Beam Design - FLOOR JOIST

Similar documents
Beam Design - Pine Tree

Beam Design - Awning

Beam Design - Shed Roof Back Wall Beam-S

Beam Design - Trotin Project

Structural Calculations For:

DES140: Designing for Lateral-Torsional Stability in Wood Members

STRENGTH AND STIFFNESS REDUCTION OF LARGE NOTCHED BEAMS

The first NDS (1944) was based on allowable stress design (ASD). Copyright American Wood Council. All rights reserved.

Introduction to Structural Member Properties

Substituting T-braces for continuous lateral braces on wood truss webs

Lecture 7 Two-Way Slabs

DESIGN EXAMPLES APPENDIX A

Wood Design. fv = shear stress fv-max = maximum shear stress Fallow = allowable stress Fb = tabular bending strength = allowable bending stress

5. What is the moment of inertia about the x - x axis of the rectangular beam shown?

Section Downloads. Section Downloads. Handouts & Slides can be printed. Other documents cannot be printed Course binders are available for purchase

2018 NDS Changes. National Design Specification for Wood Construction (STD120)

Properties of Sections

This procedure covers the determination of the moment of inertia about the neutral axis.

Physics 8 Monday, November 23, 2015

SERVICEABILITY LIMIT STATE DESIGN

Serviceability Deflection calculation

2018 North Carolina Residential Code Prescriptive Tables for Selection of Support Elements for Beams, Girders, and Headers: Example Problems

Steel Cross Sections. Structural Steel Design

CIV 207 Winter For practice

Physics 8 Wednesday, November 29, 2017

Steel Post Load Analysis

Section Downloads. Section Downloads. Handouts & Slides can be printed. Course binders are available for purchase. Download & Print. Version 2.

CH. 4 BEAMS & COLUMNS

HELIODYNE SOLAR COLLECTOR RACK STRUCTURES FOR HELIODYNE, INC. Structural calculations. Gobi 410 at 45 degrees. for WCM HELIODYNE RACK

Solution: The moment of inertia for the cross-section is: ANS: ANS: Problem 15.6 The material of the beam in Problem

FAILURE TIME OF LOADED WOODEN BEAMS DURING FIRE

Section Downloads. Design Process. Design Principles Outline. Basic Design Principles. Design Process. Section 06: Design Principles.

Allowable Design Stresses (psi)

NAME: Given Formulae: Law of Cosines: Law of Sines:


Physics 8 Monday, November 20, 2017

STRUCTURAL ANALYSIS CHAPTER 2. Introduction

Chapter (6) Geometric Design of Shallow Foundations

thirteen wood construction: column design ARCHITECTURAL STRUCTURES: FORM, BEHAVIOR, AND DESIGN DR. ANNE NICHOLS SUMMER 2017 lecture

A q u a b l u e a t t h e G o l d e n M i l e

Part 1 is to be completed without notes, beam tables or a calculator. DO NOT turn Part 2 over until you have completed and turned in Part 1.

Beam Design and Deflections

Chapter 9: Column Analysis and Design

Chapter 4 Seismic Design Requirements for Building Structures

Standard Specification for Computing Reference Resistance of Wood-Based Materials and Structural Connections for Load and Resistance Factor Design 1

MECE 3321: Mechanics of Solids Chapter 6

Homework No. 1 MAE/CE 459/559 John A. Gilbert, Ph.D. Fall 2004

PUNCHING SHEAR CALCULATIONS 1 ACI 318; ADAPT-PT

*Refer to IBC Section , applicable when fall protection is required. Glass stresses are designed for a safety factor of of 4.0 (IBC ).

1 Exterior Wall Members & Accessories

1 Exterior Wall Members & Accessories

SPECIFIC VERIFICATION Chapter 5

Samantha Ramirez, MSE

Stress Analysis Lecture 4 ME 276 Spring Dr./ Ahmed Mohamed Nagib Elmekawy

8 Deflectionmax. = 5WL 3 384EI

PURE BENDING. If a simply supported beam carries two point loads of 10 kn as shown in the following figure, pure bending occurs at segment BC.

