Problem
A beam 100 mm wide is to be loaded with 3 kN concentrated loads spaced uniformly at
0.40 m on centers throughout the 5 m span. The following data are given:
Allowable bending stress = 24 MPa
Allowable shear stress = 1.24 MPa
Allowable deflection = 1/240 of span
Modulus of elasticity = 18,600 MPa
Weight of wood = 8 kN/m3
1. Find the depth d considering bending stress only.
2. Determine the depth d considering shear stress only.
3. Calculate the depth d considering deflection only.
Solution
Unit weight of wood
γ=8kNm3×1000 N1 kN×(1 m1000 mm)3γ=8kNm3×1000 N1 kN×(1 m1000 mm)3
γ=0.000008 N/mm3γ=0.000008 N/mm3
wo=3000400+0.000008(100d)wo=3000400+0.000008(100d)
wo=7.5+0.0008d N/mmwo=7.5+0.0008d N/mm
Part 1: Based on allowable bending stress (Fb = 24 MPa)
M=woL28M=woL28
M=(7.5+0.0008d)(50002)8M=(7.5+0.0008d)(50002)8
M=3125000(7.5+0.0008d) N⋅mmM=3125000(7.5+0.0008d) N⋅mm
Fb=6Mbd2Fb=6Mbd2
24=6[3125000(7.5+0.0008d)]100d224=6[3125000(7.5+0.0008d)]100d2
d=245.2 mmd=245.2 mm answer
Part 2: Based on allowable shear stress (Fv = 1.24 MPa)
V=R=woL2V=R=woL2
V=(7.5+0.0008d)(5000)2V=(7.5+0.0008d)(5000)2
V=2500(7.5+0.0008d)V=2500(7.5+0.0008d)
Fv=3V2bdFv=3V2bd
1.24=3[2500(7.5+0.0008d)]2(100d)1.24=3[2500(7.5+0.0008d)]2(100d)
d=232.44 mmd=232.44 mm answer
Part 3: Based on allowable deflection (δ = L/240)
δ=L240=5000240δ=L240=5000240
δ=1256mmδ=1256mm
δ=5woL4384EIδ=5woL4384EI
1256=5(7.5+0.0008d)(50004)384(18600)(100d312)1256=5(7.5+0.0008d)(50004)3
84(18600)(100d312)
d=268.9 mmd=268.9 mm answer
Problem
A timber beam 4 m long is simply supported at both ends. It carries a uniform load of 10
kN/m including its own weight. The wooden section has a width of 200 mm and a depth
of 260 mm and is made up of 80% grade Apitong. Use dressed dimension by reducing
its dimensions by 10 mm.
Properties of Apitong
Bending and tension parallel to grain = 16.5 MPa
Shear parallel to grain = 1.73 MPa
Modulus of elasticity in bending = 7.31 GPa
1. What is the maximum flexural stress of the beam?
2. What is the maximum shearing stress of the beam?
3. What is the maximum deflection of the beam?
Solution
Fb=16.5 MPaFb=16.5 MPa
Fv=1.73 MPaFv=1.73 MPa
E=7.31 GPaE=7.31 GPa
Part 1: Maximum flexural stress
M=woL28M=woL28
M=10(42)8M=10(42)8
M=20 kN⋅mM=20 kN⋅m
fb=6Mbd2fb=6Mbd2
fb=6(20)(10002)190(2502)fb=6(20)(10002)190(2502)
fb=10.105 MPafb=10.105 MPa answer
fb<Fbfb<Fb (okay)
Part 2: Maximum flexural stress
V=20 kNV=20 kN
fv=3V2bdfv=3V2bd
fv=3(20)(1000)2(190)(250)fv=3(20)(1000)2(190)(250)
fv=0.6316 MPafv=0.6316 MPa answer
fv<Fvfv<Fv (okay)
Part 3: Maximum deflection
I=bd312I=bd312
I=190(2503)12I=190(2503)12
I=247395833 mm4I=247395833 mm4
δ=5woL4384EIδ=5woL4384EI
δ=5(10)(4)(10004)384(7310)(247395833)δ=5(10)(4)(10004)384(7310)(247395833)
δ=18.43 mmδ=18.43 mm answer
Notching on Beams
NSCP 2001
When rectangular shaped girders, beams or joists are notched at points of supports on
the tension side, the horizontal shear stress at such point shall not exceed:
Fv=3V2bd′(dd′)Fv=3V2bd′(dd′)
Where
dd = total depth of beam
d′d′ = actual depth of beam at notch
When girder, beams or joists with circular cross section are notched at points of
support on the tension side, the actual shear stress at such point shall not exceed:
fv=3V2An(ddn)fv=3V2An(ddn)
Where:
AnAn = cross-sectional area of notched member
dd = total depth of beam
dndn = actual depth of beam at notch
When girders, beams or joists are notched at point of support on the
compression side, the shear at such point shall not exceed:
V=23Fvb[d−d−d′d′e]V=23Fvb[d−d−d′d′e]
Where:
dd = total depth of beam
d′d′ = actual depth of beam at notch
ee = distance notch extends inside the inner edge of support
NSCP 2010
When rectangular-shaped girder, beams or joists are notched at points of
support on the tension side, they shall meet the design requirements of that
section in bending and in shear. The horizontal shear stress at such point
shall be calculated by:
fv=3V2bd′(dd′)2fv=3V2bd′(dd′)2
Where:
dd = total depth of beam.
d′d′ = actual depth of beam at notch
Safe Uniform Load for a Beam that was Notched at the Tension Fibers at
Supports
Problem
A 75 mm × 150 mm beam carries a uniform load wo over the entire span of 1.2 m.
Square notches 25 mm deep are provided at the bottom of the beam at the supports.
Calculate the safe value of wo based on shear alone.
Allowable shear parallel to grain = 1.40 MPa
Allowable shear normal to grain = 1.85 MPa
Use the NSCP 2010 provision below:
NSCP 2010 Section 616.4: Horizontal Shear in Notched Beams
When rectangular-shaped girder, beams or joists are notched at points of support
on the tension side, they shall meet the design requirements of that section in
bending and in shear. The horizontal shear stress at such point shall be calculated
by:
fv=3V2bd′(dd′)2
fv=3V2bd′(dd′)2
Where:
dd = total depth of beam.
d′d′ = actual depth of beam at notch.
Solution
V=woL2=wo(1.2)2V=woL2=wo(1.2)2
V=0.6wo kNV=0.6wo kN
Fv=fvFv=fv ← Use Fv = Allowable shear parallel to grain
1.40=3V2bd′(dd′)21.40=3V2bd′(dd′)2
1.40=3(0.6wo)2(75)(275)(300275)21.40=3(0.6wo)2(75)(275)(300275)2
wo=26.96 kN/mwo=26.96 kN/m answer
