A WF16x67 is made from A572 grade 50 steel. A ¾ in. cover plate, 9in. wide, is welded to the top flange. Find the plastic moment strength of the beam in ft-kips: (a) With no cover plate (b) If the cover plate has Fy = 50 ksi
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A WF16x67 is made from A572 grade 50 steel. A ¾ in. cover plate, 9in. wide, is welded to the top flange. Find the plastic moment strength of the beam in ft-kips:
(a) With no cover plate
(b) If the cover plate has Fy = 50 ksi
(c) If the cover plate has Fy = 75 ksi.
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- Q20 The moment connection shown generates a maximum tensile or compressive load of 222.8 kips at each flange plate connection with the W14x145 column. Determine if stiffeners are needed by checking the applicable concentrated force strengths. The column is A992 steel and the connection occurs at the middle of the 20ft tall column. Use LRFD. PL /16 × 9 W14 × 145 W24 × 76 PL /16 X 9 a) Failure by flange local bending b) Failure by web local yielding c) Failure by web local crippling d) Column is OK without stiffenersF7-21. Determine the maximum force P that can be applied to the rod if it is made of material having a yield stress of oy - 250 MPa. Consider the possibility that failure occurs in the rod and at section a-a. Apply a factor of safety of F.S. - 2 against yielding. 40 mm 50 mm 120 mim |60 mm Section a-aAn 80-foot long plate girder (see below) is fabricated from a ½-inch x 78-inch web and two 3inch x 22-inch flanges. Continuous lateral support is provided. The steel is A992. The loading consists of a uniform service dead load of 1.0 kip/ft (including the self-weight), a uniform service live load of 2.0 kips/ft, and a concentrated service live load of 500 kips at midspan. Stiffeners are placed at each end and at 4 feet, 16 feet, and 28 feet from each end. Once stiffener is placed at midspan. Determine whether the flexural strength is adequate using LRFD.
- From the Truss member shown, design all the tension members for ASD and LRFD, if the connection for the angle bars are shown. Use A36 STEEL. 100 mm 150 mm E F 2.5 m B SERVICE DEAD LOAD =300 KN SERVICE LIVE LOAD = 200 KN 2 m 2 m 2 m 2 mA girder supports concentrated dead and live loads. Determine the following required strengths for the girder based on the controlling load combinations.Use statics or Table 3-22a from the Steel Manual. The maximum bending moment, MD (kip-ft), due to dead load. The maximum bending moment, ML (kip-ft), due to live load. ASD required flexural strength, Ma (kip-ft) LRFD required flexural strength, Mu (kip-ft)Q3) The beam shown in Figure below has lateral support at the ends only. The concentrated loads are live loads. Use A992 steel and select a W shape. (Do not check deflections. Use C-1). 23 k 25 k ttst
- Current Attempt in Progress A pin-connected structure is supported and loaded as shown. Member ABCD is rigid and is horizontal before the load Pis applied. Bars (1) and (2) are both made from steel (E - 30600 ksi) and both have a cross-sectional area of 0.87 in.?. Assume L;-87 in, Lg-111 in., a-50 in., b-93 in, and c-39 in. If the normal stress in each steel bar must be limited to 19.4 ksi, determine the maximum load Pthat may be applied to the rigid bar. (2) (1) D Answer: P- kips.From the Truss member shown, design all the tension members for ASD and LRFD, if the connection for the angle bars are shown. Use A36 STEEL. 100 mm 150 mm 2.5 m B SERVICE DEAD LOAD =300 KN SERVICE LIVE LOAD = 200 KN 2 m 2 m 2 m 2 m4-47. The support consists of a solid red brass C83400 post surrounded by a 304 stainless steel tube. Before the load is applied the gap between these two parts is 1 mm. Determine the greatest axial load that can be applied to the rigid cap A without causing yielding of any one of the materials. A 0.25 m = 1 mm 60 mm 80 mm Prob. 4-47 -10 mm
- How much service live load, in kN/m, can be supported? The member weight is the only dead load. The axial compressive load consists of a service dead load of 45 kN and a service live load of 90 kN. Do not consider moment amplification. Use NSCP 2015. a. b. LRFD. ASD. wt A kN/m mm² W460x144 1.41 18400 472 13.6 283 SECTION tw bf tf Ix Sx mm mm mm mm³ 22.2 726 3080 mm mm mm 199 6.2m W460x144 Sy Ty Zx ly mm mm³ mm 83.6 591 Zy J Cw mmº mm³ mm³ 67.4 3450 906 2440 4230 mmªThe cross-sectison of a simply supported plate girder is shown in figure. The loading on the grider is symmetrical. The bearing stiffeners at supports are the sole means of providing restraint against torsion. Design the bearing stiffeners at supports. with minimum moment of inertia about the center line of web palte only as the sole design criterion. The flat section available are: 250 x 25, 250 x 32, 200 x 28, and 200 x 32 mm. Draw a sketch 500 25 8 .1445 Dimensions in mm 20 +425A plate girder must be designed for the conditions shown in Figure P10.7-4. The given loads are factored, and the uniformly distributed load includes a conservative estimate of the girder weight. Lateral support is provided at the ands and at the load points. Use LRFD for that following: a. Select the, flange and web dimensions so that intermediate stiffeners will he required. Use Fy=50 ksi and a total depth of 50 inches. Bearing stiffeners will be used at the ends and at the load points, but do not proportion them. b. Determine the locations of the intermediate stiffeners, but do not proportion them.