Previous - Next
Bending Stresses
When a sponge is put across two supports and gently pressed downwards between the supports, the pores at the top will close indicating compression, and the pores at the bottom will open wider indicating tension. Similarly, a beam of any elastic material such as wood or steel will produce a change in shape when external loads are acting on it.
Figure 4.2 Bending effects on beams.
The stresses will vary from maximum compression at the top to maximum tension at the bottom. Where the stress changes from compressive to tensile, there will be one layer which remains unstressed and this is called the neutral layer or the neutral axis (NA).
This is why beams with l-section are so effective. The main part of its material is concentrated in the flanges, away from the neutral axis. Hence, the maximum stresses occur where there is maximum material to resist them.
If the material is assumed to be elastic, then the stress distribution can be represented by two triangular shapes with the line of action of the resultant force of each triangle of stress at its centroid.
The couple produced by the compression and tension triangles of stress is the internal-reaction couple of the beam section.
The moment caused by the external loads acting on the beam will be resisted by the moment of this internal couple. Therefore:
M = MR = C (or T) x h where:
M = the external moment
MR = the internal resisting moment
C = resultant of all compressive forces on the cross sec tion of the beam
T = resultant of all tensile forces on the cross section of the beam
h = lever arm of the reaction couple
Now consider a small element with the area (R) at a distance (a) from the neutral axis (N/A).
Small element with the area (R) at a distance (a) from the neutral axis (N/A)
Note that it is common practice to use the symbol for bending stress rather than the more general symbol . Maximum compressive stress (fc) is assumed to occur in this case at the top of the beam. Therefore, by similar triangles, the stress in the chosen element is:
fa / a = fc / ymax
fa = a x (fc / ymax)
Since force = stress x area; the force on the element
= fa x R = a x (fc / ymax) x R
The resisting moment due to the small element is: force x distance (a)
= a x (fc / ymax) x R x a = Ra2 x (fc / ymax)
The total resisting moment due to all such small elements in the cross section is:
MR = S Ra2 x (fc / ymax)
But S Ra2 = 1, the moment of inertia about the neutral axis and therefore
MR = I x (fc / ymax)
Since the section modulus Zc = I / ymax
MR=fc x Zc= M;
similarly
MR=ft x Zt=M
The maximum compressive stress (fc) will occur in the cross section area of the beam where the bending moment (M) is greatest. A size and shape of cross section, i.e., is section modulus (Z), must be selected such that the fc does not exceed an allowable value. Allowable working stress values can be found in building codes or engineering handbooks.
From the diagrams below it can be seen that the concept of a "resisting" couple can be seen in many structural members and systems.
Girders and l-Beams (1/6 of web area can be added to each flange area for moment resistance).
Rectangular Reinforced-Concrete Beams (Note that the steel bars are assumed to carry the whole of the tensile forces).
In summary the following equation is used to test for safe bending:
fw ³ f = Mmax / Z
where:
fw = allowable bending stress
f = actual bending stress
Mmax = maximum bending moment
Z = section modulus
Horizontal Shear The horizontal shear force (Q) at a given cross section in a beam induces a shearing stress which acts tangentially to the horizontal cross-sectional plane. The average value of this shear stress is:
t = Q / A
A where A is the transverse cross sectional area. This average value is used when designing rivets, bolts, welded joints.
The existence of such a horizontal stress can be illustrated by bending a paper pad. The papers will slide relative to each other, but in a beam this is prevented by the developed shear stress.
Figure 4.3 Shearing effects on beams.
Rectangular Beams
However, the shear stresses are not equal across the cross section. At the top and bottom edge of the beam they must be zero, since no horizontal shear stresses can develop.
If the shear stresses at a certain distance from the neutral axis are considered, their value can be determined according to the following formula:
where:
t = shear stress
Q = shear force
D A = area for the part of the section being sheared off
y = perpendicular distance from the centroid of PA to the neutral axis
I = moment of inertia for the whole cross section
b = width of the section at the place where shear stress is being calculated.
