1. WIND FORCES

2. wind load

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

6. SEISMIC FORCES

7. seismic load

10. 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 =

11. Where are we going with all of this?

12. 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)

13. 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)

14. EXTERNAL FORCES

16. ( + ) 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

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

20. resultant force = area loading diagram resultant force = w( L)
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

21. resultant force = area loading diagram resultant force = w( L)
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

22. 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

23. ( + ) 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

24. ( + ) 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

25. ( + ) 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

26. ( + ) 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

27. ( + ) 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

28. ( + ) 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

29. ( + ) 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

30. ( + ) 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

31. ( + ) 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

32. External – for solving reactions (Applied Loading & Support Reactions)
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.

33. STRUCTURAL ANALYSIS:
INTERNAL FORCES
P V M

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

35. P + + M V +

36. - P - M V -

37. 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

38.
-10k
+20k
-10k
-10k
-20k
-10k
+20k
20k
compression

39. 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

40. 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

41. 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

42. 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

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

45. 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

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

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

48. wroof= 150 PLF
wsecond= 250 PLF

49. seismic load

52. 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 =

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

54. wroof

55. wsecond flr

56. W = wroof + wsecond flr

57. VBASE

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

59. 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

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