Steel Design (Activate Learning with these NEW titles from Engineering!)
6th Edition
ISBN: 9781337094740
Author: Segui, William T.
Publisher: Cengage Learning
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Question
Chapter 5, Problem 5.12.2P
To determine
(a)
The nominal flexural strength of the section
To determine
(b)
The percent reduction in strength of the section
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A 15" x 3/8" bar of A572 Gr. 50 steel is used as a tension member. It is connected to a gusset plate with 7/8-in diameter bolts as shown in the figure. Use s = 2.0 and g = 3.0.
a) Determine the allowable tensile strength of the section based on tensile rupture of the net area.b) Determine the allowable tensile strength of the section based on yielding of the gross area.
The angle L 8 x 8 x ½ in tension shown in the figure below must resist aservice dead load of 35 kips, live load of 50 kips, and a roof live load of 2kips. The steel used is A50 (Fy = 50 ksi, Fu = 65 ksi) and its welding length is 4”.Determine if the member has sufficient fracture and yield strength.
TENSION MEMBERS:
THE SINGLE 200 X 10 mm STEEL PLATE IS CONNECTED TO A 12 mm THICK STEEL PLATE BY FOUR 16 mm DIAMETER RIVETS AS SHOWN IN THE FIGURE. THE RIVETS
USED ARE A502 GRADE 2, HOT DRIVEN RIVETS. THE STEEL IS ASTM A36 WITH Fy = 248 MPa AND Fu = 400 MPa. DETERMINE THE VALUE OF P.
a. P BASED ON TENSION OF GROSS AREA
b. P BASED ON TENSION OF NET AREA
c. P BASED ON BEARING OF PROJECTED AREA
d. P BASED ON SHEAR RUPTURE (BLOCK SHEAR)
Chapter 5 Solutions
Steel Design (Activate Learning with these NEW titles from Engineering!)
Ch. 5 - Prob. 5.2.1PCh. 5 - Prob. 5.2.2PCh. 5 - Verify the value of Zx for a W1850 that is...Ch. 5 - Prob. 5.2.4PCh. 5 - Prob. 5.4.1PCh. 5 - Prob. 5.4.2PCh. 5 - Determine the smallest value of yield stress Fy,...Ch. 5 - Prob. 5.5.1PCh. 5 - Prob. 5.5.2PCh. 5 - Prob. 5.5.3P
Ch. 5 - Prob. 5.5.4PCh. 5 - Prob. 5.5.5PCh. 5 - Prob. 5.5.6PCh. 5 - Prob. 5.5.7PCh. 5 - Prob. 5.5.8PCh. 5 - Prob. 5.5.9PCh. 5 - If the beam in Problem 5.5-9 i5 braced at A, B,...Ch. 5 - Prob. 5.5.11PCh. 5 - Prob. 5.5.12PCh. 5 - Prob. 5.5.13PCh. 5 - Prob. 5.5.14PCh. 5 - Prob. 5.5.15PCh. 5 - Prob. 5.5.16PCh. 5 - Prob. 5.6.1PCh. 5 - Prob. 5.6.2PCh. 5 - Prob. 5.6.3PCh. 5 - Prob. 5.6.4PCh. 5 - Compute the nominal shear strength of an M107.5 of...Ch. 5 - Compute the nominal shear strength of an M1211.8...Ch. 5 - Prob. 5.8.3PCh. 5 - Prob. 5.8.4PCh. 5 - Prob. 5.10.1PCh. 5 - Prob. 5.10.2PCh. 5 - Same as Problem 5.10-2, except that lateral...Ch. 5 - Prob. 5.10.4PCh. 5 - The given beam is laterally supported at the ends...Ch. 5 - Prob. 5.10.6PCh. 5 - Prob. 5.10.7PCh. 5 - Prob. 5.11.1PCh. 5 - Prob. 5.11.2PCh. 5 - Prob. 5.11.3PCh. 5 - Prob. 5.11.4PCh. 5 - Prob. 5.11.5PCh. 5 - Prob. 5.11.6PCh. 5 - Prob. 5.11.7PCh. 5 - Prob. 5.11.8PCh. 5 - Prob. 5.11.9PCh. 5 - Prob. 5.12.1PCh. 5 - Prob. 5.12.2PCh. 5 - Prob. 5.12.3PCh. 5 - Prob. 5.13.1PCh. 5 - Prob. 5.13.2PCh. 5 - Prob. 5.14.1PCh. 5 - Prob. 5.14.2PCh. 5 - Prob. 5.14.3PCh. 5 - Prob. 5.14.4PCh. 5 - Prob. 5.15.1PCh. 5 - Prob. 5.15.2PCh. 5 - Prob. 5.15.3PCh. 5 - Prob. 5.15.4PCh. 5 - Prob. 5.15.5PCh. 5 - Prob. 5.15.6PCh. 5 - Prob. 5.15.7PCh. 5 - Same as Problem 5.15-7, except that the sag rods...
