Steel Design (Activate Learning with these NEW titles from Engineering!)
6th Edition
ISBN: 9781337094740
Author: Segui, William T.
Publisher: Cengage Learning
expand_more
expand_more
format_list_bulleted
Concept explainers
Question
Chapter 4, Problem 4.4.2P
To determine
The nominal compressive strength
Expert Solution & Answer
Trending nowThis is a popular solution!
Students have asked these similar questions
Problem2.
The compression member is shown in figure. Find the following:
a. The Euler stress Fe.
b. The buckling stress Fcr
c. The design strength
d. The allowable strength
e. Does the member satisfactorily meet the design requirements?
Why?
HSS 8x 8x4
ASTM AS00, Grade B steel
(Fy = 46 ksi)
15'
2. A column HP 14 x 102 of A572 Gr. 55 steel has a length of 15 ft is fixed at both ends. Compute
the design compressive strength for LRFD and the allowable compressive strength for ASD.
(Steel section properties are provided in the next page)
ASTM
Designation
A572
Gr. 42
Gr. 50
Gr. 55
Gr. 60⁰
Gr. 65⁰
Yield
Stress
(ksi)
42
50
55
60
65
Fu
Tensile
Stressa
(ksi)
60
65
70
75
80
Estimate the cross-sectional area of a 350S125-27 cold-formed shape.
a. If the member is tested in tension, what would be the maximum force thesample could carry before reaching the yield strength if the steel has ayield strength of 225 MPa?b. Would you expect a 2.5 m stud to carry the same load in compression?(explain)
Chapter 4 Solutions
Steel Design (Activate Learning with these NEW titles from Engineering!)
Ch. 4 - Prob. 4.3.1PCh. 4 - Prob. 4.3.2PCh. 4 - Prob. 4.3.3PCh. 4 - Prob. 4.3.4PCh. 4 - Prob. 4.3.5PCh. 4 - Prob. 4.3.6PCh. 4 - Prob. 4.3.7PCh. 4 - Prob. 4.3.8PCh. 4 - Prob. 4.4.1PCh. 4 - Prob. 4.4.2P
Ch. 4 - Prob. 4.6.1PCh. 4 - Prob. 4.6.2PCh. 4 - Prob. 4.6.3PCh. 4 - Prob. 4.6.4PCh. 4 - Prob. 4.6.5PCh. 4 - Prob. 4.6.6PCh. 4 - Prob. 4.6.7PCh. 4 - Prob. 4.6.8PCh. 4 - Prob. 4.6.9PCh. 4 - Prob. 4.7.1PCh. 4 - Prob. 4.7.2PCh. 4 - Prob. 4.7.3PCh. 4 - Use A992 steel and select a W14 shape for an...Ch. 4 - Prob. 4.7.5PCh. 4 - Prob. 4.7.6PCh. 4 - Prob. 4.7.7PCh. 4 - The frame shown in Figure P4.7-8 is unbraced, and...Ch. 4 - Prob. 4.7.9PCh. 4 - Prob. 4.7.10PCh. 4 - Prob. 4.7.11PCh. 4 - Prob. 4.7.12PCh. 4 - Prob. 4.7.13PCh. 4 - Prob. 4.7.14PCh. 4 - Prob. 4.8.1PCh. 4 - Prob. 4.8.2PCh. 4 - Prob. 4.8.3PCh. 4 - Prob. 4.8.4PCh. 4 - Prob. 4.9.1PCh. 4 - Prob. 4.9.2PCh. 4 - Prob. 4.9.3PCh. 4 - Prob. 4.9.4PCh. 4 - Prob. 4.9.5PCh. 4 - Prob. 4.9.6PCh. 4 - Prob. 4.9.7PCh. 4 - Prob. 4.9.8PCh. 4 - Prob. 4.9.9PCh. 4 - Prob. 4.9.10PCh. 4 - Prob. 4.9.11PCh. 4 - Prob. 4.9.12P
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, civil-engineering and related others by exploring similar questions and additional content below.Similar questions
- Use 2015 NSCP. A compression member shown below with Fy = 50 ksi 30′ W12 x 87 A992 steel 1. Which of the following most nearly gives the value of critical buckling stress in ksi? 2. Which of the following most nearly gives the nominal strength in kips? 3. Considering ASD. Which of the following most nearly gives the allowable strength in kips? 4. Considering LRFD. Which of the following most nearly gives the design strength in kips?arrow_forwardQuestion 1: (Use NSCP 505.3 provisions and ASEP Steel Handbook for effective length factors) Use AISC Equation E3-2 or E3-3 and determine the nominal axial compressive strength for the following cases: a. L= 15 ft b. L=20 ft W10 × 33 A992 steel FIGURE P4.3-1arrow_forwardFor the column shown what is the nominal compressive strength (Pn) ? Pn=? W12x50 10 12X50 THT X combilevered WIarrow_forward