Maximum Horizontal Shear Force in Beams
It can be shown that the maximum shear stress rmaX in a beam will occur at the neutral axis. Thus, the following relations for the maximum shear stress in beams of differ ent shapes can be deduced, assuming the maximum shear force (Q) to be the end reaction at a beam support
(column).
For rectangular sections t max = 3Q / 2bd = 3Q / 2A = 1.5 Q / A
For square sections t max = 3Q / 2a2 = 1.5 Q / A
For circular sections t max = 16Q / 3p D2 = 4Q / 3A
For l-shaped sections of steel beams, a convenient approximation is to assume that all shearing resistance is afforded by the web plus that part of the flanges that is a continuation of the web.
Thus:
For I - sections t max = Q / (d x t)
where:
d = depth of beam
t = thickness of web
If timber and steel beams with spans normally used in buildings are made large enough to resist the tensile and compressive stresses caused by bending, they are usually strong enough also to resist the horizontal shear stresses. However, the size or strength of short, heavily loaded timber beams may be limited by these stresses.
Deflection of Beams
Excessive deflections are unacceptable in building construction, since they can cause cracking of plaster in ceilings and can result in jamming of doors and windows. Most building codes limit the amount of allowable deflection as a proportion of the member's length, i.e. 1/180, 1/240 or 1/360 of the length.
For standard cases of loading the deflection formulae can be expressed as:
d max = Kc x ( WL3 / EI)
where:
d max = maximum deflection (mm)
Kc = constant depending on the type of loading and the end support conditions
W = total load (N)
L = effective span (mm)
E = modulus of elasticity (N/mm²)
I = moment of inertia (mm4)
It can be seen that deflection is greatly influenced by the span L, and that the best resistance is provided by beams which have the most depth (d), resulting in a large moment of inertia.
Note that the effective span is greater than the clear span. It is convenient to use the centre to centre distance of the supports as an approximation to the effective span.
Some of the standard cases of loading and resulting deflection for beams can be found later in this section.
Design Criteria
The design of beams is dependent upon the following factors:
Beams are designed by use of the following formulae:
1. Bending Stress
fw ³ f = Mmax / z where
fw = allowable bending stress
f = actual bending stress
Mmax = maximum bending moment
z = section modulus
This relationship derives from simpled beam theory and
Mmax / INA = fmax / ymax
and INA / ymax = Z
The maximum bending stress will be found in the section of the beam where the maximum bending moment occurs. The maximum moment can be obtained from the B.M. diagram.
2. Shear Stress
For rectangular cross-sections:
t w ³ t = (3 x Qmax) / (3 x A) = 3Qmax / 2bd
For circular cross-sections:
t w ³ t = (4 x Qmax) / (3 x A) = 16Qmax / 3p d2
For l-shaped cross-sections of steel beams
t w ³ t @ Qmax / A
t w = allowable shear stress
T = actual shear stress
Qmax = maximum shear force
A = cross-section area
Allowable shear stress like the allowable bending stress differs for different materials and can be obtained from a building code. Maximum shear force is obtained from the shear force diagram.
3. Deflection
In addition, limitations sometimes are placed on maximum deflection of the beam.
d max = Kc x ( WL3 / EI)
Example 15
Consider a floor where beams are spaced at 1200mm and have a span of 4000mm. The beams are seasoned cypress with the following properties:
fw = 8.0N/mm²,
t w = 0.7N/mm²,
E = 8.400N/mm²,
density 500 kg/m³
Loading on floor and including floor is 2.5kN/m².
Allowable deflection is L/240
Beam loading: w = 1.2m x 2.5kN/m² = 3kN/m
Assume a 100 by 250mm cross section for the beams.
