stresses in wood
DESCRIPTION
Stresses in Wood. BSE 2294 Animal Structures and Environment. Dr. Susan Wood Gay & Dr. S. Christian Mariger. Stress is the internal resistance of a material to external forces. Stresses in this barn’s roof components are resisting the external force caused by the snow load. - PowerPoint PPT PresentationTRANSCRIPT
Stresses in Wood
BSE 2294
Animal Structures and Environment
Dr. Susan Wood Gay & Dr. S. Christian Mariger
Stress is the internal resistance of a material to external forces.
Stresses in this barn’s roof components are resisting the external force caused by the snow load.
Wood members must resist five basic types of stresses.
• Tension
• Compression
• Bearing
• Shear
• BendingTesting bearing stress in a wood member.
Stress in wood are adjusted for numerous factors.
• Moisture content
• Duration of load
• Size and shape of cross-sectional area
• Fire retardant treatment
• Repetitive use
Wood roof trusses in a dairy barn.
Tension is a pulling force acting along the length of the member.
• Parallel to the grain
• Tensile stress (Ft):
– Tensile force on a member
– Divided by the cross-sectional area of member
• Ft = P/AExample of a wood member under tension.
PP
Cross-sectional area (A)
The bottom chords of triangular trusses are normally under tension.
Example of the bottom chord of a truss under tension.
PP
Tensile Stress Example
Determine the tensile stress developed in a
2 by 4 under a load of 5000 lbs.
1. Calculate the cross-sectional area (A) of the board.
A = W x L
A = 1.5 in x 3.5 in = 5.25 in2
2. Calculate the tensile stress (Ft).
Ft = P A
Ft = 5000 lbs = 952 psi 5.25 in2
Compression is a pushing force acting along the length of a member.
• Parallel to the grain
• Compressive stress (Fc):
– Compressive force on member
– Divided by the cross-sectional area of member
• Fc = P/A
Example of a wood member in compression.
PP
Cross-sectional area (A)
Posts are an example of wood members under compression.
Example of a post under compression.
P
P
Compressive Stress Example
Determine the maximum compressive load, parallel to the grain, that can be carried by a 2 by 6 No. 2, dense, Southern Pine without failure.
1. Find the maximum compressive force parallel (Fc) to the grain from the Southern Pine Use Guide.
Fc = 1750 psi
2. Calculate the cross-sectional area (A).
A = W x L
A = 1.5 in x 5.5 in = 8.25 in2
3. Calculate the maximum compressive load (P).
Fc = P A
P = Fcx A
P = 1750 lbs/in2 x 8.25 in2 = 14,440 lbs
Bearing is a pushing force transmitted across the width of a structural member.
• Perpendicular to the grain
• Bearing stress (Fc):
– Bearing force on member
– Divided by the contact area
• Fc= P/A
Example of bearing force on a wood member.
Contactarea (A)
P
Rafters often carry a bearing loads.
Example of bearing stress on a structural member – where the bottom chord of the truss meets the rafter.
P
Bearing Stress Example
Determine the bearing stress developed in the beam caused by
the loading as shown. The 2 by 6 beam is supported on
both ends by 2 by 4’s.
1000 lb
1. Calculate the contact area (A) where the bearing stress is applied.
A = W x L
A = 1.5 in x 3.5 in = 5.25 in2
2. Calculate the bearing stress (Fc).
Fc = P A
Fc = 1000 lb = 191 psi 5.25 in2
Shear force is the force that produces an opposite, but parallel, sliding motion of planes in a member.
• Parallel to grain
• Shear stress (Fv):
– Shear force on beam– Divided by the shear
area (area parallel to the load)
• Fv = P/AsExample of bearing force on a wood member.
P
P
Shear area (As)
Shear stresses can occur in either the horizontal or vertical direction.
Horizontal shear
Vertical shear
Vertical and horizontal shear in a wood member.
Shear Stress Example
Determine the horizontal shear stress in an 8 ft long, 2 by 4 caused by a shear force of 5,000 lbs.
1. Calculate the shear area (As).
As = W x L
As = 1.5 in x 8 ft(12 in/1 ft) = 144 in2
2. Calculate the shear stress (Fv).
Fv = P As
P = 5000 lb = 34.7 psi 144 in2
Bending force is a force applied to a member in such a way as cause the member to bend.
• Perpendicular to longitudinal axis
• Bending stress (Fb):
– Bending moment (M) on beam
– Divided by section modulus (bh2/6) of the beam
• Fb = 6M/bh2
Example of bending in a wood member.