- In order to reinforce a column in an existing structure, two channels are continuously welded to the column as shown below. Fy = 50 ksi for both the column and the channels. The effective length with respect to each axis is 16 feet. a) calculate the Moment of Inertia about each axis of the built-up shape. a) what is the available axial compressive strength and what is the percentage increase in strength? Use both LRFD and ASD.arrow_forward2) Find the axial stresses of menbers FD, GD, GE State if it is tensile or Compressive. 4M 3M A LE 3m G 20 RN Go KNarrow_forwardTopic: COMBINED STRESS-AXIAL TENSION AND FLEXURE BENDING: STEEL DESIGN Please solve your Solution in a handwritten Note: It should be handwritten pleaseeee Questions A Tension member with no holes is subjected to axial loads of PD=68kN & PL=64kN.It is also subjected with bending moments of MDy=40kN-m & MLy=55kN-m. Is themember adequate? Steel is of A992 Gr 50 Specs. Use LRFD. Neglect the weight of the beam. Section properties:Lp = 1.863 md=459.99mmtw=9.02mmzx= 1835x10^3 mm^4Lr = 5.305 mLb = 4.8 mbf=191.26mmtf=16mmSx=1611x10^3 mm^4 Sy=195x10^3 mm^4Zy=303x 10^3 mm^4 Cb=1.32arrow_forward
- A flexural member is fabricated from two flanges plates 1" x 20" and a web plate 1/2" x 22". The yield stress of the steel is 36 ksi. a. Compute the elastic modulus and the yield moment with respect to major principal axis. b. Compute the plastic modulus and the plastic moment with respect to major principal axis.arrow_forwardAW18 x 40 standard steel shape is used to support the loads shown on the beam. Assume P = 18 kips, w = 4.6 kips/ft, LAB = 3.4 ft, LBc = 3.4 ft, and LCD= 16.6 ft. Determine the magnitude of the maximum bending stress in the beam. A P Answer: Omax LAB = B Save for Later LBC eTextbook and Media C ksi W LCD D X Attempts: 0 of 3 used Submit Answerarrow_forwardProblem 1. The composite beam shown below carries a cantilevered load of 10 kN. The beam consists of one 30 x 124 mm plate and four 12 x 50 mm plates. They are pinned together at 120 mm intervals with round pins. The pin material has a shear strength of 159 MPa. Compute the minimum acceptable diameter for the pins. O O O O O O O -0 O 0- O 1000 mm Do O -120 mm (typ) O O O P = 10 KN 30 x 124 mm 12 x 50 mm (typ)arrow_forward
- Ce tip.instructure.com Incognito (2) A composite section will be used to support a heavy axial load of P-800 kN with a First Semester SY 2021-2. rigid plate at the top of the section. Determine the average compressive stress in the brass section. Use h=16 mm. Answer in MegaPascals. Home Announcements Brass core (E = 105 GPa) Rigid end plate count Assignments Aluminum plates (E = 70 GPa) Discussions aboard Grades People urses Pages endar Files 300 mm Syllabus |auizes box Modules tory BigBlueButton (Conferences) elp Chat 60 mm T.I.P. Manila Library Video 40 mm Presentation Canvas LMS Satisfactorvarrow_forwardAn I-section with two cover plates is used as tension member and is subjected to a load of 120 t over a length of 16 m. If the allowable tensile stress is 15000 kg/cm², is the member design safe? If 9821.6 cm4, so, find the factor of safety. Use ISWB 300@ 48.1 kg/m, A = 61.33 cm², Ixx Ivy 990.1 cm4. There are two rows of fasteners in each flange, thus making atleast two holes in one section. The fastener diameter is 22 mm, cover plate size 260 mm x 8 mm, flange thickness of l-section = 10 mm. The center of fastener is located at 60 mm from center line of the l-section. = =arrow_forwardDetermine the safe load of the column section shown, if it has a yield strength of 25 MPa. E = 200000 MPa. Use NSCP Specifications. Fyz248 mpa Properties of Channel Section d = 305 mm t₂ = 7.2 mm A = 3929 mm² t₁ = 12.7 mm Ix=53.7 x 10mm¹ x = 117 mm Properties of W 460 x 74 A = 9450 mm² b = 190 mm ly= 1.61 x 10 mm x = 17.7 mm tw = 9.0 mm rx = 188 mm ry = 41.9 mm d = 457 mm tr = 14.5 mm Ix = 333 x 10 mm Iy = 16.6 x 10mm* 7.21 When the height of column is 6 m. When the height of column is 10 m. Assume K= 1.0 457 CIVIL ENGINEERING- STEEL DESIGNarrow_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