BSE 2294 Animal Structures and Environment 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 Bending Testing 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/A Cross-sectional area (A) P Example of a wood member under tension.

The bottom chords of triangular trusses are normally under tension. P P Example of the bottom chord of a truss under tension.

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 Cross-sectional area (A) P P Example of a wood member in compression.

Posts are an example of wood members under compression. Example of a post under compression.

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 = Fc|| x A P = 1750 lbs/in2 x 8.25 in2 = 14,440 lbs

Example of bearing force on a wood member. 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 P Contactarea (A) Example of bearing force on a wood member.

Rafters often carry a bearing loads. P Example of bearing stress on a structural member – where the bottom chord of the truss meets the rafter.

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

Example of bearing force on a wood member. 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/As Shear area (As) P P Example of bearing force on a wood member.

Vertical and horizontal shear in a wood member. 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

Example of bending in a wood member. 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 P Example of bending in a wood member.

The moment of inertia of a rectangle. 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 h b The moment of inertia of a rectangle.

Cantilever beam, concentrated load. The modulus of elasticity (E) is the amount a member will deflect in proportion to an applied load. Stiffness of material Effect of cross-sectional shape on resistance to bending Slope of stress-strain curve, E (lb/in2) Yield point Stress E Strain Cantilever beam, concentrated load.

The equations for a simply supported beam, concentrated load are: Bending moment, M = PL 4 Deflection, Δ = PL3 48EI Note: L, b, and h are in inches. P L Simply supported, concentrated load.

The equations for a simply supported beam, distributed load are: Bending moment, M = WL2 8 Deflection, Δ = 5WL4 384EI L W (lb/ft) Simply supported, distributed load.

The equations for a cantilever beam, concentrated load are: Bending moment, M = PL Deflection, Δ = PL3 3EI P L Cantilever beam, concentrated load.

The equations for a cantilever beam, distributed load are: Bending moment, M = WL2 2 Deflection, Δ = WL4 8EI L W (lb/ft) Cantilever beam, distributed load.

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 6 in x 6 in rough-cut post 15 ft

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

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

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

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)

Δactual < Δmax , design is adequate 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

2. Calculate the bending stress (Fb). Fb = 6M bh2 Fb = (6)(23,638 in-lb) = 747 psi (1.5 in)(11.25 in)2

Fbactual < Fbmax , design is adequate 3. Check design criteria. Fbmax = 1250 psi Fbactual = 747 psi Fbactual < Fbmax , design is adequate