P
The moment of inertia (I) is related to how an object’s mass is distributed as it rotates around an axis.
• For a rectangle, I = bh3 (in4)
12
• b = beam thickness
• h = beam width
The moment of inertia of a rectangle.
b
h
The modulus of elasticity (E) is the amount a member will deflect in proportion to an applied load.
Cantilever beam, concentrated load.
• Stiffness of material
• Effect of cross-sectional shape on resistance to bending
• Slope of stress-strain curve, E (lb/in2)
Strain
Str
ess
Yieldpoint
E
The equations for a simply supported beam, concentrated load are:
P
L
Simply supported, concentrated load.
• Bending moment, M = PL 4
• Deflection, Δ = PL3
48EI
• Note: L, b, and h are in inches.
The equations for a simply supported beam, distributed load are:
L
W (lb/ft)
Simply supported, distributed load.
• Bending moment, M = WL2
8
• Deflection, Δ = 5WL4
384EI
The equations for a cantilever beam, concentrated load are:
P
L
Cantilever beam, concentrated load.
• Bending moment, M = PL
• Deflection, Δ = PL3
3EI
The equations for a cantilever beam, distributed load are:
Cantilever beam, distributed load.
L
W (
lb/ft
)
• Bending moment, M = WL2
2
• Deflection, Δ = WL4
8EI
Bending Stress Example #1
Determine the quality (No. 1, 2, or 3) of Southern pine required
for a post that would be used for the application in the figure
below.
200 lb
15 ft6 in x 6 inrough-cut
post
1. Calculate the bending moment (M).
M = P x L (in)
M = 200 lb x 15 ft(12 in/ft) = 36,000 in-lb
2. Calculate the bending stress (Fb).
Fb= 6M bh2
Fb = (6)(36,000 in-lb) = 1000 psi (6 in) (6 in)2
2. Find the grade of lumber from Table #2 of the Southern Pine Use Guide.
Fb > 1000 psi for No. 1
Deflection Example #1
Determine the deflection of the post used in the previous problem caused by the 200 lb load.
1. Calculate the moment of inertia (I).
I = bh3
12
I = (6 in) (6 in)3 = 108 in4
12
2. Calculate deflection (Δ).
Δ = PL3
3EI
Δ = (200 lb) [(15 ft)(12in/ft)]3 = 2.4 in (3)(1500000 lb/in2)(108 in4)
Deflection Example #2
Determine the bending stress and the deflection if the post is a dressed 6 by 6.
1. Calculate the bending stress (Fb).
Fb= 6M bh2
Fb = (6)(36,000 in-lb) = 1298 psi (5.5 in)(5.5 in)2
2. Calculate the moment of inertia (I).
I = bh3
12
I = (5.5 in)(5.5 in)3 = 76.3 in4
12
3. Calculate deflection (Δ).
Δ = PL3
3EI
Δ = (200 lb) [(15 ft)(12in/ft)]3 = 3.4 in (3)(1500000 lb/in2)(76.3 in4)
Deflection Example #3
Determine the deflection in a 14-ft long, 2 by 12 floor joist caused by a total load of 60 psf.
The joists are on 16” on center (OC)
E = 1,200,000 psi.
The design criteria are Δmax = L/360.
1. Calculate the load on one floor joist.
W = (60 lb/ft2)(ft/12 in)2 x (16 in) = 6.7 lb/in
2. Calculate the moment of inertia (I).
I = bh3
12
I = (1.5 in)(11.25 in)3 = 178 in4
12
3. Calculate the deflection (Δ).
Δ = 5WL4
384EI
Δ = (5)(6.67 lb/in)(168 in)4 = 0.33 in (384)(1,200,000 lb/in2)(178 in4)
4. Check design criteria.
Δmax = L/360
Δmax = 168 in/360 = 0.47 in
Δactual = 0.33 in
Δactual < Δmax , design is adequate
Bending Stress Example #2
Would the floor joist from the previous problem work if the design is based on stress? For No. 1 Southern pine, Fb =1250 psi.
1. Calculate the bending moment (M).
M = WL2
8
M = (6.7 lb/in)(168 in)2 = 23,638 in-lb 8
2. Calculate the bending stress (Fb).
Fb= 6M bh2
Fb = (6)(23,638 in-lb) = 747 psi (1.5 in)(11.25 in)2
3. Check design criteria.
Fbmax = 1250 psi
Fbactual = 747 psi
Fbactual < Fbmax , design is adequate