Evaluation of the Mechanical Properties of Douglas-fir Lumber and It s Structural Glulam by Non-destructive Techniques

Design of Reinforced Concrete Structures (II)

THEORETICAL DESIGN OF A NAILED OR BOLTED JOINT UNDER LATERAL LOAD 1. Summary

MTE 119 STATICS FINAL HELP SESSION REVIEW PROBLEMS PAGE 1 9 NAME & ID DATE. Example Problem P.1

5 G R A TINGS ENGINEERING DESIGN MANUAL. MBG Metal Bar Grating METAL BAR GRATING MANUAL MBG METAL BAR GRATING NAAMM

ε t increases from the compressioncontrolled Figure 9.15: Adjusted interaction diagram

Company : May 17, 2015 Designer : 9:34 PM Job Number : Checked By:

Failure in Flexure. Introduction to Steel Design, Tensile Steel Members Modes of Failure & Effective Areas

3.032 Problem Set 1 Fall 2007 Due: Start of Lecture,

Lab Exercise #5: Tension and Bending with Strain Gages

TABLE OF CONTANINET 1. Design criteria. 2. Lateral loads. 3. 3D finite element model (SAP2000, Ver.16). 4. Design of vertical elements (CSI, Ver.9).

Lecture-05 Serviceability Requirements & Development of Reinforcement

FLOW CHART FOR DESIGN OF BEAMS

MAXIMUM SUPERIMPOSED UNIFORM ASD LOADS, psf SINGLE SPAN DOUBLE SPAN TRIPLE SPAN GAGE

[8] Bending and Shear Loading of Beams

Mechanics of Solids notes

CHAPTER 5. T a = 0.03 (180) 0.75 = 1.47 sec 5.12 Steel moment frame. h n = = 260 ft. T a = (260) 0.80 = 2.39 sec. Question No.

Case Study in Reinforced Concrete adapted from Simplified Design of Concrete Structures, James Ambrose, 7 th ed.

Symmetric Bending of Beams

Structural Calculations

Method of elastic line

Longitudinal strength standard

4.3 Moment Magnification

Failure modeling of sawn lumber with a fastener hole

2018 WFCM Changes. Wood Frame Construction Manual for One- and Two-Family Dwellings (STD350)

Flexure: Behavior and Nominal Strength of Beam Sections

Shear Forces And Bending Moments

Mechanics of Materials

CHAPTER 4: BENDING OF BEAMS

Chapter Objectives. Copyright 2011 Pearson Education South Asia Pte Ltd

Roadway Grade = m, amsl HWM = Roadway grade dictates elevation of superstructure and not minimum free board requirement.

Chapter 7: Bending and Shear in Simple Beams

Entrance exam Master Course

24' SEE DWG FOR SCREWING 2 PLY TOGETHER NOTE: FLOOR TRUSSES 19.2" O.C. **ALL FLOOR TRUSSES ARE THE SAME** 1'5"6 1'7"3 1'7"3 1'7"3 1'7"3 1'7"3

PROBLEM 5.1 SOLUTION. Reactions: Pb L Pa L. From A to B: 0 < x < a. Pb L Pb L Pb L Pbx L. From B to C: a < x < L Pa L. Pa L. L Pab At section B: M = L

DL CMU wall = 51.0 (lb/ft 2 ) 0.7 (ft) DL beam = 2.5 (lb/ft 2 ) 18.0 (ft) 5

DEFLECTION CALCULATIONS (from Nilson and Nawy)

QUESTION BANK SEMESTER: III SUBJECT NAME: MECHANICS OF SOLIDS

Characteristics of a Force Loads on Structures. Dead Load. Load Types Dead Live Wind Snow Earthquake. Load Combinations ASD LRFD

Design of Reinforced Concrete Beam for Shear

Beam Bending Stresses and Shear Stress

QUESTION BANK DEPARTMENT: CIVIL SEMESTER: III SUBJECT CODE: CE2201 SUBJECT NAME: MECHANICS OF SOLIDS UNIT 1- STRESS AND STRAIN PART A

