + RY2 – 21,120 lb = 0

RY1 + 10,560 lb – 21,120 lb = 0

RY1 = 10,560 lb

( +) FX = 0

RX1 = 0

= 880 lb/ft

24 ft 10,560 lb

0 lb

10,560 lb

RY1 RY2

RX1

24 ft

resultant force - equivalent total load that is a result of a distributed line load

resultant force = area loading diagram

resultant force = ( L)

= 880 lb/ft(24ft) = 21,120 lb12 ft 12 ft

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)

V

+P + +

M

V

- -

M

-P

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

+20k

- 0 +

-20k 0 compression

-10k

-20k

-10k

-10k

+20k

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

= - 880 lb/ft

24 ft

+10.56k

0 lb

10,560 lb

P

V

Area of Loading Diagram

-0.88k/ft * 24ft = -21.12k

10.56k + -21.12k = -10.56k

-10.56k

0 0

0 0

-880 plf = slope

10,560 lb

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

= - 880 lb/ft

24 ft

+10.56k

0 lb

10,560 lb

P

V

M

Area of Loading Diagram

-0.88k/ft * 24ft = -21.12k

10.56k + -21.12k = -10.56k

-10.56k

0 0

0 0

-880 plf = slope

10,560 lb

0 0

Slope initial = +10.56k

Area of Shear Diagram

(10.56k )(12ft ) 0.5 = 63.36 k-ft

pos.

slope

zero slope 63.36k’ neg. slope

(-10.56k)(12ft)(0.5) = -63.36 k-ft

W2 = 30 PSF

W1 = 20 PSF

Wind Loading

W2 = 30 PSF

W1 = 20 PSF

1/2 LOAD

SPAN

SPAN

1/2 LOAD

1/2 +

1/2 LOAD

Wind Load spans to each level

10 ft

10 ft

roof= (30 PSF)(5 FT)

= 150 PLF

Total Wind Load to roof level

second= (30 PSF)(5 FT) + (20 PSF)(5 FT)

= 250 PLF

Total Wind Load to second floor level

second= 250 PLF

roof= 150 PLF

seismic load

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

(w h)

( VBASE )

(Cs)(W)VBASE =

Fx =

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

Fx =

Froof

Fsecond flr

VBASE (wx) (hx)

(w h)

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 :

R = 1+1+1+1 = 4

Px = ( Rx / R ) (Ptotal)

PMF1 = 1/4 Ptotal