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STEEL DESIGN
FLEXURAL MEMBERS
Power Exercises 5.0 1.0 Find the allowable bending stress for the following shapes. Assume adequate lateral support for the compression flange. a) W30 x 132 132 of A36 Steel Steel Properties of W30 x 132 section 2 A = 25,0 25,097 97 mm tw = 15.6 mm Sx = 6227 6227.0 .08 8 x 10 10 3 mm3 bf = 267.8 mm Ix = 2401 2401.6 .66 6 x 10 10 6 mm4 Sy = 609. 609.60 60 x 103 mm3 d = 769.9 mm Iy = 81.58 1.58 x 10 106 mm4 tf = 25.4 mm w = 1.93 kN/m b) W12 x 65 of of A242 Steel Steel 2
A = 12,3 12,323 23 mm bf = 304.8 mm d = 307.8 mm tf = 15.4 mm
Properties of W12 x 65 section tw = 9.9 mm Ix = 221.85 x 10 6 mm4 Iy = 72.42 x 10 6 mm4 w = 0.95 kN/m
Sx = 1440.42 x 10 3 mm3 Sy = 476.86 x 10 3 mm3
Ans.
a) 163.68 163.68 MPa MPa ; b) 224.23 224.23 MPa MPa
2.0
A W21 x 50 spans 11.0 m on a simple span. The compression flange is laterally supported at the third points. points. A36 Steel is used. Properties of W21 x 50 section 2 3 3 A = 9,48 9,484 4 mm tw = 9.7 mm Sx = 1548 1548.5 .58 8 x 10 10 mm 6 4 3 3 bf = 165.9 mm Ix = 409. 409.57 57 x 10 mm Sy = 125. 125.20 20 x 10 mm 6 4 d = 529.1 mm Iy = 10.36 0.36 x 10 10 mm tf = 13.6 mm w = 0.73 kN/m a) Determine the allowable bending stress of this beam. b) Determine the allowable superimposed uniformly distributed load be place on this beam. Ans. a) 118.93 118.93 MPa MPa ; b) 11.4 11.45 5 kN/m kN/m
3.0
Select the lightest W shape beam shown in the figure below. Assume full lateral support and A36 steel. Consider moment only. P = 26.7 kN
W = 43.8 kN/m
8.5 m Ans.
4.0
W24 x ___
A steel beam having a simple span of 8 m is subjected to to a counterclockwise moment at at the left end and 25% of M (counterclockwise) at the right end. The steel section has the following properties. bf = 0.210 m rt = 0.053 m 3 Sx = 0.00206 m tf = 0.016 m d = 0.533 m Fy = 248 MPa a. Determine the maximum slenderness slenderness ratio which the beam beam would be considered long. long. b. Compute the allowable bending bending stress. c. Compute the moment at the the right end. Ans. a) 146.14 146.14 ; b) 98.2 98.21 1 MPa; MPa; c) 50.5 50.58 8 kN.m kN.m
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STEEL DESIGN
FLEXURAL MEMBERS
5.0
A cantilever beam has a span of 4 m and carries a uniformly distributed load throughout its span. bf = 0.210 m rt = 0.053 m 3 Sx = 0.002077 m tf = 0.016 m d = 0.533 m Fy = 248 MPa a. Compute the allowable bending stress. b. Compute the maximum uniform load that can be carried by the beam. Ans. a) 132.13 MPa; b) 274.43 kN.m
6.0
A simple supported beam has a span of 10 m. The beam used is a W33 x 240 with an A242 steel with Fy = 345 MPa. Compute the safe uniformly distributed load that can be safely carried by the beam. Properties of W33 x 240 section 2 3 3 A = 45548 mm tw = 21.08 mm Sx = 13322.68 x 10 mm bf = 402.97 mm Ix = 5660.75 x 10 6 mm4 Sy = 1933.67 x 10 3 mm3 d = 850.9 mm Iy = 388.34 x 10 6 mm4 tf = 35.56 mm w = 3.51 kN/m a. If the compression flange is braced laterally. b. If the compression flange has lateral support only at its ends. c. If the compression flange has lateral supports only at its ends and the midspan. Ans. a) 239.18 kN/m ; b) 145.0 kN/m; c) 217.11 kN/m
7.0
Select the lightest W shape beam shown in the figure below. The depth of the beam is limited to 900 mm (36 in) maximum.
