Structural Steel Design (6th Edition)
6th Edition
ISBN: 9780134589657
Author: Jack C. McCormac, Stephen F. Csernak
Publisher: PEARSON
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Question
Chapter 4, Problem 4.24PFS
To determine
The size of a round rod using A36 steel.
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What size threaded round rod is required for member AC shown in Fig. P4-26? The given load is a service live load. Use A36 steel.
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Chapter 4 Solutions
Structural Steel Design (6th Edition)
Ch. 4 - Prob. 4.1PFSCh. 4 - Prob. 4.2PFSCh. 4 - Prob. 4.3PFSCh. 4 - Prob. 4.4PFSCh. 4 - Prob. 4.5PFSCh. 4 - Prob. 4.6PFSCh. 4 - Prob. 4.7PFSCh. 4 - Prob. 4.8PFSCh. 4 - Prob. 4.9PFSCh. 4 - Prob. 4.10PFS
Ch. 4 - Prob. 4.11PFSCh. 4 - Prob. 4.12PFSCh. 4 - Prob. 4.13PFSCh. 4 - Prob. 4.14PFSCh. 4 - Prob. 4.15PFSCh. 4 - Prob. 4.16PFSCh. 4 - Prob. 4.17PFSCh. 4 - Prob. 4.18PFSCh. 4 - Prob. 4.19PFSCh. 4 - Prob. 4.20PFSCh. 4 - Prob. 4.21PFSCh. 4 - Prob. 4.22PFSCh. 4 - Prob. 4.23PFSCh. 4 - Prob. 4.24PFSCh. 4 - Prob. 4.25PFSCh. 4 - Prob. 4.26PFS
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- Give me right solution according to the question. The tension member L 5 x 3 x 1/2 is used to carry the deadload of 42 kips and liveload of 23 kips with an 7/8 inch diameter attached. A36 is used. Determine its adequacy based on: a.LRFD b. ASDarrow_forwardDetermine the safe load P that the member below can carry using ASD and LRFD. The column is W18x60, A-50 steel. The column has an intermediate brace in the weak direction as shown. W18x60: A = 11400mm^2Rx = 189.64mmRy = 42.82mmFy = 350 MPaarrow_forwardFor the flat plate shown in the figure below (edge beam are not used), design the hatchedframe (middle and column strips) in the N-S direction.Given:- fc = 28MPa , f y = 420MPa- Super imposed dead load = 4 kN/m^2- Live Load = 3 kN/m^2arrow_forward
- Design a round spiral column to support an axial dead load of 450kN and an axial live load of 756kN. Assume 2.5% longitudinal steel is desired. Use fc' = 27.6MPa, fy = 276MPa (ties), fy = 414MPa (main bars - 20mm dia.)arrow_forwardA plate is used as a bracket and is attached to a column flange, as shown in Figure P8.2-3. Use an elastic analysis and compute the maximum bolt shear force.arrow_forwardFor a 6.56ft- long simply supported W shape A36-Steel Beam having a total concentrated-load of 18.9-K located at 2.187-ft from one of its supports (take It from right-side or left-side), you need to: 1. Find Fbx, Fv,(Sx) req and use the AISC Manual to find the designed-section (e.g.: W14×43) of steel beam? USE: W-shape Steel Beam, Steel A36 (Fy=36 ksi and Fu =58 ksi): P =18.9-K (Kips): and L =6.56ftarrow_forward
- Determine: a. max allowable Pu if the shearing if 0.5in bolts dia. governs b. max allowable Pu if the bearing of the tension member governs c. max allowable Pu if all "actions" are considered d. number of anchor bolts needed if an A992 steel bolt is used instead of A325. Assume force is equally shared by the bolts, and the anchor bolts are properly bonded on the ceilingarrow_forwardUse load and resistance factor design and select a W shape with a nominal depth of 10 inches (a W10) to resist a dead load of 175 kips and a live load of 175 kips. The connection will be through the flanges with two lines of 1 1/4–inch-diameter bolts in each flange, as shown. Each line contains more than two bolts. The length of the member is 30 feet. Use A242 steel.arrow_forwardA tension member with a length of 6 feet must resist a service dead load of 22 kips and a service live load of 55 kips. Select a member with a rectangular cross section. Use A36 steel and assume a connection with one line of 7/8-inch-diameter bolts. Use load combination Pu = 1.2D + 1.6L for LRFD.arrow_forward
- Consider R= 433.5 Calculate the factored load capacity of a double channel section member of A36 steel according to AISC LRFD Specification. The member is 4m long and consists of 2Cs 180x19.5, with flanges turned out with clear gap of 90mm. Assume that there can be as many as two 14mm rivets at any one cross-section (one in each flange). U = 0.80.arrow_forwardSelect the lightest avaliable W-10 section to support the axial loads PD= 90 kips, PL= 120 kips, Lc= 15ft, Fy= 50 ksi Answer is W10 x 45 LRFD and ASD using the colum table in Aisc manual especially part 4 please do a step by step solution I have many problems to work on and I want to understand the entire process.arrow_forwardCalculate the factored load capacity of a double channel section member of A36 steel according to AISC LRFD Specification. The member is 4m long and consists of 2Cs 180x18.2, with flanges turned out with clear gap of 90mm. Assume that there can be as many as two 14mm rivets at any one cross-section (one in each flange). U = 0.80. Ag= 2x2320= 4640mm2 U=0.80 L= 4m Clear gap= 90mm Tf= 9.3 Bf= 55 Tw=8.0arrow_forward
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