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Wood Beam Design Based On NDS 2018: Input Data & Design Summary

This document summarizes the design of a 16x24 wood beam with a 10 foot span. It includes input parameters such as load values and material properties, as well as calculations for reactions, bending moment, shear force, deflections, and stress checks. The beam design was found to be adequate with bending and shear stresses below allowable values, deflections within limits, and combined stresses satisfying requirements.

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Ziyad Monier
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242 views2 pages

Wood Beam Design Based On NDS 2018: Input Data & Design Summary

This document summarizes the design of a 16x24 wood beam with a 10 foot span. It includes input parameters such as load values and material properties, as well as calculations for reactions, bending moment, shear force, deflections, and stress checks. The beam design was found to be adequate with bending and shear stresses below allowable values, deflections within limits, and combined stresses satisfying requirements.

Uploaded by

Ziyad Monier
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as XLSX, PDF, TXT or read online on Scribd
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PROJECT : PAGE :

CLIENT : DESIGN BY :
JOB NO. : DATE: REVIEW BY :
Wood Beam Design Based on NDS 2018
INPUT DATA & DESIGN SUMMARY
MEMBER SIZE 16 x 24 Select Structural, Douglas Fir-Larch

MEMBER SPAN L= 10 ft

UNIFORMLY DISTRIBUTED DEAD LOAD wD = 112 lbs / ft


UNIFORMLY DISTRIBUTED LIVE LOAD wL = 0
lbs / ft
CONCENTRATED DEAD LOADS PD1 = 1500 lbs
(0 for no concentrated load) L1 = 4 ft
PD2 = 3936 lbs
L2 = 5 ft
DEFLECTION LIMIT OF LIVE LOAD D L = L / 360 Camber => 0.01 inch
DEFLECTION LIMIT OF LONG-TERM DKcr D + L = L / 180
THE BEAM DESIGN IS ADEQUATE.
Does member have continuous lateral support by top diaphragm ?
(1= yes, 0= no) 0 No

Code Duration Factor, CD Condition Code Designation


1 0.90 Dead Load 1 Select Structural, Douglas Fir-Larch
2 1.00 Occupancy Live Load 2 No. 1, Douglas Fir-Larch
3 1.15 Snow Load 3 No. 2, Douglas Fir-Larch
4 1.25 Construction Load 4 Select Structural, Southern Pine
5 1.60 Wind/Earthquake Load 5 No. 1, Southern Pine
6 2.00 Impact Load 6 No. 2, Southern Pine
Choice => 4 Construction Load Choice => 1

ANALYSIS
DETERMINE REACTIONS, MOMENT, SHEAR
wSelf Wt = 79 lbs / ft RLeft = 3.82 kips RRight = 3.53 kips
VMax = 3.44 kips, at 23.5 inch from left end MMax = 15.22 ft-kips, at 5.01 ft from left end

DETERMINE SECTION PROPERTIES & ALLOWABLE STRESSES


b = 15.50 in E'min = 580 ksi E = Ex = 1600 ksi Fb* = 1856.0862 psi
d = 23.50 in FbE = 28784 psi Fb = 1,600 psi F = FbE / Fb* = 15.51
A = 364.3 in2 I = 16,763 in4 Fv = 170 psi Fb' = 1,850 psi
Sx = 1426.6 in3 RB = 4.917 < 50 E' = 1,600 ksi Fv' = 213 psi
lE = 20.6 (ft, Tab 3.3.3 footnote 1)
CD CM Ct Ci CL CF CV Cc Cr
1.25 1.00 1.00 1.00 1.00 0.93 1.00 1.00 1.00

CHECK BENDING AND SHEAR CAPACITIES


fb = MMax / Sx = 128 psi < Fb = 1850 psi [Satisfactory]
fv' = 1.5 VMax / A = 14 psi < Fv ' [Satisfactory]

CHECK DEFLECTIONS
D(L, Max) = 0.00 in, at 5.000 ft from left end, < DL = L / 360 [Satisfactory]
D(Kcr D + L , Max) = 0.01 in, at 4.900 ft from left end < DKcr D + L = L / 180 [Satisfactory]
Where Kcr = 1.50 , (NDS 3.5.2)

DETERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT)


D(1.5D, Max) = 0.01 in, at 4.900 ft from left end
CHECK THE BEAM CAPACITY WITH AXIAL LOAD
AXIAL LOAD F = 8.1 kips

THE ALLOWABLE COMPRESSIVE STRESS IS


Fc' = Fc CD CP CF = 1558 psi
Where Fc = 1100 psi
CD = 1.60
CF = 0.93 (Lumber only)
CP = (1+F) / 2c - [((1+F) / 2c)2 - F / c]0.5 = 0.954
Fc* = Fc CD CF = 1633 psi
Le = Ke L = 1.0 L = 120 in
b = 15.5 in
SF = slenderness ratio = 7.7 < 50 [Satisfies NDS 2018 Sec. 3.7.1.4]
FcE = 0.822 E'min / SF2 = 7954 psi
E'min = 580 ksi
F = FcE / Fc* = 4.870
c = 0.8
THE ACTUAL COMPRESSIVE STRESS IS
fc = F / A = 22 psi < Fc ' [Satisfactory]

THE ALLOWABLE FLEXURAL STRESS IS


Fb' = 2368 psi, [ for CD = 1.6 ]

THE ACTUAL FLEXURAL STRESS IS


fb = (M + Fe) / S = 216 psi < Fb ' [Satisfactory]

CHECK COMBINED STRESS [NDS 2018 Sec. 3.9.2]

(fc / Fc' )2 + fb / [Fb' (1 - fc / FcE)] = 0.092 < 1 [Satisfactory]

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