Bundle: Steel Design, Loose-leaf Version, 6th + Mindtap Engineering, 1 Term (6 Months) Printed Access Card
6th Edition
ISBN: 9781337761505
Author: William T. Segui
Publisher: Cengage Learning
expand_more
expand_more
format_list_bulleted
Question
Chapter 3, Problem 3.4.3P
To determine
(a)
The design strength.
To determine
(b)
The allowable strength.
Expert Solution & Answer
Trending nowThis is a popular solution!
Students have asked these similar questions
A double-angle shape is shown in the figure. The steel is
A36, and the holes are for 2-inch-diameter bolts.
Assume that A₂ = 0.75An.
Ae
1
2
Determine the design tensile strength for LRFD.
Determine the allowable strength for ASD.
CIVIL ENGINEERING
- STEEL DESIGN
AL
Section
2L5 x 3 x 516 LLBB
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 design tensile strength of the section based on yielding of the gross area.b) Determine the critical net area of the connection shown.
Question 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=
section
Chapter 3 Solutions
Bundle: Steel Design, Loose-leaf Version, 6th + Mindtap Engineering, 1 Term (6 Months) Printed Access Card
Ch. 3 - Prob. 3.2.1PCh. 3 - Prob. 3.2.2PCh. 3 - Prob. 3.2.3PCh. 3 - Prob. 3.2.4PCh. 3 - Prob. 3.2.5PCh. 3 - Prob. 3.2.6PCh. 3 - Prob. 3.3.1PCh. 3 - Prob. 3.3.2PCh. 3 - Prob. 3.3.3PCh. 3 - Prob. 3.3.4P
Ch. 3 - Prob. 3.3.5PCh. 3 - Prob. 3.3.6PCh. 3 - Prob. 3.3.7PCh. 3 - Prob. 3.3.8PCh. 3 - Prob. 3.4.1PCh. 3 - Prob. 3.4.2PCh. 3 - Prob. 3.4.3PCh. 3 - Prob. 3.4.4PCh. 3 - Prob. 3.4.5PCh. 3 - Prob. 3.4.6PCh. 3 - Prob. 3.5.1PCh. 3 - Prob. 3.5.2PCh. 3 - Prob. 3.5.3PCh. 3 - Prob. 3.5.4PCh. 3 - Prob. 3.6.1PCh. 3 - Prob. 3.6.2PCh. 3 - Prob. 3.6.3PCh. 3 - Select an American Standard Channel shape for the...Ch. 3 - Prob. 3.6.5PCh. 3 - Use load and resistance factor design and select a...Ch. 3 - Select a threaded rod to resist a service dead...Ch. 3 - Prob. 3.7.2PCh. 3 - Prob. 3.7.3PCh. 3 - Prob. 3.7.4PCh. 3 - Prob. 3.7.5PCh. 3 - Prob. 3.7.6PCh. 3 - Prob. 3.8.1PCh. 3 - Prob. 3.8.2PCh. 3 - Prob. 3.8.3PCh. 3 - Prob. 3.8.4PCh. 3 - Prob. 3.8.5P
Knowledge Booster
Similar questions
- A double-angle shape is shown in the figure. The steel is A36, and the holes are for 1/2-inch-diameter bolts. Assume the Ae = 0.75An. Section 215 x 3 x %6 LLBB Determine the design tensile strength for LRFD. Select one: а. 66 b. 156 c. 78 d. 132arrow_forwardA C8 × 11.5 is connected to a gusset plale with 22 mm diameter bolts as shown. The steel is A572 Grade 50. If the member is subjected to dead load and live load only, what is the total scrvice load capacity if the live-to-dead load ratio is 3? Assume that Ae = 0.85Ap. a. Use LRFD. b. Use ASD. Properties of A572 Grade 50: Fy = 345 MPa C8 × 11.5 Fu = 448 MPa Properties of C8 x 11.5: Ag = 2,181 mm? br = 57.4 mm d= 203.2 mm Lw = 5.6 mm lf = 9.9 mmarrow_forwardThe given angle bar L125x75x12 with Ag = 2,269 sq.mm. is connected to a gusset plate using 20 mm diameter bolts as shown in the figure. Using A36 steel with Fy = 248 MPa and Fu = 400 MPa, determine the following: 2. Determine the nominal tensile strength of the 12 mm thick, A36 angle bar shown based on: a. Gross yielding b. Tensile rupture Bolts used for