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Published byJack Prentis Modified about 1 year ago

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WIND FORCES

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wind load

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

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SEISMIC FORCES

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seismic load

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Seismic Load is generated by the inertia of the mass of the structure : V BASE Redistributed (based on relative height and weight) to each level as a ‘Point Load’ at the center of mass of each floor level: F X V BASE w x h x (w h) ( V BASE ) (Cs)(W)V BASE = F x =

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Where are we going with all of this?

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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)

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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)

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EXTERNAL FORCES

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( + ) M 1 = 0 0= -200 lb(10 ft) + R Y2 (15 ft) R Y2 (15 ft) = 2000 lb-ft R Y2 = 133 lb ( +) F Y = 0 R Y1 + R Y lb = 0 R Y lb lb = 0 R Y1 = 67 lb ( +) F X = 0 R X1 = 0 R Y1 200 lb R Y2 R X1 10 ft5 ft 67 lb 200 lb 0 lb 10 ft5 ft 133 lb

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= 880lb/ft R Y2 24 ft R Y1 R X1

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R Y2 24 ft resultant force - equivalent total load that is a result of a distributed line load resultant force = area loading diagram resultant force = ( L) R Y1 R X1

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R Y1 R Y2 R X1 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 lb 12 ft

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( + ) M 1 = 0 R Y1 R Y2 R X1 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 lb 12 ft

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( + ) M 1 = 0 –21,120 lb(12 ft) + R Y2 (24 ft) = 0 R Y1 R Y2 R X1 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 lb 12 ft

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( + ) M 1 = 0 –21,120 lb(12 ft) + R Y2 (24 ft) = 0 R Y2 (24 ft) = 253,440 lb-ft R Y1 R Y2 R X1 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 lb 12 ft

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( + ) M 1 = 0 –21,120 lb(12 ft) + R Y2 (24 ft) = 0 R Y2 (24 ft) = 253,440 lb-ft R Y2 = 10,560 lb R Y1 R Y2 R X1 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 lb 12 ft

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( + ) M 1 = 0 –21,120 lb(12 ft) + R Y2 (24 ft) = 0 R Y2 (24 ft) = 253,440 lb-ft R Y2 = 10,560 lb ( +) F Y = 0 R Y1 R Y2 R X1 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 lb 12 ft

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( + ) M 1 = 0 –21,120 lb(12 ft) + R Y2 (24 ft) = 0 R Y2 (24 ft) = 253,440 lb-ft R Y2 = 10,560 lb ( +) F Y = 0 R Y1 + R Y2 – 21,120 lb = 0 R Y1 R Y2 R X1 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 lb 12 ft

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( + ) M 1 = 0 –21,120 lb(12 ft) + R Y2 (24 ft) = 0 R Y2 (24 ft) = 253,440 lb-ft R Y2 = 10,560 lb ( +) F Y = 0 R Y1 + R Y2 – 21,120 lb = 0 R Y1 + 10,560 lb – 21,120 lb = 0 R Y1 R Y2 R X1 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 lb 12 ft

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( + ) M 1 = 0 –21,120 lb(12 ft) + R Y2 (24 ft) = 0 R Y2 (24 ft) = 253,440 lb-ft R Y2 = 10,560 lb ( +) F Y = 0 R Y1 + R Y2 – 21,120 lb = 0 R Y1 + 10,560 lb – 21,120 lb = 0 R Y1 = 10,560 lb R Y1 R Y2 R X1 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 lb 12 ft

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( + ) M 1 = 0 –21,120 lb(12 ft) + R Y2 (24 ft) = 0 R Y2 (24 ft) = 253,440 lb-ft R Y2 = 10,560 lb ( +) F Y = 0 R Y1 + R Y2 – 21,120 lb = 0 R Y1 + 10,560 lb – 21,120 lb = 0 R Y1 = 10,560 lb ( +) F X = 0 R X1 = 0 R Y1 R Y2 R X1 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 lb 12 ft

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( + ) M 1 = 0 –21,120 lb(12 ft) + R Y2 (24 ft) = 0 R Y2 (24 ft) = 253,440 lb-ft R Y2 = 10,560 lb ( +) F Y = 0 R Y1 + R Y2 – 21,120 lb = 0 R Y1 + 10,560 lb – 21,120 lb = 0 R Y1 = 10,560 lb ( +) F X = 0 R X1 = 0 = 880 lb/ft 24 ft 10,560 lb 0 lb 10,560 lb R Y1 R Y2 R X1 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 lb 12 ft

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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.

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STRUCTURAL ANALYSIS: INTERNAL FORCES P V M

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

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V + P + + M

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V - - M - P

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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

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-10k +20k k 0 compression -10k -20k -10k +20k

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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

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= lb/ft 24 ft k 0 lb 10,560 lb P V k Area of Loading Diagram -0.88k/ft * 24ft = k 10.56k k = k k plf = slope 10,560 lb

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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

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= lb/ft 24 ft k 0 lb 10,560 lb P V M k Area of Loading Diagram -0.88k/ft * 24ft = k 10.56k k = k k plf = slope 10,560 lb 00 Slope initial = k Area of Shear Diagram (10.56k )(12ft ) 0.5 = k-ft k-ft pos. slope zero slope 63.36k’ k-ft neg. slope (-10.56k)(12ft)(0.5) = k-ft

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

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W 2 = 30 PSF W 1 = 20 PSF 1/2 LOAD SPAN 1/2 LOAD 1/2 + 1/2 LOAD Wind Load spans to each level 10 ft

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

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

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second = 250 PLF roof = 150 PLF

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seismic load

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Seismic Load is generated by the inertia of the mass of the structure : V BASE 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 : F X V BASE w x h x (w h) ( V BASE ) (Cs)(W)V BASE = F x =

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Total Seismic Loading : V BASE = 0.3 W W = W roof + W second

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w roof

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w second flr

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W = w roof + w second flr

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V BASE

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Redistribute Total Seismic Load to each level based on relative height and weight F x = F roof F second flr V BASE (w x ) (h x ) (w h)

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Load Flow to Lateral Resisting System : Distribution based on Relative Rigidity Assume Relative Rigidity : Single Bay MF : Rel Rigidity = Bay MF : Rel Rigidity = Bay MF : Rel Rigidity = 3

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Distribution based on Relative Rigidity : R = = 4 P x = ( R x / R ) (P total ) P MF1 = 1/4 P total

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