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Chapter Objectives
To develop methods for designing beams to resist both bending and shear loads.
To develop methods for determining the deflection of beams subject to bending and shear loads.
To use beam deflection methods to solve statically indeterminate equilibrium problems.
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In-class Activities
Prismatic beam design
Fully stressed beam design
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PRISMATIC BEAM DESIGN
Basis of beam design
Strength concern (i.e. provide safety margin to normal/shear stress limit)
Serviceability concern (i.e. deflection limit – See Chapter 12)
Section strength requirement
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4. PRISMATIC BEAM DESIGN (cont)
Choices of section:
Steel sections e.g. AISC standard
W 460 X 68
(height = 459 ≈ 460 mm
weight = 0.68 kN/m)
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5. PRISMATIC BEAM DESIGN (cont)
Wood sections
Nominal dimensions (in multiple of 25mm) e.g. 50 (mm) x 100 (mm) actual or “dressed” dimensions are smaller, e.g. 50 x 100 is 38 x 89.
Built-up sections
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6. PRISMATIC BEAM DESIGN (cont)
Procedures:
Shear and Moment Diagram
Determine the maximum shear and moment in the beam. Often this is done by constructing the beam’s shear and moment diagrams.
For built-up beams, shear and moment diagrams are useful for identifying regions where the shear and moment are excessively large and may require additional structural reinforcement or fasteners.
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7. PRISMATIC BEAM DESIGN (cont)
Normal Stress
If the beam is relatively long, it is designed by finding its section modulus using the flexure formula, Sreq’d = Mmax/σallow.
Once Sreq’d is determined, the cross-sectional dimensions for simple shapes can then be computed, using Sreq’d = I/c.
If rolled-steel sections are to be used, several possible values of S may be selected from the tables in Appendix B. Of these, choose the one having the smallest cross-sectional area, since this beam has the least weight and is therefore the most economical.
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8. PRISMATIC BEAM DESIGN (cont)
Normal Stress (cont)
Make sure that the selected section modulus, S, is slightly greater than Sreq’d, so that the additional moment created by the beam’s weight is considered.
Shear Stress
Normally beams that are short and carry large loads, especially those made of wood, are first designed to resist shear and then later checked against the allowable-bending-stress requirements.
Using the shear formula, check to see that the allowable shear stress is not exceeded; that is, use τallow ≥ Vmax Q/It.
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9. PRISMATIC BEAM DESIGN (cont)
Shear Stress (cont)
If the beam has a solid rectangular cross section, the shear formula becomes τallow ≥ 1.5(Vmax/A), Eq.7-5, and if the cross section is a wide flange, it is generally appropriate to assume that the shear stress is constant over the cross-sectional area of the beam’s web so that τallow ≥ Vmax/Aweb, where Aweb is determined from the product of the beam’s depth and the web’s thickness.
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10. PRISMATIC BEAM DESIGN (cont)
Adequacy of Fasteners
The adequacy of fasteners used on built-up beams depends upon the shear stress the fasteners can resist. Specifically, the required spacing of nails or bolts of a particular size is determined from the allowable shear flow, qallow = VQ/I, calculated at points on the cross section where the fasteners are located.
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EXAMPLE 1
A beam is to be made of steel that has an allowable bending stress of σallow = 170 MPa and an allowable shear stress of τallow = 100 MPa. Select an appropriate W shape that will carry the loading shown in Fig. 11–7a.
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EXAMPLE 1 (cont)
Solution
The required section modulus for the beam is determined from the flexure formula,
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EXAMPLE 1 (cont)
Solution
The beam having the least weight per foot is chosen, W460 x 60
The beam’s weight is
From Appendix B, for a W460 x 60, d = 455 mm and tw = 8 mm.
Thus
Use a W460 x 60.
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EXAMPLE 2
The laminated wooden beam shown in Fig. 11–8a supports a uniform distributed loading of 12 kN/m. If the beam is to have a height-to-width ratio of 1.5, determine its smallest width. The allowable bending stress is 9 MPa and the allowable shear stress is 0.6 MPa. Neglect the weight of the beam.
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EXAMPLE 2 (cont)
Solution
Applying the flexure formula,
Assuming that the width is a, the height is 1.5a.
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EXAMPLE 2 (cont)
Solution
Applying the shear formula for rectangular sections,
Since the design fails the shear criterion, the beam must be redesigned on the basis of shear.
This larger section will also adequately resist the normal stress.
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EXAMPLE 3
The wooden T-beam shown in Fig. 11–9a is made from two 200mm x 30mm boards. If the allowable bending stress is 12 MPa and the allowable shear stress is 0.8 MPa, determine if the beam can safely support the loading shown. Also, specify the maximum spacing of nails needed to hold the two boards together if each nail can safely resist 1.50 kN in shear.
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EXAMPLE 3 (cont)
Solution
The reactions on the beam are shown, and the shear and moment diagrams are drawn in Fig. 11–9b.
The neutral axis (centroid) will be located from the bottom of the beam.
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EXAMPLE 3 (cont)
Solution
Thus
Since c = m,
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EXAMPLE 3 (cont)
Solution
Maximum shear stress in the beam depends upon the magnitude of Q and t.
We will use the rectangular area below the neutral axis to calculate Q, rather than a two-part composite area above this axis, Fig. 11–9c
So.
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EXAMPLE 3 (cont)
Solution
From the shear diagram it is seen that the shear varies over the entire span.
Since the nails join the flange to the web, Fig. 11–9d, we have
The shear flow for each region is therefore
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EXAMPLE 3 (cont)
Solutions
One nail can resist 1.50 kN in shear, so the maximum spacing becomes
For ease of measuring, use
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FULLY STRESSED BEAMS
The strength requirements are not the same for every cross sections of the beam.
To optimize the design so as to reduce the beam’s weight, non- prismatic beam is a choice.
A beam designed in this manner is called a fully stressed beam; e.g.
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EXAMPLE 4
Determine the shape of a fully stressed, simply supported beam that supports a concentrated force at its center, Fig a.The beam has a rectangular cross section of constant width b, and the allowable stress is σallow.
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EXAMPLE 4 (cont)
Solution
The internal moment in the beam, Fig. 11–11b, expressed as a function of position,
The required section modulus is
For a cross-sectional area h by b we have
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EXAMPLE 4 (cont)
Solution If
By inspection, the depth h must therefore vary in a parabolic manner with the distance x.
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