the connection are 20 mm in diameter. O O O O O O O Effective net area of the tension member if the shear lag factor is 0.80. Select the correct response: 1,516.1 1,354.4 1,431.2 1,221.6arrow_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_forwardA bolted lap joint is shown in the figure below. The bolts are 19mm in diameter. Effective diameter of the holes is 23mm. The plates are 12 mm thick, x1 = 33 mm, x2 = 76 mm, x3 = 50 mm. Yield strength of plate, Fy = 248 MPa Ultimate tensile strength of plate, Fu = 400 MPa Required: Using allowable strength design: 1. Find the safe load “P” based on gross area yielding. 2. Find the safe load “P” based on net area rupture. 3. Find the safe load “P” based on block shear.arrow_forward3. 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_forward
- 4. Calculate the design strength (ocPn) of W24X76 with length of 12 ft. and pinned ends. A572 Grade50 steel is used. E=29x103 ksi. Show your work in detail. ASTM Classification A36 A572 Grade 50 A992 Grade 50 A500 Grade B (HSS rect, sq) A500 Grade B (HSS round) A53 Grade B Yield Strength F, (ksi) 36 50 50 46 42 35 Ultimate Strength F (ksi) 58 65 65 58 58 60arrow_forwardA double-angle shape is shown in the figure. The steel is A36, and the holes are for 1/2-inch-diameter bolts. Assume the Ae = 0.75An.a) Determine the allowable tensile strength for ASD.b) Determine the design tensile strength for LRFD.arrow_forwardAn A36 tension member shown in the figure is connected with three 19 mm diameter bolts. Dia. Of bolt hole = 22 mm. %3D 50, 100 100 99.45 50.55 100 Properties of 150 mm x 100 mm x 12.5 mm angle A - 3065 mm- y - 50.55 mm Fy - 248 MPa Fu 400 MPa Determine the design strength due to yielding in the gross section. (kN)arrow_forward
- Two plates each with thickness t = 16 mm are bolted together with 6 – 22 mm diameter bolts forming a lap connection. Bolt spacing are as follows: S1 = 40 mm, S2 = 80 mm, S3 = 100 mm. Bolt hole diameter = 25 mm. Using A36 steel having Fy = 248 MPa and Fu = 400 MPa. P P S - P S2 S2 DETERMINE: 1. Determine the allowable load Pa based on bearing at bolt holes. (ASD) 2. Determine the ultimate load Pu based on block shear.arrow_forwardA double-angle shape, 2L7×4×3/s , is used as a tension member. The two angles are connected to a gusset plate with 7/8-inch- diameter bolts through the 7-inch legs, as shown in Figure P3.4-5. A572 Grade 50 steel is used. a. Compute the design strength. b. Compute the allowable strength. 2/2" 7" 3" 1/2" 2L7 × 4 × 3/s FIGURE P3.4-5arrow_forwardA PL 38 X 6 tension member is welded to a gusset plate as shown. The steel is A36 (Fy = 36ksi, Fu = 58ksi). a. The design strength, Pu based on gross area b. The design strength, Pu based on effective area PL % x 6 3/8" 6" Cross Sectional area of PL3/8x6 a) Blank 1 b) Blank 2 Blank 1 Add your answer Blank 2 Add your answerarrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Steel Design (Activate Learning with these NEW ti...Civil EngineeringISBN:9781337094740Author:Segui, William T.Publisher:Cengage Learning
Steel Design (Activate Learning with these NEW ti...
Civil Engineering
ISBN:9781337094740
Author:Segui, William T.
Publisher:Cengage Learning