W = 102 kN/m
10 m
8.0
Consider the following cases: a. A36 steel, Lb = 1.2 m b. A36 steel, Lb = 5 m c. A441 steel, Lb = 6 m The W18 x 50 beam shown in the figure below has been designed foe moment. The uniform load includes the beam’s weight. Assume A36 steel and full lateral support. 89 kN
267 kN
223 kN
W=45 kN/m
0.9 m 1.2 m
1.2 m
0.9 m
3m
Calculate the following: a. Actual web shear stress b. Allowable shear stress Ans.
a) 84.59 MPa; b) 99.20 MPa
22
STEEL DESIGN
FLEXURAL MEMBERS
Properties of W18 x 50 section A = 9,484 mm tw = 9.0 mm Sx = 1456.81 x 10 3 mm3 bf = 190.5 mm Ix = 332.99 x 10 6 mm4 Sy = 175.34 x 10 3 mm3 d = 456.9 mm Iy = 16.69 x 10 6 mm4 tf = 14.5 mm w = 0.73 kN/m Calculate the value of C b on all unbraced length. The bracings are located at the reactions and at each applied concentrated load. 2
9.0
26.7 kN
2.4 m
8.9 kN
26.7 kN
2.4 m
2.4 m
2.4 m
Calculate the following: a. Actual web shear stress b. Allowable shear stress Ans.
10.0
1.75; 1.06; 2.26; 1.75 (in order form left to right)
Refer to figure on the next page. The beams are used to carry the dead loads and live loads coming from the slab having thickness of 150 mm. All beams are assumed to have full lateral support. Given:
S 1 = S 2 = S 3 = S 4 = 7.0 m; Superimposed dead load Live load
S 5 = S 6 = S 7 = S 8 = 2.8 m = 6.0 kPa = 4.8 kPa
Beam Properties (W12 x 65) A = 12,323 mm tw = 9.9 mm Sx = 1440.42 x 10 3 mm3 bf = 304.8 mm Ix = 221.85 x 10 6 mm4 Sy = 476.86 x 10 3 mm3 d = 307.8 mm Iy = 72.42 x 10 6 mm4 tf = 15.4 mm w = 0.95 kN/m Assume that the beam-girder joints and girder column joints are simple supports. 2
a) b) c) d) e) f) g) h) Ans.
What is the maximum dead load shear on beam BGMRW caused by the loads? What is the maximum live load shear on beam BGMRW caused by the loads? Apply moving loads What is the design shear of beam BGMRW? What is the maximum dead load moment on beam BGMRW caused by the loads? What is the maximum live load moment on beam BGMRW caused by the loads? Apply moving loads What is the design moment of beam BGMRW? Calculate the shear capacity of the beam BGMRW. Calculate the moment capacity of the beam BGMRW.
a) -117.44 kN; b) -58.33 kN; c) 175.77 kN; d) -145.04 kN.m; e) -79.36 kN.m; f) -224.40 kN.m; g) 302.28 kN; h ) 235.77 kN.m
23
STEEL DESIGN
FLEXURAL MEMBERS
C
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O
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FRAMING PLAN Determine the flexural strength of a W14 x 68 of A242 steel subject to Properties of W14 x 68 section 2 A = 45548 mm tw = 21.08 mm Sx = 13322.68 x 10 3 mm3 bf = 402.97 mm Ix = 5660.75 x 10 6 mm4 Sy = 1933.67 x 10 3 mm3 d = 850.9 mm Iy = 388.34 x 10 6 mm4 tf = 35.56 mm w = 3.51 kN/m a. Continuous lateral support. b. An unbraced length of 6 m with C b = 1.0. c. An unbraced length of 9 m with C b = 1.0. Ans. a) 384.33 kN.m ; b) 304.24 kN.m; c) 202.83 kN.m