Beam mass = 0.1 x 0.25 x 500 x 9.81 = 122.6N/m =0.12kN/m
Total w =3+012=3.12kN/m
2 Calculate reactions and draw shear-force and bending moment diagrams
Calculate reactions and draw shear-force and bending moment diagrams
3 Calculate maximum bending moment (Mmax) using equation for a simple beam, uniformly loaded. See table 4.4
Mmax = wL2 / 8 = (3.12 x 42) / 8 = 6.24kNm = 6.24 x 106/Nmm
4 Find the required section modulus (Z)
Zreq = Mmax /fw = (6.24 x 106) / 8 = 0.78 x 106mm³
5. Find a suitable beam depth assuming 100mm breaths:
From table 4.3 the section modulus for a rectangular shape isZ=1/6 x bd2
Choose a 100 by 225mm timber. The timber required is a little less than that assumed. No recalculations are required unless it is estimated that a smaller size timber would be adequate if a smaller size had been assumed initially.
6 Check for shear loading:
t = 3Qmax / 2A = (3 x 6.24 x 103) / (2 x 200 x 225) = 0.42N/mm²
Since the safe for the timber is 0.7N/mm², the section is adequate in resistance to horizontal shear.
7 Check deflection to ensure less than 1/240 of the span. From table 4.4
d max = -5 / 384 = ( WL3 / EI) where
E = 8400N/mm²
I = bd3 / 12 = (100 x 2253) / 12 = 95 x 106mm4
W = 3.12kN/m x 4m = 12.48kN = 12.48 x 103N
L = 4 x 103mm
d max = (-5 / 384) x [(12.48 x 103 x 43 x 109) / (8400 x 95 x 106)
The allowable deflection, 400/240 = 16.7 >13 ·. beam is satisfactory.
Bending Moments Caused by Askew Loads
If beam is loaded so that the resulting bending moment is not about one of the main axes, the moment has to be resolved into components acting about the main axes. The stresses are then calculated separately relative to each axis and the total stress is found by adding the stresses caused by the components of the moment.
Example 16
Design a timber purlin, which will span rafters 2.4m on center. The angle of the roof slope is 30° and the purlin will support a vertical dead load of 250N/m and a wind load of 200N/m acting normal to the roof. The allowable bending stress (fw) for the timber used is 8N/mm². The timber density is 600 kg/m³.
1 Assume a purlin cross-section size of 50 x 125mm. Find an estimated self-load.
W=0.05 x 0.125 x 600 x 9.81 =37N/m
The total dead load becomes 250 + 37 = 287N/m
2 Find the components of the loads relative to the main axes.
Wx = 200N/m + 287N/m x cos 30° = 448.5N/m
Wy = 287N/m x sin 30° = 143.5N/m
3 Calculate the bending moments about each axis for a uniformly distributed load. The purlin is assumed to be a simple beam.
Mmax =WL/8 = wL2/8
Mmax x = (wx x L2) / 8 = (448.5 x 2.42) / 8 = 323 x 103Nmm
Mmax y = (wy x L2) / 8 = (143.5 x 2.42) / 8 = 103 x 103Nmm
4 The actual stress in the timber must be no greater than the allowable stress.
f= Mmax x / Zx + Mmax y / Zy £ fw
5 Try the assumed purlin size of 50 x 150mm.
Zx = bd2 / 6 = (50 x 1252) / 6 = 130 x 103mm³
Zy = db2 / 6 = (125 x 502) / 6 = 52 x 103mm³
f = (320 x 103) / (130 x 103) + (130 x 103) / (52 x 103) = 2.5 + 2 = 4.5N/mm²
This size is safe, but a smaller size may be satisfactory. Try 50 x I00mm.
Zx = bd2 / 6 = (50 x 1002) / 6 = 83 x 103mm³
Zy = db2 / 6 = (100 x 502) / 6 = 42 x 103mm³
f = (323 x 103) / (83 x 103) + (103 x 103) / (42 x 103) = 3.9 + 2.5 = 6.4N/mm²
This is much closer to the allowable stress. To save money, also 5O x 75mm should be tried. In this case f > fw and therefore 50 x 100mm is chosen.
Universal Steel Beams
Steel beams of various cross-sectional shapes are commercially available. Even though the properties of their cross sections can be calculated with the formulae given in the section "Design of members in direct stress" it is easier to obtain them from hand book tables. These tables will also take into consideration the effect of rounded edges, welds, etc.
