WIND FORCES wind load Wind Load is an

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 = (Cs)(W) ( VBASE ) Redistributed (based on relative height and weight) to each level as a ‘Point Load’ at the center of mass of each Ffloor x = level: 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 move nearly imperceptibly boundary conditions: fixed, pin, or roller idealize member supports & connections external forces: are applied to beams & columns as concentrated point loads & linear loads categories of external loading: DL, LL, W, E, S, H (fluid pressure) reactions: we use three equations of equilibrium to calculate these

internal forces: axial, shear, bending/flexure internal stresses: tension stress, compression stress shear stress, bending stress, stability, slenderness, and allowable compression stress member sizing for flexure member sizing for combined flexure and axial stress (Proj. 2)

EXTERNAL FORCES

200 lb ( + ) SM 1 = 0 0= -200 lb(10 ft) + RY 2(15 ft) = 2000 lb-ft RX 1 RY 1 10 ft 5 ft RY 2 = 133 lb ( +) SFY = 0 RY 1 + RY 2 - 200 lb = 0 RY 1 + 133 lb - 200 lb = 0 RY 1 = 67 lb 200 lb ( +) SFX = 0 RX 1 = 0 0 lb 67 lb 10 ft 5 ft 133 lb

w = 880 lb/ft RX 1 RY 1 24 ft RY 2

resultant force - equivalent total load that is a result of a distributed line load resultant force = area loading diagram resultant force = w( L) RX 1 RY 1 24 ft RY 2

resultant force - equivalent total load that is a result of a distributed line load resultant force = area loading diagram RX 1 RY 1 resultant force = w( L) 12 ft 24 ft RY 2 = 880 lb/ft(24 ft) = 21, 120 lb

resultant force - equivalent total load that is a result of a distributed line load resultant force = area loading diagram RX 1 RY 1 resultant force = w( L) 12 ft 24 ft RY 2 = 880 lb/ft(24 ft) = 21, 120 lb ( + ) SM 1 = 0

resultant force - equivalent total load that is a result of a distributed line load resultant force = area loading diagram RX 1 RY 1 resultant force = w( L) 12 ft 24 ft RY 2 = 880 lb/ft(24 ft) = 21, 120 lb ( + ) SM 1 = 0 – 21, 120 lb(12 ft) + RY 2(24 ft) = 0

resultant force - equivalent total load that is a result of a distributed line load resultant force = area loading diagram RX 1 RY 1 resultant force = w( L) 12 ft 24 ft RY 2 = 880 lb/ft(24 ft) = 21, 120 lb ( + ) SM 1 = 0 – 21, 120 lb(12 ft) + RY 2(24 ft) = 0 RY 2(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 RX 1 RY 1 resultant force = w( L) 12 ft 24 ft RY 2 = 880 lb/ft(24 ft) = 21, 120 lb ( + ) SM 1 = 0 – 21, 120 lb(12 ft) + RY 2(24 ft) = 0 RY 2(24 ft) = 253, 440 lb-ft RY 2 = 10, 560 lb

resultant force - equivalent total load that is a result of a distributed line load resultant force = area loading diagram RX 1 RY 1 resultant force = w( L) 12 ft 24 ft RY 2 = 880 lb/ft(24 ft) = 21, 120 lb ( + ) SM 1 = 0 – 21, 120 lb(12 ft) + RY 2(24 ft) = 0 RY 2(24 ft) = 253, 440 lb-ft RY 2 = 10, 560 lb ( +) SFY = 0

resultant force - equivalent total load that is a result of a distributed line load resultant force = area loading diagram RX 1 RY 1 resultant force = w( L) 12 ft 24 ft RY 2 = 880 lb/ft(24 ft) = 21, 120 lb ( + ) SM 1 = 0 – 21, 120 lb(12 ft) + RY 2(24 ft) = 0 RY 2(24 ft) = 253, 440 lb-ft RY 2 = 10, 560 lb ( +) SFY = 0 RY 1 + RY 2 – 21, 120 lb = 0

