Find: Lightest I Beam fy=50,000 [lb/in2] unfactored axial

Slides:



Advertisements
Similar presentations
Beam Design Beam Design Civil Engineering and Architecture
Advertisements

Design of Steel Flexural Members
Limit States Flexure Shear Deflection Fatigue Supports Elastic Plastic
Beams Stephen Krone, DSc, PE University of Toledo.
REVIEW OF STEEL DESIGN KNOWLEDGE BASE REQUIRED: STRENGTH OF MATERIALS
Reinforced Concrete Flexural Members
Chp-6:Lecture Goals Serviceability Deflection calculation
Chp.12 Cont. – Examples to design Footings
Design of Tension Members
Shear Design of Beams CE Steel Design Class By, Amit H. Varma.
Shear Design of Beams CE Steel Design Class By: Amit H. Varma.
CTC 422 Design of Steel Structures
Design of Tension Members
Distributed Loads Reference Chapter 4, Section 9
BOLTS TENSION MEMBERS AISC / LRFD WILD CARD
SAFE 605: Application of Safety Engineering Principles Strength of Materials.
Strength of Materials I EGCE201 กำลังวัสดุ 1 Instructor: ดร. วรรณสิริ พันธ์อุไร ( อ. ปู ) ห้องทำงาน : 6391 ภาควิชาวิศวกรรมโยธา
Lecture # 2 Allowable Stress Objective:
Tension Members Last Time
CTC / MTC 222 Strength of Materials
Combined Bending & Axial Forces (BEAM – COLUMNS)
EXAMPLE 9.2 – Part IV PCI Bridge Design Manual
Dr. Ali I. Tayeh First Semester
Footings.
University of Palestine
Ship Strength Stress & Strain Bending & Shear Moment of Inertia & Section Modulus.
FOOTINGS. FOOTINGS Introduction Footings are structural elements that transmit column or wall loads to the underlying soil below the structure. Footings.
LRFD – Floor beam Unbraced top flange. Lateral Torsion Buckling  We have to check if there is plastic failure (yielding) or lateral-torsion buckling.
LRFD-Steel Design Dr. Ali Tayeh Second Semester
Driven Pile Design George Goble. Basic LRFD Requirement η k Σ γ ij Q ij ≤ φ g R ng η k – factor for effect of redundancy, ductility and importance γ ij.
CTC / MTC 222 Strength of Materials
Footing.
Beam Design Beams are designed to safely support the design loads.
Mechanics of Elastic Materials. Why study mechanics? Useful for the analysis and design of load-bearing structures, such as: buildings bridges space shuttles.
STEEL DESIGN KNOWLEDGE BASE REQUIRED: STRENGTH OF MATERIALS STEEL DESIGN REVIEW OF TIMBER DESIGN BEARING PROBLEM TIMBER DESIGN SOIL MECHANICS REVIEW OF.
General Comparison between AISC LRFD and ASD
Structures Agenda: Forces & Architectural Form - review
Analysis and Design of Beams for Bending
Analysis and Design of Beams for Bending
ERT352 FARM STRUCTURES TRUSS DESIGN
62323: Architectural Structures II
Design of Beams for Flexure
Pure Bending.
UNIT-IV SHEAR,TORSION AND BOND.
contents Design of beams (week 11,12,13), (10,17,24 Nov.)
RELIABILITY IN DESIGN Prof. Dr. Ahmed Farouk Abdul Moneim.
CE Steel Design Class By, Amit H. Varma
Design of Beams - Limit States
STRESS DUE TO BENDING.
Structure I Course Code: ARCH 208 Dr. Aeid A. Abdulrazeg
Find: The Lightest I Beam
Chapter 13 – Buckling of Columns
Structural Member Properties
Analysis and Design of Beams for Bending
Find: Lightest Angled Beam
Units of N m.
Find: Ae [in2] dbolt=1/2 [in] tdamaged=1/32 [in] 2 [in] A) 0.3 plate
Find: ρc [in] from load after 2 years
Find: Fcr [lb/in2] Fcr E = 2.9 * 107 [lb/in2] fy = 36,000 [lb/in2]
REVIEW OF STEEL DESIGN KNOWLEDGE BASE REQUIRED: STRENGTH OF MATERIALS
Find: λc P P 1[in] fixed y 8 [in] A A’ 14[in] x 1[in] L z z fixed L/2
Analysis and Design of Beams for Bending
Find: φcPn [lb] column in compression E = 2.9 * 107 [lb/in2]
Find: Bearing Capacity, qult [lb/ft2]
Find: φ * Rn [lb] dhole=19/16 [in] fy= 36,000 [lb/in2]
Find: L [ft] L L d A) 25 B) 144 C) 322 D) 864 T = 13 [s] d = 20 [ft]
Find: M [k*ft] at L/2 A B w 5 w=2 [k/ft] 8 21 L=10 [ft] 33 L
KNOWLEDGE BASE REQUIRED:
Civil Engineering and Architecture
Deflection of Beams In addition to being safe, the structure must be serviceable ; Serviceable structure is one that performs satisfactorily, not causing.
Presentation transcript:

