Building Structures Session 2 Loads – Tributary Areas, Load Flow Learning Objec=ves • Calculate the tributary area for distributed loads • Trace the flow of loads through deck and panel loaded structures Idealized Deck Loaded Structures • Frequently structures have a skin of horizontal or ver=cal spanning structures that actually support the loads Hibbeler 2-­‐1 Idealized Deck Loaded Structures • Examples: roofs, floors, walls • Distributed loads act over an area and are transmiPed to suppor=ng structural members • Tributary area is the area that contributes loads to a par=cular member either directly or through a series of intermediate members • We idealize load flow from a slab to a member as either – 1 way (loads transmiPed in one direc=on only) – 2 way (loads transmiPed in two direc=ons) Hibbeler 2-­‐1 One-­‐way and Two-­‐way Systems • Deck behavior depends on construc=on methods/ materials and geometry • Construc=on methods (1 way) • Concrete with rebar in one direc=on • Corrugated steel deck • Thin plywood (why?) • Construc=on methods (2 way) •Concrete with rebar in two direc=ons •Waffle slabs •Flat steel plates •Thick plywood Hibbeler 2-­‐1 One-­‐way System • Geometry – L2 ≥ 2L1 model as one-­‐way Hibbeler 2-­‐1 Two-­‐way System • Geometry – L2 < 2L1 model as two-­‐way – Bisect corners to distribute load Hibbeler 2-­‐1 Two-­‐way System (cont) Hibbeler 2-­‐1 Learning Objec=ves ✔
• Calculate the tributary area for distributed loads • Trace the flow of loads through deck and panel loaded structures 6 in (0.5 ft)
Mul=-­‐level Load Paths 4@16 ] 3@4 ] Live Loads 07 ASCE 7-­‐10 Table 4-­‐1 Minimum Live Loads.pdf Pedestrian walkway live load 60 psf Given: Interior pedestrian walkway Dead Loads • Concrete deck 6 in (.
5]) thick, 150lb/]3 • Stringers 20 lb/] • Beams 30 lb/] • Girders 50 lb/] • Example W16x50 Exterior Interior 16 ] Stringers 2 ] WDead =(150pcf)(2])(.5]) + 20 lb/] WDead = 170 lb/] Wlive =(60psf)(2])= 120 lb/] End Reac=ons (by symmetry) RD = ½ (170 lb/])(16 ]) = 1360 lb RL = ½ (120 lf/])(16 ]) = 960 lb 16 ] 4 ] WDead =(150pcf)(4])(.5]) + 20 lb/] WDead = 320 lb/] Wlive =(60psf)(4]) = 240 lb/] End Reac=ons (by symmetry) RD = ½ (320 lb/])(16 ]) = 2560 lb RL = ½ (240 lf/])(16 ]) = 1920 lb Exterior Beams 3@4 ] 3@4 ] 960lb 1920lb 1920lb 960 lb 1360lb 2560lb 2560lb 1360 lb 30 lb/] RL = 2880 lb RD = 4100 lb Interior RL = 2880 lb RD = 4100 lb End Reac=ons (by symmetry) RL = ½ [(2)(960lb) + (2)(1920lb)] RL = 2880 lb RD = ½ [(2)(1360lb) + (2)(2560lb) + (30lb/])(12])] RD = 4100 lb 960lb 1920lb 1920lb 960 lb 960lb 1920lb 1920lb 960 lb 1360lb 2560lb 2560lb 1360 lb 1360lb 2560lb 2560lb 1360 lb 30 lb/] RL = 5760 lb RD = 8020 lb RL = 5760 lb RD = 8020 lb End Reac=ons (by symmetry) RL = ½ [(4)(960lb) + (4)(1920lb)] RL = 5760 lb RD = ½ [(4)(1360lb) + (4)(2560lb) + (30lb/])(12])] RD = 8020 lb Girders Exterior Only 4@16 ] 2880lb 5760lb 5760lb 5760lb 2880lb 4100lb 8020lb 8020lb 8020lb 4100lb 50 lb/] RL = 11520 lb RD = 17730 lb End Reac=ons (by symmetry) RL = ½ [(2)(2880lb) + (3)(5760lb)] RL = 11520 lb RD = ½ [(2)(4100lb) + (3)(8020lb) + (50lb/])(4)(16])] RD = 17730 lb RL = 11520 lb RD = 17730 lb In Class Problem The steel framework is used to support a 4in reinforced
concrete slab that is used for an office. Sketch the loading
that acts along members BE and FD. Take a = 7.5 ft and
b = 10 ft, unit weight of concrete is 150pcf and uniform
live load of an office is 500psf. Ignore the weight of the
steel beam.
Idealized Panel Loaded Structure • Ideal trusses are only loaded at panel points (joints) • Example problem -­‐ Roof Loads (wind & snow) transferred to roof trusses Plan View 32 ] Tributary Area Exterior Truss 15 ] Tributary Area Interior Truss Tributary Area Exterior Truss 15 ] Wind Loads (normal to surface) Wind Loads Exterior Truss 5 ft
90 mph +20psf C 10 ft
2@6] 5 ft
B 5 4 A -­‐5 psf D 3 F 4@8] E Wind Loads 750lb 187.5lb C 2@6] 1500lb 5 750lb A 4 B D 3 Wind Loads Exterior 375lb Truss 187.5lb F 4@8] E 1500lb 375lb C 2@6] 3000lb B 5 1500lb A 4 D 3 Wind Loads Interior 750lb Truss 375lb F 4@8] E +40psf Live Loads Ver=cal 2@6] 5 Snow Loads Exterior Truss C B D 3 4 F A 4@8] 4 ft
8 ft
8 ft
E 8 ft
4 ft
2400lb Live Loads Ver=cal 2@6] 2400lb B 5 1200lb A C 4 D Snow Loads Exterior 2400lb Truss 3 1200lb F 4@8] E 4,800lb C 2@6] +20psf 4,800lb B 5 2,400lb A 4 D Snow Loads Interior 4,800lb Truss 3 2,400lb F 4@8] E Learning Objec=ves ✔
• Calculate the tributary area for distributed loads • Trace the flow of loads through deck and ✔
panel loaded structures