resultant force - equivalent total load that is a result of a distributed line load resultant force = area loading diagram RX 1 RY 1 resultant force = w( L) 12 ft 24 ft RY 2 = 880 lb/ft(24 ft) = 21, 120 lb ( + ) SM 1 = 0 – 21, 120 lb(12 ft) + RY 2(24 ft) = 0 RY 2(24 ft) = 253, 440 lb-ft RY 2 = 10, 560 lb ( +) SFY = 0 RY 1 + RY 2 – 21, 120 lb = 0 RY 1 + 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 RX 1 RY 1 resultant force = w( L) 12 ft 24 ft RY 2 = 880 lb/ft(24 ft) = 21, 120 lb ( + ) SM 1 = 0 – 21, 120 lb(12 ft) + RY 2(24 ft) = 0 RY 2(24 ft) = 253, 440 lb-ft RY 2 = 10, 560 lb ( +) SFY = 0 RY 1 + RY 2 – 21, 120 lb = 0 RY 1 + 10, 560 lb – 21, 120 lb = 0 RY 1 = 10, 560 lb

resultant force - equivalent total load that is a result of a distributed line load resultant force = area loading diagram RX 1 RY 1 resultant force = w( L) 12 ft 24 ft RY 2 = 880 lb/ft(24 ft) = 21, 120 lb ( + ) SM 1 = 0 – 21, 120 lb(12 ft) + RY 2(24 ft) = 0 RY 2(24 ft) = 253, 440 lb-ft RY 2 = 10, 560 lb ( +) SFY = 0 RY 1 + RY 2 – 21, 120 lb = 0 RY 1 + 10, 560 lb – 21, 120 lb = 0 RY 1 = 10, 560 lb ( +) SFX = 0 RX 1 = 0

resultant force - equivalent total load that is a result of a distributed line load resultant force = area loading diagram RX 1 RY 1 resultant force = w( L) 12 ft 24 ft RY 2 = 880 lb/ft(24 ft) = 21, 120 lb ( + ) SM 1 = 0 – 21, 120 lb(12 ft) + RY 2(24 ft) = 0 RY 2(24 ft) = 253, 440 lb-ft RY 2 = 10, 560 lb ( +) SFY = 0 RY 1 + RY 2 – 21, 120 lb = 0 w = 880 lb/ft RY 1 + 10, 560 lb – 21, 120 lb = 0 RY 1 = 10, 560 lb 24 ft 10, 560 lb ( +) SFX = 0 RX 1 = 0

SIGN CONVENTIONS (often confusing and 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)

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

-10 k 0 -10 k -20 k -10 k +20 k -20 k 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 0 +10. 56 k +10. 56 k V 10, 560 lb Area of Loading Diagram -880 plf = slope -0. 88 k/ft * 24 ft = -21. 12 k 0 +10. 56 k P 24 ft -10. 56 k + -21. 12 k = 10. 56 k

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. 56 k +10. 56 k slo s. po M 0 -0. 88 k/ft * 24 ft = -21. 12 k 63. 36 k’ +10. 56 k 0 zero slope 10. 56 k + -21. 12 k = 10. 56 k -10. 56 k ne g. slo pe -63. 36 k-ft 0 pe V Area of Loading Diagram -880 plf = slope +63. 36 k-ft P 10, 560 lb 0 Slope initial = +10. 56 k Area of Shear Diagram (10. 56 k )(12 ft ) 0. 5 = 63. 36 k-ft (-10. 56 k)(12 ft)(0. 5) = -63. 36 k-

Wind Loading W 2 = 30 PSF W 1 = 20 PSF

Wind Load spans to each level 1/2 LOAD W 2 = 30 PSF SPAN 10 ft 1/2 + 1/2 LOAD SPAN W 1 = 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 = (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) PMF 1 = 1/4 Ptotal