Simple Beam Bending
This page provides the sections on simple beam bending from the "Stress Analysis Manual," Air Force Flight Dynamics Laboratory, October 1986.
Other related chapters from the Air Force "Stress Analysis Manual" can be seen to the right.
 Simple Beam Bending
 Shear Web Beam Bending
 Partial Tension Beam Bending
 Beam Forces & Moments
 Beam Columns
 Beam Torsion
Nomenclature
A  =  crosssectional area 
b  =  width of section 
c  =  distance from neutral axis to extreme fiber 
E  =  modulus of elasticity 
F_{ty}  =  yield stress in tension 
f_{b}  =  calculated primary bending stress 
f_{cr}  =  calculated critical compressive stress 
G  =  modulus of elasticity in shear 
h  =  height or depth 
I  =  moment of inertia 
J  =  torsion constant 
L  =  length 
L'  =  effective length of beam 
M  =  applied bending moment 
M_{cr}  =  critical moment 
M_{fp}  =  fully plastic bending moment 
M_{y}  =  bending moment at the onset of yielding 
P  =  applied concentrated load 
Q  =  statical moment of cross section, \( \int_{A_1} y ~dA \) 
q  =  shear flow 
s  =  distance from centroidal axis to point of application of load 
t_{f}  =  flange thickness 
t_{w}  =  web thickness 
V  =  shear force 
x, y, z  =  rectangular coordinates 
y  =  deflection of beam due to bending 
ρ  =  radius of gyration 
θ  =  slope of beam 
1.3.1 Simple Beams in Bending
Simple beams in elastic and plastic bending are treated in Sections 1.3.1.1 and 1.3.1.3, respectively, while the possibility of lateral instability of deep beams in bending is treated in Section 1.3.1.5.
1.3.1.1 Simple Beams in Elastic Bending
This section treats simple beams in bending for which the maximum stress remains in the elastic range.
The maximum bending stress in such a beam is given by the formula
while the shear flow is given by
where \( Q = \int_{A_1} y~dA \). The use of these equations is illustrated in Section 1.3.2.2.
The vertical and angular displacements of a simple beam in elastic bending are given by Equations (13) and (14), respectively, where A and B are constants of integration.
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1.3.1.2 Sample Problem  Simple Beams in Elastic Bending
Given: The cantilever beam shown in Figure 11.
Find: The maximum bending and shear stresses.
Solution: From the equations of statics, the shear and moment diagrams in Figure 12 may be obtained.
Since c and I are constant along the beam, the maximum bending stress occurs at the point of maximum bending moment; and from Equation (11),
$$ f_b = { Mc \over I } = { 1000(0.5) \over 0.0833 } = 6,000~ \text{psi} $$Q may be computed at a distance y_{1} from the neutral axis by considering the beam cross section shown in Figure 13:
$$ Q = \int_{A_1} y~ dA = \int_{y_1}^1 y ~(1) ~dy = {1 \over 2}  { y_1^2 \over 2 } $$Q is maximum at y_{1} = 0 where Q = 1/2. Thus, the maximum shear flow occurs at the neutral axis and is given by Equation (12) as
$$ q = { VQ \over I } = { 50 (0.5) \over 0.0833 } = 300 ~\text{ lb/in } $$The maximum shear stress is thus,
$$ { 300 ~\text{ lb/in } \over 1 ~\text{in} } = 300~ \text{ lb/in }^2 $$Need a Beam Calculator?
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1.3.1.3 Simple Beams in Plastic Bending
In some cases, yielding of a beam in bending is permissible. If the beam material may be considered to be elasticperfectly plastic, the bending moment at failure is given by
where M_{y} is the moment that causes initial yielding of the extreme fibers and K is the shape factor given in Table 11.
Section  
K  1.0  1.5  2.0  1.7 
$$ 1.27 \left( 1  {t \over r} \right) $$

$$ { 32 D_o (D_o^3  D_i^3) \over 3\pi (D_o^4  D_i^4) } $$

$$ {3h \over 2} \left({ bh^2  2 b_1 h_1^2 \over b h^3  2 b_1 h_1^3 }\right) $$

1.3.1.4 Sample Problem  Simple Beams in Plastic Bending
Given: The simply supported beam shown in Figure 14.
Find: The load, P, that causes fully plastic bending.
Solution: Rearranging Equation (11) and replacing the bending stress with the yield stress gives
$$ M_y = { F_y ~I \over c } = { 55000 (0.666) \over 1.0 } = 36,600 ~\text{in/lb} $$Inserting the value of K from Table 11 into Equation (15) gives
$$ M_{fp} = K ~M_y = 1.5 (36,600) = 54,900 ~\text{in/lb} $$From statics, the maximum moment on the bar is 10P. Thus, for fully plastic bending,
$$ P = { M_{fp} \over 10 } = 5,490 ~\text{lb} $$1.3.1.5 Intoduction to Lateral Instability of Deep Beams in Bending
Beams in bending under certain conditions of loading and restraint can fail by lateral buckling in a manner similar to that of columns loaded in axial compression. However, it is conservative to obtain the buckling load by considering the compression side of the beam as a column since this approach neglects the torsional rigidity of the beam.
In general, the critical bending moment for the lateral instability of the deep beam, such as that shown in Figure 15, may be expressed as
where J is the torsion constant of the beam and K is a constant dependent on the type of loading and end restraint. Thus, the critical compressive stress is given by
where c is the distance from the centroidal axis to the extreme compression fibers. If this compressive stress falls in the plastic range, an equivalent slenderness ratio may be calculated as
The actual critical stress may then be found by entering the column curves of Chapter 2 at this value of (L'/ρ). This value of stress is not the true compressive stress in the beam, but is sufficiently accurate to permit its use as a design guide.
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1.3.1.6 Lateral Instability of Deep Rectangular Beams in Bending
The critical moment for deep rectangular beams loaded in the elastic range loaded along the centroidal axis is given by
where K_{u} is presented in Table 12, and b, h, and L are as shown in Figure 15. The critical stress for such a beam is
where K_{u} is presented in Table 12.
If the beam is not loaded along the centroidal axis, as shown in Figure 16, a corrected value K'_{f} is used in place of K_{f} in Equation (110). This factor is expressed as
where n is a constant defined below:
 For simply supported beams with a concentrated load at midspan, n = 2.84.
 For cantilever beams with a concentrated end load, n = 0.816.
 For simply supported beams under a uniform load, n = 2.52.
 For cantilever beams under a uniform load, n = 0.725.
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1.3.1.7 Lateral Instability of Deep I Beams
Figure 17 shows a deep I beam.
The critical stress of such a beam in the elastic range is given by
where K_{I} may be obtained from Table 13, and a is given by
where J is the torsion constant of the I beam. This constant may be approximated by
This method can be applied only if the load is applied at the centroidal axis.