1. BSE 2294 Animal Structures and Environment
Stresses in Wood
BSE 2294
Animal Structures and Environment
Dr. Susan Wood Gay & Dr. S. Christian Mariger

2. 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.

3. Wood members must resist five basic types of stresses.
Tension
Compression
Bearing
Shear
Bending
Testing bearing stress in a wood member.

4. 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.

5. 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.

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

7. Tensile Stress Example
Determine the tensile stress developed in a
2 by 4 under a load of 5000 lbs.

8. 1. Calculate the cross-sectional area (A) of the board.
A = W x L
A = 1.5 in x 3.5 in = in2

9. 2. Calculate the tensile stress (Ft).
Ft = P A
Ft = lbs = 952 psi
5.25 in2

10. 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.

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

12. Compressive Stress Example
Determine the maximum compressive load, parallel to the grain, that can be carried by a by 6 No. 2, dense, Southern Pine without failure.

13. 1. Find the maximum compressive force parallel (Fc||) to the grain from the Southern Pine Use Guide.
Fc|| = psi

14. 2. Calculate the cross-sectional area (A).
A = W x L
A = 1.5 in x 5.5 in = in2

15. 3. Calculate the maximum compressive load (P).
Fc|| = P A
P = Fc|| x A
P = lbs/in2 x in2 = 14,440 lbs

16. 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.

17. 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.

18. 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

19. 1. Calculate the contact area (A) where the bearing stress is applied.
A = W x L
A = 1.5 in x 3.5 in = in2

20. 2. Calculate the bearing stress (Fc^).
Fc^ = P A
Fc^ = lb = 191 psi
5.25 in2

21. 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.

22. 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.

23. 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.

24. 1. Calculate the shear area (As).
As = W x L
As = 1.5 in x 8 ft(12 in/1 ft) = 144 in2

25. 2. Calculate the shear stress (Fv).
Fv = P As
P = lb = psi
144 in2

26. 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.

27. 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.

28. 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.

29. 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.

30. 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.

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

32. 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.

33. 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

34. 1. Calculate the bending moment (M).
M = P x L (in)
M = 200 lb x 15 ft(12 in/ft) = 36,000 in-lb

35. 2. Calculate the bending stress (Fb).
Fb = 6M
bh2
Fb = (6)(36,000 in-lb) = psi
(6 in) (6 in)2

36. 2. Find the grade of lumber from Table #2 of the Southern Pine Use Guide.
Fb > 1000 psi for No. 1

37. Deflection Example #1
Determine the deflection of the post used in the previous problem caused by the lb load.

38. 1. Calculate the moment of inertia (I).
I = bh3 12
I = (6 in) (6 in)3 = 108 in4

39. 2. Calculate deflection (Δ).
Δ = PL3
3EI
Δ = (200 lb) [(15 ft)(12in/ft)] = 2.4 in
(3)( lb/in2)(108 in4)

40. Deflection Example #2
Determine the bending stress and the deflection if the post is a dressed 6 by 6.

41. 1. Calculate the bending stress (Fb).
Fb = 6M
bh2
Fb = (6)(36,000 in-lb) = psi
(5.5 in)(5.5 in)2

42. 2. Calculate the moment of inertia (I).
I = bh3 12
I = (5.5 in)(5.5 in)3 = in4

43. 3. Calculate deflection (Δ).
Δ = PL3
3EI
Δ = (200 lb) [(15 ft)(12in/ft)] = 3.4 in
(3)( lb/in2)(76.3 in4)

44. 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.

45. 1. Calculate the load on one floor joist.
W = (60 lb/ft2)(ft/12 in)2 x (16 in) = 6.7 lb/in

46. 2. Calculate the moment of inertia (I).
I = bh3 12
I = (1.5 in)(11.25 in)3 = 178 in4

47. 3. Calculate the deflection (Δ).
Δ = 5WL4
384EI
Δ = (5)(6.67 lb/in)(168 in) = in
(384)(1,200,000 lb/in2)(178 in4)

48. Δactual < Δmax , design is adequate
4. Check design criteria.
Δmax = L/360
Δmax = 168 in/360 = in
Δactual = in
Δactual < Δmax , design is adequate

49. 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.

50. 1. Calculate the bending moment (M).
M = WL2 8
M = (6.7 lb/in)(168 in)2 = 23,638 in-lb

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

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