STRUCTURAL ANALYSIS load flow external forces actions 1

STRUCTURAL ANALYSIS : load flow

external forces (actions) : 1. gravitational pull of the earth on the mass of the structure, its contents and the things that move through it. 2. thrust from retained earth or water. 3. pressure and suction of wind. 4. inertial force due to earthquakes (ground acceleration). 5. impact from explosions or flying objects

GRAVITY LOADS. dead load (DL) permanent self-weight including structure, enclosure, finishes and appendages. live load (LL) transitory loads based upon use and occupancy including people, cars and furniture. snow load (S) semi-permanent load based on local snowfall statistics. snow load may have a lateral component.

live loading

dead + live loading

area loads account for loads which are spread out uniformly over an area. line loads account for loads which occur along a line which may be uniform or vary. point loads are discrete forces which occur at a specific location.

AREA LOADS

LINE LOAD.

POINT LOADS.

tuned mass damping Taipei 101 Building in Taiwan. Completed 2003 -2004. Overall height 509 m (1671 ft). 728 -ton counterweight ball hung at 87 th/88 th floor Behavior of this damper during the 2005 Szechuan earthquake: https: //www. youtube. com/watch? v=NYSgd 1 XSZ Xc

LIVE LOADS

Recall the method of solving for reactions when you have a point load: 200 lb ( + ) SM 1 = 0 -200 lb(10 ft) + RY 2(15 ft) = 0 RX 1 RY 2(15 ft) = 2000 lb-ft 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

Now what if you have a uniformly distributed load? w = 2000 lb/ft RX 1 RY 1 20 ft RY 2

resultant force - equivalent total load that is a result of a distributed line load resultant force = “area” of loading diagram resultant force = w( L) RX 1 RY 1 20 ft RY 2

resultant force - equivalent total load that is a result of a distributed line load resultant force = area of loading diagram RX 1 RY 1 resultant force = w( L) 10 ft 20 ft RY 2 = 2000 lb/ft(20 ft) = 40, 000 lb * This resultant force is located at the centroid of the distributed load.

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) 10 ft 20 ft RY 2 = 2000 lb/ft(20 ft) = 40, 000 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) 10 ft 20 ft RY 2 = 2000 lb/ft(20 ft) = 40, 000 lb ( + ) SM 1 = 0 -40, 000 lb(10 ft) + RY 2(20 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) 10 ft 20 ft RY 2 = 2000 lb/ft(20 ft) = 40, 000 lb ( + ) SM 1 = 0 -40, 000 lb(10 ft) + RY 2(20 ft) = 0 RY 2(20 ft) = 400, 000 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) 10 ft 20 ft RY 2 = 2000 lb/ft(20 ft) = 40, 000 lb ( + ) SM 1 = 0 -40, 000 lb(10 ft) + RY 2(20 ft) = 0 RY 2(20 ft) = 400, 000 lb-ft RY 2 = 20, 000 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) 10 ft 20 ft RY 2 = 2000 lb/ft(20 ft) = 40, 000 lb ( + ) SM 1 = 0 -40, 000 lb(10 ft) + RY 2(20 ft) = 0 RY 2(20 ft) = 400, 000 lb-ft RY 2 = 20, 000 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) 10 ft 20 ft RY 2 = 2000 lb/ft(20 ft) = 40, 000 lb ( + ) SM 1 = 0 -40, 000 lb(10 ft) + RY 2(20 ft) = 0 RY 2(20 ft) = 400, 000 lb-ft RY 2 = 20, 000 lb ( + ) SFY = 0 RY 1 - 40, 000 lb + RY 2 = 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) 10 ft 20 ft RY 2 = 2000 lb/ft(20 ft) = 40, 000 lb ( + ) SM 1 = 0 -40, 000 lb(10 ft) + RY 2(20 ft) = 0 RY 2(20 ft) = 400, 000 lb-ft RY 2 = 20, 000 lb ( +) SFY = 0 RY 1 - 40, 000 lb + RY 2 = 0 RY 1 - 40, 000 lb + 20, 000 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) 10 ft 20 ft RY 2 = 2000 lb/ft(20 ft) = 40, 000 lb ( + ) SM 1 = 0 -40, 000 lb(10 ft) + RY 2(20 ft) = 0 RY 2(20 ft) = 400, 000 lb-ft RY 2 = 20, 000 lb ( + ) SFY = 0 RY 1 - 40, 000 lb + RY 2 = 0 RY 1 - 40, 000 lb + 20, 000 lb = 0 RY 1 = 20, 000 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) 10 ft 20 ft RY 2 = 2000 lb/ft(20 ft) = 40, 000 lb ( + ) SM 1 = 0 -40, 000 lb(10 ft) + RY 2(20 ft) = 0 RY 2(20 ft) = 400, 000 lb-ft RY 2 = 20, 000 lb ( +) SFY = 0 RY 1 - 40, 000 lb + RY 2 = 0 RY 1 - 40, 000 lb + 20, 000 lb = 0 RY 1 = 20, 000 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) 10 ft 20 ft RY 2 = 2000 lb/ft(20 ft) = 40, 000 lb ( + ) SM 1 = 0 40, 000 lb(10 ft) - RY 2(20 ft) = 0 RY 2(20 ft) = 400, 000 lb-ft RY 2 = 20, 000 lb ( +) SFY = 0 RY 1 + RY 2 - 40, 000 lb = 0 w = 2000 lb/ft RY 1 + 20, 000 lb - 40, 000 lb = 0 RY 1 = 20, 000 lb 20 ft 20, 000 lb ( +) SFX = 0 RX 1 = 0

framing diagram plan vertical load flow : deck > beam > girder > column

UNIFORM LOAD SPAN 1/2 LOA D

LENGTH Tributary Width is the width of area that contributes load to a specific spanning element. assumed distributed line load from deck: w BEAM Area Load = DL+LL w = (Trib Width) (Area Load)

Tributary Area is the area that contributes load to a specific structural element. PBEAM = (Trib Area BEAM) (Area Load) PBM PBM

Tributary Area to the column includes loads from the girders and beams which frame directly to the columns PCOL = (Trib Area) (Area Load)

foundation bearing pressure

foundation bearing pressure

retainin g height (H) 1/3 H lateral earth pressure (30 -120 lb/ft 3)(H) earth pressure load

wind load

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

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 https: //www. youtube. com/watch? v=snfcs. Lu S(w LY 88 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