Find: Lightest I Beam fy=50,000 [lb/in2] unfactored axial tensile loads: DL=600,000 [lb] LL = 60,000 [lb] A) W 8 x 58 B) W10x60 C) W12x65 D) W10x68 Find the lightest I beam. [pause] In this problem, an I beam is subjected to a given dead load, and, --- assume yield failure

Find: Lightest I Beam fy=50,000 [lb/in2] unfactored axial tensile loads: DL=600,000 [lb] LL = 60,000 [lb] A) W 8 x 58 B) W10x60 C) W12x65 D) W10x68 live load, which is an axial force, in tension. The yield strength --- assume yield failure

Find: Lightest I Beam fy=50,000 [lb/in2] unfactored axial tensile loads: DL=600,000 [lb] LL = 60,000 [lb] A) W 8 x 58 B) W10x60 C) W12x65 D) W10x68 of the steel is provided, and the problem states to assume the I beam will experience --- assume yield failure

Find: Lightest I Beam fy=50,000 [lb/in2] unfactored axial tensile loads: DL=600,000 [lb] LL = 60,000 [lb] A) W 8 x 58 B) W10x60 C) W12x65 D) W10x68 a yield failure. [pause] Looking ahead, we’ll first observe our 4 choices --- assume yield failure

Find: Lightest I Beam fy=50,000 [lb/in2] unfactored axial tensile loads: DL=600,000 [lb] LL = 60,000 [lb] A) W 8 x 58 B) W10x60 C) W12x65 D) W10x68 of I beams to choose from. We’ll use I beam W 8 by --- assume yield failure

Find: Lightest I Beam fy=50,000 [lb/in2] assume yield failure unfactored axial tensile loads: DL=600,000 [lb] LL = 60,000 [lb] A) W 8 x 58 B) W10x60 C) W12x65 D) W10x68 58 as an example. Here, the W stands for --- W 8 x 58

Find: Lightest I Beam fy=50,000 [lb/in2] assume yield failure unfactored axial tensile loads: DL=600,000 [lb] LL = 60,000 [lb] A) W 8 x 58 B) W10x60 C) W12x65 D) W10x68 wide flange, which means this type of I beam will have an increased section modulus, --- wide flange W 8 x 58

Find: Lightest I Beam fy=50,000 [lb/in2] assume yield failure unfactored axial tensile loads: DL=600,000 [lb] LL = 60,000 [lb] A) W 8 x 58 B) W10x60 C) W12x65 D) W10x68 area moment of inertia, and other paramers related to section geometry. First number after --- wide flange W 8 x 58

Find: Lightest I Beam fy=50,000 [lb/in2] assume yield failure unfactored axial tensile loads: DL=600,000 [lb] LL = 60,000 [lb] A) W 8 x 58 B) W10x60 C) W12x65 D) W10x68 the W represents the depth of the beam, in inches, which is the distance from the top of the top flange, --- wide flange W 8 x 58 depth [in]

Find: Lightest I Beam fy=50,000 [lb/in2] assume yield failure depth unfactored axial tensile loads: DL=600,000 [lb] LL = 60,000 [lb] A) W 8 x 58 B) W10x60 C) W12x65 D) W10x68 to the bottom of the bottom flange. And the last number represents ---- wide flange W 8 x 58 depth [in]

Find: Lightest I Beam fy=50,000 [lb/in2] assume yield failure depth unfactored axial tensile loads: DL=600,000 [lb] LL = 60,000 [lb] A) W 8 x 58 B) W10x60 C) W12x65 D) W10x68 linear the weight per length, of the beam, in pounds per foot. Therefore, --- wide flange weight [lb/ft] W 8 x 58 depth [in]

Find: Lightest I Beam fy=50,000 [lb/in2] assume yield failure depth unfactored axial tensile loads: DL=600,000 [lb] LL = 60,000 [lb] A) W 8 x 58 B) W10x60 C) W12x65 D) W10x68 linear the linear weight of the 4 possible I beams to choose from ranges from --- wide flange weight [lb/ft] W 8 x 58 depth [in]

Find: Lightest I Beam fy=50,000 [lb/in2] assume yield failure depth unfactored axial tensile loads: DL=600,000 [lb] LL = 60,000 [lb] A) W 8 x 58 B) W10x60 C) W12x65 D) W10x68 linear 58 pounds per foot, to 68 pounds per foot. Since we’re trying to find --- wide flange weight [lb/ft] W 8 x 58 depth [in]

