Structural Steel Design (6th Edition)
6th Edition
ISBN: 9780134589657
Author: Jack C. McCormac, Stephen F. Csernak
Publisher: PEARSON
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A W14X120 is used as a tension member in atruss. The flanges of the member are connected to a gusset plate by ¾ inch boltas shown below. Use A36 steel with Fy=36 ksi and Fu=58 ksi
Determine the Yielding Capacity of the section based on LRFD (kips)
Determine the Tensile Rupture capacity of the section based on LRFD
Determine the Demand to Governing Capacity Ratio (based on yielding and rupture only) if the Demand load carried by the section are DL=200 kips LL=400 kips use LRFD
4.3-4
Determine the available strength of the compression member shown in Figure P4.3-4.
in each of the following ways:
a. Use AISC Equation E3-2 or E3-3. Compute both the design strength for LRFD and
the allowable strength for ASD.
15
HSS 10x6x
ASTM A500, Grade B steel
(Fy=46 ksi)
2/3
7.9-2
A structural tee bracket is attached to a column flange with six bolts as
shown in Figure P7.9-2. All structural steel is A992. Check this connec-
tion for compliance with the AISC Specification. Assume that the bearing
strength is controlled by the bearing deformation strength of 2.4dtF.
a. Use LRFD.
b. Use ASD.
D = 30 k
L=70 k
WWw.ma
W12 X 65
34" diameter Group A
bearing-type bolts
with threads in shear
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- A 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.arrow_forwardIf the beam in Problem 5.5-9 i5 braced at A, B, and C, compute for the unbr Cb aced length AC (same as Cb for unbraced length CB). Do not include the beam weight in the loading. a. Use the unfactored service loads. b. Use factored loads.arrow_forwardA built-up section was made using PL414x12mm thk plates as shown in the figure below. It is pinned at both ends with additional support against weak axis at middle point. Assume A50 steel. PL414x12 DO Section W16x67 L x-axis a) Calculate moment of inertia at both axes in mm*. b) Determine the design compressive strength in kN if L-3m. c) Find the design compressive strength in kN if L=18m. Elevation y-axisarrow_forward
- A W14X120 is used as a tension member in atruss. The flanges of the member are connected to a gusset plate by 3/4 inch boltas shown below. Use A36 steel with Fy-36 ksi and Fu=58 ksi Determine the Yielding Capacity of the section based on LRFD (kips) Determine the Tensile Rupture capacity of the section based on LRFD Determine the Demand to Governing Capacity Ratio (based on yielding and rupture only) if the Demand load carried by the section are DL=200 kips LL=400 kips use LRFD Properties and Dimension Ag=35.30 in^2 x = 6.24 in ry= 3.74 in d=14.5 in tf=0.94 in bf=14.7 in tw=0.59 in k=1.54 d=14.5 Y k1=1.5 bf=14.7 tf-0.94 X -tw=0.59 Harrow_forwardQ2) The members of the truss structure shown below is plain concrete. The compressive strength of the concrete is 25 MPa. Compute the maximum load P that can be carried by the structure. (Cross section of each member of the truss is 200 x 200 mm and don't use material factors and do not consider slenderness) Comment on your results briefly. P A& 2m SC 2 m 1380 2m Darrow_forwardThree plates (14 mm and 16 mm in thickness) are welded to a W10 x 49 to form a built-up shape as shown. K.L= K.L= 7.6 m and F= 345 MPa, compute the design strength for LRFD. Mm x300mm WIOX49 14mm Piate Platearrow_forward
- 1.Use LRFD and design the tension members of the roof truss shown in Figure below. Use double-angle shapes throughout and assume 10-mm-thick gusset plates and welded connections. Assume a shear lag factor of U = 0.80. The trusses are spaced at 9 meters. Use A36 steel and design forthe following loads.Metal deck : 190 Pa of roof surfaceBuilt-up roof : 575 Pa of roof surfacePurlins : 145 Pa of roof surface (estimated)Roof Live Load : 960 Pa of horizontal projectionTruss weight : 240 Pa of horizontal projection (estimated) 2. Use A36 steel and design sag rods for the truss of Problem 1. Assume that, once attached, the metal deck will provide lateral support for the purlins; therefore, the sag rods need to be designed for the purlin weight only.a. Use LRFD.b. Use ASD.arrow_forwardCompute the nominal compressive strength of the member shown in Figure . Use AISC Equation E3-2 or E3-3.arrow_forwardW3D 25 Y= 155 %31 1oow= 2500 KN 27=310mm X= 475 Z= 60 %3D 102=600mmarrow_forward
- Design the reinforcements of the given T beam below. bf=900mm bw=420mm tf-120mm d=550mm d'=80mm fc'=34MPa fy=415MPa USE NSCP 2015 A.Mu = 1300kN-m, As = mm2 B.Mu = 1800kN-m, As = mm2, As' = mm2arrow_forwardThe truss shown is composed of steel sections. If the 2L150 x 90 x 8 A36 steel section was used in member CG, is it satisfactory? Use 2 lines of 4-20mm fasteners spaced 75 mm on center for all joints. The loads applied in the truss are. service loads. Use ASD only. You can use ASEP Manual. Multiply the loads by 10. 50 kN 30 kN- 3 m G 3 m 3 m 30 kN- D 30 kN- A m E H m Carrow_forwardН WWWN 160kN 140kN 120kN 100kN 130kN 150kN 8 panels @ 6m ea. = 48m 10.8m 8.1m D Determine the force in the members and identify each if (T) - tension or (C) - compression. HI= DG= DE= IJ = 419.444kN T HJ= НК - KJ = JM = 425kN T KL - EG GF EF FI = 419.444kN T FH = GH = || = 9.6m 10.5marrow_forward
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