Sections of steel beams are indicated with a combination of letters and a number, where the letters represent the shape of the section and the number is the dimension, usually height, of the section in millimetres, e.g., IPE 100. In the case of HE sections, the number is followed by a letter indicating the thickness of the web and flanges, e.g., HE 180B.
An example of an alternative method of notation is 305 x 102 UB 25, i.e., 305 by 102mm universal beam weighing 25 kg/m.
The following example demonstrates another method of taking into account the self-weight of the structural member being designed.
Example 17
A steel beam, used as a lintel over a door opening, is required to span 4.0m between centres of simple supports. The beam will be carrying a 220mm thick and 2.2m high brick wall, weighing 20kN/ m³. Allowable bending stress is 165N/mm².
Uniformly distributed load caused by brickwork is 0.22 x 2.2 x 4 0 x 20 = 8.7kN.
Assumed self weight for the beam is 1.5kN.
(Note: triangular load distribution for bricks above lintel would result in a slightly lower value of load).
Total uniformly distributed load W = 38.7 + 1.5 = 40.2 kN
Mmax = WL/8 = (40.2 x 4.0) / 8 = 20.1kNm=20.1 x 106Nmm
Zreq = (20.1 x 106) /165 = 0.122 x 106mm³ = 122cm² (as shown in tables)
Suitable sections as found in a handbook would be:
| Section | Zx-x | Mass |
| INP 160 | 117 cm³ | 17.9 kg/m |
| IPE 180 | 146cm³ | 18.8 kg/m |
| HE 140A | 155 cm³ | 24.7 kg/m |
| HE 120A | 144 cm³ | 26.7 kg/m |
Choose INP 160 because it is closest to the required section modulus and has the lowest weight. Then recalculate the required Z using the INP 160 weight: 4.0 x 17.9 x 9.81 = 702N which is less than the assumed self weight of l.5kN. A recheck on Z required reveals a value of 119cm³ which is close enough.
Continuous Beams
A single continuous beam extending over a number of supports will safely carry a greater load than a series of simple beams extending from support to support. Consider the Shear Force and Bending Movement diagrams for the following two beam loadings:
Although the total value of the load has increased, the maximum shear force remains the same but the maximum bending is reduced when the beam is cantilevered over the supports.
Although continuous beams are statically indeterminate and the calculations are complex, approximate values can be found with simplified equations. Conservative equations for two situations are as follows:
Load concentrated between supports: BM = WL / 6
Load uniformly distributed: BM = WL / 12
It is best to treat the two end sections as simple beams.
Standard Cases of Beam Loading
A number of beam loading cases occur frequently and it is useful to have standard expressions for these available. Several of these will be found in Table 4.4.
In small-scale buildings the spans are relatively small, and with normal loading, solid rectangular or square sections are generally the most economical beams. But where members beyond the available sizes and/or length of solid timber are required, one of the following combinations may be chosen:
Built-up Timber Beams
When designing large members, there are advantages in building up solid sections from smaller pieces since these are less expensive and easier to obtain. Smaller pieces also season properly without checking. The composite beams may be built up in ways that minimize warping and permit rigid connections between columns and beams. At the same time the importance of timber defects is decreased, since the load is distributed to several pieces, not all having defects in the same area.
Figure 4.4 Built-up beams and trusses.
Built-up solid beams are normally formed by using vertical pieces nailed or bolted together, nailing being satisfactory for beams up to about 250mm in depth, although these may require the use of bolts at the ends if the shear stresses are high. Simply multiplying the strength of one beam by the number is satisfactory as long as the staggered joints occur over supports.
Built-up sections with the members spaced apart are used mainly where the forces are tensile, such as in the bottom chords of a truss. Where used in beams designed to resist bending buckling of the individual members may have to be considered if those members have a large depth to width ratio. This can, however, be avoided by appropriate spacing of stiffeners which connect the spaced members at intervals.
Where the loading is heavy, the beam will require considerable depth, resulting in a large section modulus to keep the stresses within the allowable value. If sufficient depth cannot be obtained in one member, it may be achieved by combining several members, as, for example, where the members are glued together to form a laminate.