Find: Lightest I Beam fy=50,000 [lb/in2] assume yield failure depth unfactored axial tensile loads: DL=600,000 [lb] LL = 60,000 [lb] A) W 8 x 58 B) W10x60 C) W12x65 D) W10x68 linear the lightest I beam. We’ll choose the beam with the smallest linear weight, which can withstand --- wide flange weight [lb/ft] W 8 x 58 depth [in]

Find: Lightest I Beam fy=50,000 [lb/in2] assume yield failure depth unfactored axial tensile loads: DL=600,000 [lb] LL = 60,000 [lb] A) W 8 x 58 B) W10x60 C) W12x65 D) W10x68 linear the axial loading provided in the problem statement. [pause] Using LRFD, the maximum factored load, --- wide flange weight [lb/ft] W 8 x 58 depth [in]

Find: Lightest I Beam γ * Q ≤ φ * R DL=600,000 [lb] LL = 60,000 [lb] yield failure A) W 8 x 58 B) W10x60 C) W12x65 D) W10x68 gamma Q, shall not exceed the design strength of the steel member, ---

Find: Lightest I Beam γ * Q ≤ φ * R DL=600,000 [lb] LL = 60,000 [lb] factored design load strength yield failure A) W 8 x 58 B) W10x60 C) W12x65 D) W10x68 phi R. [pause] The maximum factored tensile load upon the member, ---

Find: Lightest I Beam γ * Q ≤ φ * R γ * Q = max +0.5 * (Lr or S or R) factored DL=600,000 [lb] load γ * Q ≤ φ * R LL = 60,000 [lb] 1.4 * (D+F) γ * Q = max 1.2 * (D+F+T)+1.6 * (L+H) +0.5 * (Lr or S or R) is the maximum computed load, from various load combinations. In these equations, D equals --- …

Find: Lightest I Beam γ * Q ≤ φ * R γ * Q = max +0.5 * (Lr or S or R) factored DL=600,000 [lb] dead load γ * Q ≤ φ * R load LL = 60,000 [lb] 1.4 * (D+F) live load γ * Q = max 1.2 * (D+F+T)+1.6 * (L+H) +0.5 * (Lr or S or R) the dead load, and L equals the live load. None of the other loads --- …

Find: Lightest I Beam … γ * Q ≤ φ * R γ * Q = max factored DL=600,000 [lb] dead load γ * Q ≤ φ * R load LL = 60,000 [lb] fluid 1.4 * (D+F) live load load γ * Q = max 1.2 * (D+F+T)+1.6 * (L+H) +0.5 * (Lr or S or R) soil are mentioned in the problem statement. Although there are 5 other --- load … rain roof snow load live load load

Find: Lightest I Beam γ * Q ≤ φ * R γ * Q = max +0.5 * (Lr or S or R) factored DL=600,000 [lb] dead load γ * Q ≤ φ * R load LL = 60,000 [lb] 1.4 * (D+F) live load γ * Q = max 1.2 * (D+F+T)+1.6 * (L+H) +0.5 * (Lr or S or R) load combination equations, we only have to consider the first 2 because they are the most conservative --- 1.2 * D +1.6 *(Lr or S or R) +(f1 * L or 0.8 * W) 1.2 * D +1.0 * E + f1*L + f2*S

Find: Lightest I Beam γ * Q ≤ φ * R γ * Q = max +0.5 * (Lr or S or R) factored DL=600,000 [lb] load γ * Q ≤ φ * R LL = 60,000 [lb] 1.4 * (D+F) γ * Q = max 1.2 * (D+F+T)+1.6 * (L+H) +0.5 * (Lr or S or R) when only dead loads and live loads are considered. After plugging in these loads, --- 1.2 * D +1.6 *(Lr or S or R) +(f1 * L or 0.8 * W) 1.2 * D +1.0 * E + f1*L + f2*S

Find: Lightest I Beam γ * Q ≤ φ * R γ * Q = max +0.5 * (Lr or S or R) factored DL=600,000 [lb] load γ * Q ≤ φ * R LL = 60,000 [lb] 1.4 * (D+F) γ * Q = max 1.2 * (D+F+T)+1.6 * (L+H) +0.5 * (Lr or S or R) for D and L, the factored load equals the maximum of ---

Find: Lightest I Beam γ * Q ≤ φ * R γ * Q = max γ * Q = max factored DL=600,000 [lb] load γ * Q ≤ φ * R LL = 60,000 [lb] 1.4 * (D+F) γ * Q = max 1.2 * (D+F+T)+1.6 * (L+H) +0.5 * (Lr or S or R) 840,000 pounds and 816,000 pounds, which is, --- 840,000 [lb] γ * Q = max 816,000 [lb]

