1. STRUCTURAL ANALYSIS :
load flow

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

3. GRAVITY LOADS. dead load (DL) live load (LL) snow load (S)
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.

4. live loading

5. dead + live loading

6. 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.

7. AREA LOADS

8. LINE LOAD.

9. POINT LOADS.

11. tuned mass damping
Taipei 101 Building in Taiwan.
Completed
Overall height 509 m (1671 ft).
728-ton counterweight ball hung at
87th/88th floor
Behavior of this damper during the 2005 Szechuan earthquake:

12. LIVE LOADS

13. Recall the method of solving for reactions when you have a point load:
200 lb
( ) SM1 = 0
-200 lb(10 ft) + RY2(15 ft) = 0
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

14. Now what if you have a uniformly distributed load?
w = 2000 lb/ft
RX1
RY1
RY2
20 ft

15. resultant force = “area” of loading diagram resultant force = w( L)
resultant force - equivalent total load that is a result of a distributed line load
resultant force = “area” of loading diagram
resultant force = w( L)
RX1
RY1
RY2
20 ft

16. resultant force = area of loading diagram resultant force = w( L)
resultant force - equivalent total load that is a result of a distributed line load
resultant force = area of loading diagram
resultant force = w( L)
= 2000 lb/ft(20ft) = 40,000 lb
RX1
10 ft
10 ft
RY1
RY2
20 ft
* This resultant force is located at the centroid of the distributed load.

17. resultant force - equivalent total load that is a result of a distributed line load
resultant force = area loading diagram
resultant force = w( L)
= 2000 lb/ft(20ft) = 40,000 lb
RX1
10 ft
10 ft
RY1
RY2
20 ft
( ) SM1 = 0

18. ( + ) SM1 = 0 -40,000 lb(10 ft) + RY2(20 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)
= 2000 lb/ft(20ft) = 40,000 lb
RX1
10 ft
10 ft
RY1
RY2
20 ft
( ) SM1 = 0
-40,000 lb(10 ft) + RY2(20 ft) = 0

19. ( + ) SM1 = 0 -40,000 lb(10 ft) + RY2(20 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)
= 2000 lb/ft(20ft) = 40,000 lb
RX1
10 ft
10 ft
RY1
RY2
20 ft
( ) SM1 = 0
-40,000 lb(10 ft) + RY2(20 ft) = 0
RY2(20 ft) = 400,000 lb-ft

20. ( + ) SM1 = 0 -40,000 lb(10 ft) + RY2(20 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)
= 2000 lb/ft(20ft) = 40,000 lb
RX1
10 ft
10 ft
RY1
RY2
20 ft
( ) SM1 = 0
-40,000 lb(10 ft) + RY2(20 ft) = 0
RY2(20 ft) = 400,000 lb-ft
RY2 = 20,000 lb

21. ( + ) SM1 = 0 -40,000 lb(10 ft) + RY2(20 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)
= 2000 lb/ft(20ft) = 40,000 lb
RX1
10 ft
10 ft
RY1
RY2
20 ft
( ) SM1 = 0
-40,000 lb(10 ft) + RY2(20 ft) = 0
RY2(20 ft) = 400,000 lb-ft
RY2 = 20,000 lb
( ) SFY = 0

22. ( + ) SM1 = 0 -40,000 lb(10 ft) + RY2(20 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)
= 2000 lb/ft(20ft) = 40,000 lb
RX1
10 ft
10 ft
RY1
RY2
20 ft
( ) SM1 = 0
-40,000 lb(10 ft) + RY2(20 ft) = 0
RY2(20 ft) = 400,000 lb-ft
RY2 = 20,000 lb
( ) SFY = 0
RY1 - 40,000 lb + RY2 = 0

23. ( + ) SM1 = 0 -40,000 lb(10 ft) + RY2(20 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)
= 2000 lb/ft(20ft) = 40,000 lb
RX1
10 ft
10 ft
RY1
RY2
20 ft
( ) SM1 = 0
-40,000 lb(10 ft) + RY2(20 ft) = 0
RY2(20 ft) = 400,000 lb-ft
RY2 = 20,000 lb
( ) SFY = 0
RY1 - 40,000 lb + RY2 = 0
RY1 - 40,000 lb + 20,000 lb = 0

24. ( + ) SM1 = 0 -40,000 lb(10 ft) + RY2(20 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)
= 2000 lb/ft(20ft) = 40,000 lb
RX1
10 ft
10 ft
RY1
RY2
20 ft
( ) SM1 = 0
-40,000 lb(10 ft) + RY2(20 ft) = 0
RY2(20 ft) = 400,000 lb-ft
RY2 = 20,000 lb
( ) SFY = 0
RY1 - 40,000 lb + RY2 = 0
RY1 - 40,000 lb + 20,000 lb = 0
RY1 = 20,000 lb

25. ( + ) SM1 = 0 -40,000 lb(10 ft) + RY2(20 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)
= 2000 lb/ft(20ft) = 40,000 lb
RX1
10 ft
10 ft
RY1
RY2
20 ft
( ) SM1 = 0
-40,000 lb(10 ft) + RY2(20 ft) = 0
RY2(20 ft) = 400,000 lb-ft
RY2 = 20,000 lb
( ) SFY = 0
RY1 - 40,000 lb + RY2 = 0
RY1 - 40,000 lb + 20,000 lb = 0
RY1 = 20,000 lb
( ) SFX = 0
RX1 = 0

26. ( + ) SM1 = 0 40,000 lb(10 ft) - RY2(20 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)
= 2000 lb/ft(20ft) = 40,000 lb
RX1
10 ft
10 ft
RY1
RY2
20 ft
( ) SM1 = 0
40,000 lb(10 ft) - RY2(20 ft) = 0
RY2(20 ft) = 400,000 lb-ft
RY2 = 20,000 lb
( ) SFY = 0
RY1 + RY2 - 40,000 lb = 0
RY1 + 20,000 lb - 40,000 lb = 0
RY1 = 20,000 lb
( ) SFX = 0
RX1 = 0
w = 2000 lb/ft
0 lb
20 ft
20,000 lb
20,000 lb

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

28. UNIFORM LOAD
SPAN
1/2 LOAD
1/2 LOAD

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

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

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

32. foundation bearing pressure

33. foundation bearing pressure

35. lateral earth pressure (30-120 lb/ft3)(H)
retaining height (H)
1/3 H
lateral earth pressure ( lb/ft3)(H)
earth pressure load

36. wind load

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

40. Wind Loading
W2 = 30 PSF
W1 = 20 PSF

41. 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

42. Total Wind Load to roof level
wroof= (30 PSF)(5 FT)
= 150 PLF

43. Total Wind Load to second floor level
wsecond= (30 PSF)(5 FT) + (20 PSF)(5 FT)
= 250 PLF

44. wroof= 150 PLF
wsecond= 250 PLF

45. seismic load

48. 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 =

49. Total Seismic Loading : VBASE = 0.3 W
W = Wroof + Wsecond

50. wroof

51. wsecond flr

52. W = wroof + wsecond flr

53. VBASE

54. Redistribute Total Seismic Load to each level based on relative height and weight
Froof
Fsecond flr
VBASE (wx) (hx)
S (w h)
Fx =

55. 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

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