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