Bending
General Discussion | Analysis Example
- In general, bending induces compression and tension axial stresses through resolution of the internal bending moments
into a force couple as shown in the figure given below.
- These axial stresses act normal to the beam cross-section.
- In ordinary wood beams, these axial stresses are acting parallel to the grain of the member.
- The distribution of axial stresses is linear with maximum stresses at the extreme fibers (outer edges of beam) and zero stress at the beam's neutral axis.
- For simply supported members with normal downward external loads, the beam experiences a positive internal bending moment condition; resulting in compression in the top fibers above the neutral axis and tension in the bottom fibers below the neutral axis.
- The more common loading condition is to stress the beam about its strong axis, (also known here as the x-axis),
taking advantage of the larger moment of inertia that works to reduce bending stresses. This is the condition
shown in the following figure.
- In beam design, one important criterion is to examine the beam's critical bending stress conditions.
- This is done by ensuring that the actual critical stress, (e.g. fb), developed as a function of loading, support conditions, span, and cross-section is £ the factored allowable stress, (e.g. Fb').
- The following equation has been specialized for examining the maximum stress situation (at the extreme fibers)
for rectangular-shaped members.
- Since much of the bending problem, except for the adjustment factors, is a review from strength of materials, we will
jump right into a bending example problem.
- For this and the following topics, the provided example problem is an analytical one, and not a design problem.
- This is done to build your confidence with the basic stress techniques prior to the selecting and optimizing tasks that are done in design.
- The design aspects will be covered under the Design Example topic area of this Design module.
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Analysis Example: Determine if the given beam is adequate in bending.
The general solution procedure is to test if the following is true:
- The steps for getting the actual stress, fb, include:
- Loading diagram.
- Determine the critical bending moment.
- Calculate the actual beam stress.
- The loading diagram:
Incorporate beam self weight of approximately 6.2 plf.
For ease of analysis, this beam could be thought of as two separate beams:
- Critical bending moment:
Through the superposition* of maximum bending moments from a single point load case and a uniformly distributed load, the critical moment at mid-span is:
* Superposition is valid when the maximum moments for the two different load cases are located at the same place along the beam span, as is true here.
- Actual bending stress:
Where Sx is the section modulus with respect to the strong axis and equals:
Therefore, the maximum actual bending stress is:
- The allowable bending stress, Fb':
- The allowable bending stress is obtained by adjusting the tabulated base design values by the appropriate adjustment
factors as shown below:
Fb' = Fb CD CM Ct CL CF Cr Cfu- CD: Load duration factor
CM: Wet service facore
Ct: Temperature factor
CL: Beam stability factor
CF: Size factor
Cr: Repetitive member factor
Cfu: Flat-use factor
- Fb and adjustment factors, except CL
- Fb = 1000 psi.
- 1997 NDS Table 4A for a No. 1 DF-L member.
- CD = 1.25
- The governing load combination consists of dead and roof live loads.
- CM = 1.0
- In-service moisture content less than 19%
- Ct = 1.0
- In-service temperature is within ordinary ranges (less than 100 degrees F)
- CF = 1.3
- 1997 NDS Table 4A.
- Cr = 1.0
- Members spaces more than 24" o.c.
- Cfu = 1.0
- Member loaded in edge-wise bending.
- Fb = 1000 psi.
- Bending Stability Factor, CL.
- The beam stability factor equations follow the same general form of the column buckling expressions known
as the Ylinen column equations.
since point load condition dominates:
The adjusted allowable stress becomes:
Fb' = 1000 (1.25)(1.3)(.59)=959 psi
- The beam stability factor equations follow the same general form of the column buckling expressions known
as the Ylinen column equations.
- The allowable bending stress is obtained by adjusting the tabulated base design values by the appropriate adjustment
factors as shown below:
- Compare actual stress to allowable stress to determine if beam is adequate in bending:
fb = 1987 psi
Fb'= 959 psi
- Since fb > Fb', this beam is not adequate in bending.
- What could you do to change this design to make this beam work?