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- A steel plate is to be attached to a support with three bolts. The cross-sectional area of the plate is 800 mm² and the yield strength of the steel is 260 MPa. The ultimate shear strength of the bolts is 570 MPa. A factor of safety of 1.67 with respect to yield is required for the plate. A factor of safety of 4.0 with respect to the ultimate shear strength is required for the bolts. Determine the minimum bolt diameter required to develop the full strength of the plate. Note: consider only the gross cross-sectional area of the plate-not the net area. Support Plate ↓ Parrow_forward2. A steel plate is 360 mm wide and 20 mm thick with four bolts hole into the place as shown in the figure. Compute the following: a. Critical net area required by the NSCP specs. b. Max. critical net area required by the NSP specs. Note that, Max. net area is 85% of the critical net area c. Capacity of the joint if the allowable tensile stress is 0.75Fy. Use A36 steel Fy=248 MPa 90m²m 90 mm 90mm 45 45mm In my comin P Scanned with CamScanner CSarrow_forwardTopic:Bolted Steel Connection - Civil Engineering -Steel Design *Use latest NSCP/NSCP 2015 formula to solve this problem *Please use hand written to solve this problem A 12.5 mm x150 mm plate is connected to a gusset plate having a thickness of 9.5 mm.Diameter of bolt = 20 mm Both the tension member and the gusset plate are of A 36 steel. Fy = 248 MPa, FU = 400 MPa. Shear stress of bolt Fnv = 300MPa. Questions: a.Determine the shear strength of the connection. b.Determine the bearing strength of the connection. c. Determine the block shear strength of the connection.arrow_forward
- 2. A steel plate is 360 mm wide and 20 mm thick with four bolts hole into the place as shown in the figure. Compute the following: a. Critical net area required by the NSCP specs. b. Max. critical net area required by the NSP specs. Note that, Max. net area is 85% of the critical net area Capacity of the joint if the allowable tensile stress is 0.75Fy. Use A36 steel Fy=248 MPa 25 mm= DIAMETER OF BOLTS 45 45mm fmmy C. 90mm 90mm 90mm Comm Parrow_forwardA 15" x 3/8" bar of A572 Gr. 50 steel is used as a tension member. It is connected to a gusset plate with 7/8-in diameter bolts as shown in the figure. Use s = 2.0 and g = 3.0. Determine the design tensile strength of the section based on yielding of the gross area. Answer: |0 O O О оarrow_forwardDetermine the design tensile strength of the 12 in. x 1/2 in. steel plate shown in the figure. The bolts are 3/4 in. diameter. The steel is A572 Gr. 50. Check yielding and fracture. Check Block Shear. T 3in. 73im 13in 1 3in tarrow_forward
- 3. A plate with width of 300mm and thickness of 20mm is to be connected to two plates of the same width with half the thickness by 24mm diameter bolts, as shown. The rivet holes have a diameter of 2mm larger than the rivet diameter. The plate is A36 steel with yield strength F,-248MPa and ultimate strength F,-400MPa. a. Determine the design strength of the section. b. Determine the allowable strength of the section 24mm 30mmarrow_forwardQuestion 4 Complete the table below for the steel sections shown. All sections are bending about their horizontal axis. Section Location hep Ney Slenderness (show calculations) of element bf Flange te ld ++ tw be = 800 mm Web tf = 20 mm d = 1440 mm tw = 16 mm Grade 300, heavily 入。= Whole Asp= Asy= welded plates section br Flange tr d Web bf = 800 mm te = 20 mm d = 1420 mm tw = 16 mm Grade 300, heavily welded plates 入。= Whole Asp= Asy= sectionarrow_forwardDetermine the design tensile strength of the plate (200 × 100) mm connected to a 12 mm thick gussets, using 20 mm bolts shown below, if the yield and ultimate stress of the steel used are 250 MPa and 420 MPa respectively, fy= 250 MPa fu= 450 MPaarrow_forward
- An A36 tension member shown in the figure is connected with three 19 mm diamter bolts. Diameter of hole = 22mm. Properties of 150 mm x 100 mm x 12.5 mm angle A = 3065 mm2 y = 50.55 mm Fy = 248 MPa Fu = 400 MPa Determine the design strength due to yielding in the gross section. (kN)arrow_forwardSituation 2. Two plates each with thicknessF16mm are bolted together with6 -22mm dimater bolts forming a lap conne ction. Bolt spacing are as follows: S1 = 40mm, S2 = 80mm, Sa = 100mm. Bolt hole diameter=25 mm 50 250mm 30 30 60 75 Allowable stress: Tensile stess on gross area of the plate=0.60 Fy Tensile stress on net area of the plate=0.5Fy Shear Stress of the bolt Fv=120MPA Bearing Stress of the bot Fp=1.2 Fu Calculate the permiss ible tensile load P under the following Conditions: 4. Based on shear capacity of bolts 5. Based on bearing capacity of bolts 6. Based on block shear strength (taking into consideration the failure path given in the figure below) 40 80 16 mm 40 180 mm 40 Bearing Failure Path #1 140 209 Bearing Failure Path #2arrow_forwardDefine and describe local buckling. Enumerate and differentiate classifications of structural steel sections considering local buckling. I give upvote :)arrow_forward
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