MECHANICS OF MATERIALS Sample Problem 4.2


Transcription:

Beam Design - FLOOR JOIST 1. Beam Data Load Type: Uniform Dist. Load Support: Simple Beam Beam Type: Sawn Lumber Species: Douglas Fir-Larch Grade: DF No.2 Size: 2 x 10 Design Span (L): 11.83 ft. Clear Span: 11.67 ft. Total Span: 12.00 ft. Bearing (l b ): 2 in. Quantity (N): 1 3. Design Options Lateral Support: braced Defl. Limits: 360 240 Load Duration: 1.15 Exposure: dry Temperature: T <= 100 F Orientation: Vertical Incised Lumber: No Rep. Members: Yes 2. Design Loads Live Load: 40 plf Dead Load: 12 plf Selfweight: 39.0 lbs Dist. Selfweight: 3.30 plf Total Weight: 39.5 lbs 4. Design Assumptions and Notes Code Standard: IBC 2015, NDS 2015 Bending Stress: Parallel to Grain Notes: 5. Adjustment Factors Factor Description F b F t F v F c F c E/E min C D Load Duration Factor 1.15 1.15 1.15 1.15 - - C M Wet Service Factor 1 b 1 1 1 c 1 1 C t Temperature Factor 1 1 1 1 1 1 C L Beam Stability Factor 1 - - - - - C F Size Factor 1.1 1.1-1 - - C fu Flat Use Factor 1.2 d - - - - - C i Incising Factor 1 1 1 1 1 1 C r Repetitive Member Factor 1.15 - - - - - a) Adjustment factors per AWC NDS 2015 and NDS 2015 Supplement. b) When (F b)(c F) 1,150 psi, C M = 1.0. c) When (F c)(c F) 750 psi, C M = 1.0. d) Only applies when sawn lumber or glulam beams are loaded in bending about the y-y axis. ph. (425) 741-5555 www.medeek.com 1

6. Beam Calculations Determine reference design values, sectional properties and self weight of beam: A = b x d,, b = Breadth of rectangular beam in bending (in.) d = Depth of rectangular beam in bending (in.) A = Cross sectional area of beam (in. 2 ) S x = Section modulus about the X-X axis (in. 3 ) S y = Section modulus about the Y-Y axis (in. 3 ) I x = Moment of inertia about the X-X axis (in. 4 ) I y = Moment of inertia about the Y-Y axis (in. 4 ) b = 1.500 in. d = 9.250 in. A = 1.500 x 9.250 = 13.88 in. 2 S x = (1.500)(9.250) 2 /6 = 21.39 in. 3 S y = (1.500) 2 (9.250)/6 = 3.47 in. 3 I x = (1.500)(9.250) 3 /12 = 98.93 in. 4 I y = (1.500) 3 (9.250)/12 = 2.60 in. 4 Reference Design Values from Table 4A NDS Supplement (Reference Design Values for Visually Graded Dimension Lumber, 2" - 4" thick). Species & Grade F b F t F v F c F c E Emin G DF No.2 900 575 180 625 1350 1600000 580000 0.5 The following formula shall be used to determine the density of wood (lbs/ft 3. (NDS Supplement Sec. 3.1.3) ρ w = Density of wood (lbs/ft 3 G = Specific gravity of wood (dimensionless) m.c. = Moisture content of wood (percentile) G = 0.5 m.c. = 19 % (Max. moisture content at dry service conditions) ph. (425) 741-5555 www.medeek.com 2