Find: Lightest I Beam γ * Q ≤ φ * R γ * Q = max γ * Q = max factored DL=600,000 [lb] load γ * Q ≤ φ * R LL = 60,000 [lb] 1.4 * (D+F) γ * Q = max 1.2 * (D+F+T)+1.6 * (L+H) +0.5 * (Lr or S or R) 840,000 pounds. [pause] The problem states to assume --- 840,000 [lb] γ * Q = max 816,000 [lb] γ * Q = 840,000 [lb]

Find: Lightest I Beam γ * Q ≤ φ * R factored DL=600,000 [lb] load LL = 60,000 [lb] 840,000 [lb] design fy=50,000 [lb/in2] strength yield failure A) W 8 x 58 B) W10x60 C) W12x65 D) W10x68 a yield-type failure, which means the design strength, equals ---

Find: Lightest I Beam γ * Q ≤ φ * R 0.9 factored DL=600,000 [lb] load LL = 60,000 [lb] resistance 840,000 [lb] fy=50,000 [lb/in2] yield failure A) W 8 x 58 B) W10x60 C) W12x65 D) W10x68 0.9 for phi, and the resistance term, R, equals, ---

Find: Lightest I Beam γ * Q ≤ φ * R 0.9 factored DL=600,000 [lb] load LL = 60,000 [lb] 840,000 [lb] fy=50,000 [lb/in2] fy * Agross yield failure A) W 8 x 58 B) W10x60 C) W12x65 D) W10x68 the yield strength of the steel, f y, times the gross area of the steel beam, ---

Find: Lightest I Beam γ * Q ≤ φ * R 0.9 factored DL=600,000 [lb] load LL = 60,000 [lb] 840,000 [lb] fy=50,000 [lb/in2] fy * Agross yield failure yield gross strength area A) W 8 x 58 B) W10x60 C) W12x65 D) W10x68 A gross. From the problem statement, we know the yield stress of the steel, equals, ---

Find: Lightest I Beam γ * Q ≤ φ * R 0.9 factored DL=600,000 [lb] load LL = 60,000 [lb] 840,000 [lb] fy=50,000 [lb/in2] fy * Agross yield failure yield gross strength area A) W 8 x 58 B) W10x60 C) W12x65 D) W10x68 50,000 pounds per square inch. Therefore, if we simplify this inequality, ---

Find: Lightest I Beam γ * Q ≤ φ * R 0.9 factored DL=600,000 [lb] load LL = 60,000 [lb] 840,000 [lb] fy=50,000 [lb/in2] fy * Agross yield failure yield gross strength area and solve for A gross, we learn that the gross area of the I beam, --- 840,000 [lb] ≤0.9 * 50,000 [lb/in2] * Agross

Find: Lightest I Beam γ * Q ≤ φ * R 0.9 factored DL=600,000 [lb] load LL = 60,000 [lb] 840,000 [lb] fy=50,000 [lb/in2] fy * Agross yield failure yield gross strength area must be at least 18.7 inches squared. [pause] Looking back at --- 840,000 [lb] ≤0.9 * 50,000 [lb/in2] * Agross 18.7 [in2] ≤ Agross

Find: Lightest I Beam DL=600,000 [lb] A) W 8 x 58 B) W10x60 C) W12x65 D) W10x68 LL = 60,000 [lb] fy=50,000 [lb/in2] yield failure the 4 possible solutions, we can look up the gross area of each I beam, --- 840,000 [lb] ≤0.9 * 50,000 [lb/in2] * Agross 18.7 [in2] ≤ Agross

Find: Lightest I Beam Agross= 17.1 [in2] A) W 8 x 58 B) W10x60 C) W12x65 D) W10x68 Agross= 17.6 [in2] Agross= 19.1 [in2] Agross= 20.0 [in2] and determine the lightest I beam with a gross area at least 18.7 inches squared, is --- 840,000 [lb] ≤0.9 * 50,000 [lb/in2] * Agross 18.7 [in2] ≤ Agross

Find: Lightest I Beam Agross= 17.1 [in2] A) W 8 x 58 B) W10x60 C) W12x65 D) W10x68 Agross= 17.6 [in2] Agross= 19.1 [in2] Agross= 20.0 [in2] beam W 12 by 65. [pause] Therefore --- 840,000 [lb] ≤0.9 * 50,000 [lb/in2] * Agross 18.7 [in2] ≤ Agross

Find: Lightest I Beam Agross= 17.1 [in2] A) W 8 x 58 B) W10x60 C) W12x65 D) W10x68 Agross= 17.6 [in2] Agross= 19.1 [in2] Agross= 20.0 [in2] the answer is C. 840,000 [lb] ≤0.9 * 50,000 [lb/in2] * Agross 18.7 [in2] ≤ Agross answerC