][ STEEL
STRUCTURES ][
][ STEEL
STRUCTURES ][
Beam Design Shear P&A 290 Shear is seldom a size determining factor for steel beams. Only critical if: =========== short span with high load large load near supports small moment but high shear S shapes MSC 2-78 Allowable uniform load MSC 1-36 Dimensions MSC 1-37 Properties for W- S- and C-shaped beams the shear stress is assumed to be developed by the the beam web only. A = d x tw W = parallel flange surfaces wider flanges only HP Shapes are wider MSC 1-10...33 I S = American Standard Beams inner flange surface slopes 16 2/3% 2:12 slope greater web thickness than W shape narrower flange than W shape bottom flange has holes for fasteners MSC 1-36 ....37 C = American Standard Chanels inner flange surface slopes 16 2/3% 2:12 slope [ - chanel shape bottom flange has hole for fastener MSC 1-40 ...45 Shear stress in web only Fv = 0.40 Fy Fv = 14.5 ksi for A36 P&A 266 fv = V / A w fv = V / d x tw fv = actual unit shear stress V = maximum vertical shear Aw = gross area of the beam web d = overall depth of the beam tw = thickness of the beam web
SHEAR EXAMPLE-PROBLEM : A simple beam of A 36 steel is 6 ft long and has a concentrated load of 36 kips applied 1 ft from the left end. It is found that a W 8 x 24 is large enough to sustain the resulting bending moment. Investigate the beam for shear.
A. GIVEN: ======== simple beam A 36 steel 6 ft long concentrated load of 36 kips applied 1 ft from the left end a = 1 ft b = 5 ft W 8 x 24
B. GRAPH:
========
36 kips
1ft I 5 ft
V
---------------------
v V
R 1 R2
6 ft
C. ASKED: ======== Investigate the beam for shear.
D. CALCULATIONS: ================ 1. P&A 109 Reactions R 1 = left side R 1 = P x b /L R 1 = 36 kips x 5ft / 6 ft R 1 = 30 kips R 2 = right side R 2 = P x a / L R 2 = 36 kips x 1 ft/ 6 ft R 2 = 6 kips 2. Shear The maximum vertical shear is equal to the value of the larger reaction - the usual case for a simple span beam. P&A 109 V = P x b / L V = 36 kips x 5ft / 6 ft V = 30 kips 3. P&A 172 Table 4.1 Properties of W-Shapes for given W 8 x 24 member d = 7.93 inches (overall depth of beam including flanges) t w = 0.245 web thickness 4. Gross Area of the Beam Web A w = d x tw A w = 7.93 in x 0.245 in A w = 1.94 sq in 5. Shear Stress fv = V / A w fv = 30 kips/ 1.94 sqin fv = 15.5 ksi fv = actual unit shear stress V = maximum vertical shear Aw = gross area of the beam web d = overall depth of the beam tw = thickness of the beam web 6. Allowable Shear Stress P&A 266 Table 8.1 at not reduced section F v = 0.40 Fy F v = 0.40 x 36 F v = 14.4 Ksi calculation F v = 14.5 Ksi from table 8.1 7. Shear Stress 15.5 > 14.5 fv > Fv actual > allowable actual value exceeds allowable W 8 x 24 is not acceptable. 8. We may want to consider S shapes with greater web thickness than W shapes MSC 1 - 36 S 8 x 23 depth d = 8 inches web thickness tw = 0.441 9. Area A w = 8 x 0.441 A w = 3.528 inches 10. Shear fv = V / Aw fv = 30 / 3.528 fv = 8.50 ksi 8.50 < 14.5 actual smaller allowable o.k. 11. To determine min. gross area of beam web Aw= V / Fv Aw = 30/ 14.5 Aw = 2.0689655 sq inches Aw = 2.069 sq inches 12. Select another W member P&A 172 Table 4.1 W 8 x 28 d = 8.06 in tw = 0.285 in Aw = 8.06 x 0.285 Aw = 2.2971 2.2971 > 2.069 o.k.
W 8 x 24 is not acceptable The following sections will be acceptable S 8 x 23 W 8 x 28
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©Dr. Gruenwald 1996, 1997