= 34.20 lbs/ft 3 Volume total = N[A x (L + l b )] = 1 x [13.88 x (142.00 + 2)] x (12 in./ft.) 3 = 1.16 ft 3 Volume span = N[A x L] = 1 x [13.88 x 142.00] x (12 in./ft.) 3 = 1.14 ft 3 Total Weight (W T ) = ρ w x Volume total = 34.20 x 1.16 = 39.5 lbs Self Weight (W S ) = ρ w x Volume span = 34.20 x 1.14 = 39.0 lbs Distributed Self Weight (w s ) = = 3.30 plf Load, Shear and Moment Diagrams: 12.00 ft. 40 plf 12 plf w s = 3.30 plf 11.67 ft. 11.83 ft. 331.76 lbs 331.76 lbs ph. (425) 741-5555 www.medeek.com 3

1.) Bending: Members subject to bending stresses shall be proportioned so that the actual bending stress or moment shall not exceed the adjusted bending design value: f b F b ' (NDS Sec. 3.3.1) f b = M / S F b ' = F b (C D )(C M )(C t )(C L )(C F )(C i )(C r ) Beam is braced laterally along its compression edge. Laterial stability is not a consideration: C L = Beam Stability Factor = 1.0 F bx ' = (900)(1.15)(1)(1)(1)(1.1)(1)(1.15) = 1309.3 psi f b = = 542.9 psi f b = 542.9 psi < F bx ' = 1309.3 psi (CSI = 0.41) ph. (425) 741-5555 www.medeek.com 4

2.) Shear: Members subject to shear stresses shall be proportioned so that the actual shear stress parallel to grain or shear force at any cross section of the bending member shall not exceed the adjusted shear design value: f v F v ' (NDS Sec. 3.4.1) f v = F v ' = F v (C D )(C M )(C t )(C i ) F vx ' = (180)(1.15)(1)(1)(1) = 207.00 psi Shear Reduction: For beams supported by full bearing on one surface and loads applied to the opposite surface, uniformly distributed loads within a distance, d, from supports equal to the depth of the bending member shall be pemitted to be ignored. For beams supported by full bearing on one surface and loads applied to the opposite surface, concentrated loads within a distance equal to the depth of the bending member from supports shall be permitted to be multiplied by x/d where x is the distance from the beam support face to the load. See NDS 2015, Figure 3C. f v * = = 30.76 psi f v * = 30.76 psi < F vx ' = 207.00 psi (CSI = 0.15) No Reduction in Shear (conservative): f v = = 35.37 psi f v = 35.37 psi < F vx ' = 207.00 psi (CSI = 0.17) 3.) Deflection: Bending deflections calculated per standard method of engineering mechanics for live load and total load: LL Allowable: L/360 TL Allowable: L/240 E x ' = E x (C M )(C t )(C i ) = 1600000(1)(1)(1) = 1600000 psi ph. (425) 741-5555 www.medeek.com 5

Δ LL = = 0.11 in. (L/d) LL = 142.00 / 0.11 = 1274 Δ LL = 0.11 in = L/1274 < L/360 Δ TL = = 0.15 in. (L/d) TL = 142.00 / 0.15 = 921 Δ TL = 0.15 in = L/921 < L/240 4.) Bearing: Members subject to bearing stresses perpendicular to the grain shall be proportioned so that the actual compressive stress perpendicular to grain shall be based on the net bearing area and shall not exceed the adjusted compression design value perpendicular to grain: f c F c ' (NDS Sec. 3.10.2) f c = F c ' = F c (C M )(C t )(C i ) F c x ' = (625)(1)(1)(1) = 625.00 psi A b = b x l b = 1.5 x 2 = 3.00 in 2 f c = = 110.6 psi f c = 110.6 psi < F c x ' = 625.00 psi (CSI = 0.18) *Disclaimer: The calculations produced herein are for initial design and estimating purposes only. The calculations and drawings presented do not constitute a fully engineered design. All of the potential load cases required to fully design an actual structure may not be provided by this calculator. For the design of an actual structure, a registered and licensed professional should be consulted as per IRC 2012 Sec. R802.10.2 and designed according to the minimum requirements of ASCE 7-10. The beam calculations provided by this online tool are for educational and illustrative purposes only. Medeek Design assumes no liability or loss for any designs presented and does not guarantee fitness for use. ph. (425) 741-5555 www.medeek.com 6