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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
