- 5.1 Introduction
- 5.2 Pure Bending of Beams of Symmetrical Cross Section
- 5.3 Pure Bending of Beams of Asymmetrical Cross Section
- 5.4 Bending of A Cantilever of Narrow Section
- 5.5 Bending of a Simply Supported Narrow Beam
- 5.6 Elementary Theory of Bending
- 5.7 Normal and Shear Stresses
- 5.8 Effect of Transverse Normal Stress
- 5.9 Composite Beams
- 5.10 Shear Center
- 5.11 Statically Indeterminate Systems
- 5.12 Energy Method for Deflections
- 5.13 Elasticity Theory
- 5.14 Curved Beam Formula
- 5.15 Comparison of the Results of Various Theories
- 5.16 Combined Tangential and Normal Stresses
- References
- Problems
This chapter is from the book
This chapter is from the book
This chapter is from the book
5.16 Combined Tangential and Normal Stresses
Curved beams are often loaded so that there is both an axial force and a moment on the cross section. The tangential stress given by Eq. (5.70) may then be algebraically added to the stress due to an axial force P acting through the centroid of cross-sectional area A. For this simple case of superposition, the total stress at a point located at distance r from the center of curvature O may be expressed as follows:
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As before, a negative sign would be associated with a compressive load P.
Of course, the theory developed in this section applies only to the elastic stress distribution in curved beams. Stresses in straight members under various combined loads are discussed in detail throughout this text.
The following problems illustrate the application of the formulas developed to statically determinate and statically indeterminate beams under combined loadings. In the latter case, the energy method (Section 10.4) facilitates the determination of the unknown, redundant moment in the member.
CASE STUDY 5.1 STRESSES IN A STEEL CRANE HOOK BY VARIOUS METHODS
A load P is applied to the simple steel hook having a rectangular cross section, as illustrated in Fig. 5.27a. Find the tangential stresses at points A and B, using (a) the curved beam formula, (b) the flexure formula, and (c) elasticity theory.
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Figure 5.27. Case Study 5.1. A crane hook of rectangular cross section.
Given: P = 6 kN, 
, b = 25 mm, and h = 32 mm.
Solution
Curved Beam Formula. For the given numerical values, we obtain (Fig. 5.27b):
Then, Eqs. (5.73) and (5.69) result in
To maintain applied force P in equilibrium, there must be an axial tensile force P and a moment
at the centroid of the section (Fig. 5.27c). Then, by Eq. (5.74), the stress at the inner edge (r = ri) of the section A–B isLikewise, the stress at the outer edge (r = r°) is
The negative sign of (σθ)B means a compressive stress is present. The maximum tensile stress is at A and equals 97 MPa.
Comment The stress due to the axial force,
is negligibly small compared to the combined stresses at points A and B of the cross section.
Flexure Formula. Equation (5.5), with
, givesElasticity Theory. Using Eq. (5.66) with a = ri = 34 mm and b = r0 = 66 mm, we find
)
)
)
)
Superposition of −P/A and the second of Eqs. (5.67) with t = 25 mm at r = a leads to
)
Similarly, at r = b, we find (σθ)B = − 7.5 + 58.1 = 50.6 MPa.
Comments The results obtained with the curved-beam formula and with elasticity theory are in good agreement. In contrast, the flexure formula provides a result with unacceptable accuracy for the tangential stress in this nonslender curved beam.
EXAMPLE 5.14 RING WITH A DIAMETRAL BAR
A steel ring of 350-mm mean diameter and of uniform rectangular section 60 mm wide and 12 mm thick is shown in Fig. 5.28a. A rigid bar is fitted across diameter AB, and a tensile force P is applied to the ring as shown. Assuming an allowable stress of 140 MPa, determine the maximum tensile force that can be carried by the ring.
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Figure 5.28. Example 5.14. (a) Ring with a bar AB is subjected to a concentrated load P; (b) moment at a section.
Solution Let the thrust induced in bar AB be denoted by 2F. The moment at any section a–a (Fig. 5.28b) is then
)
Note that before and after deformation, the relative slope between B and C remains unchanged. Therefore, the relative angular rotation between B and C is zero. Applying Eq. (5.32), we obtain
where 
is the length of beam segment corresponding to dθ. Substituting in Eq. (a), this becomes, after integrating,
)
This expression involves two unknowns, MB and F. Another expression in terms of MB and F is found by recognizing that the deflection at B is zero. By applying Castigliano’s theorem,
where U is the strain energy of the segment. This expression, upon introduction of Eq. (a), takes the form
)
After integration,
)
Solution of Eqs. (b) and (c) yields
and 
. Substituting Eq. (a) gives
Thus, MC > MB.
Since 
, the simple flexure formula offers the most efficient means of computation. The maximum stress is found at points A and B:
)
Similarly, at C and D,
Hence σθC > σθB and we have σmax = 140×106 = 18,411P. The maximum tensile load is therefore P = 7.604 kN.