Beam Problem 3 - Loads on Lintels

Beam Problem 3 - Loads on Lintels first created 31/03/08 - last modified 31/03/08 Page Author: Ty Harness

A lintel is a beam spanning an opening such as a door or a window. Lintels will often support brick work which is interesting because an arch is formed over the lintel. It's common practice [1] to assume a isosceles triangle where 2 angles are 60 degrees as shown in figure 1. The triangular area of brick work is the load applied to the lintel as shown in figure 1. If the lintel was removed then the bricks in the triangular area would fall to the floor.

Figure 1- A typical lintel supporting a 9 inch thick brick wall.

This is only the case if side abutments are greater than L/2 because the arch will produce side thrust forces which may break either or both the abutments. For cases where the abutments of either or both sides are less than L/2 then the lintel may need to support more load to reduce the risk of the abutments failing.

You can normally find out the weight of bricks or brick work ( which includes the mortar) from the manufacturer and sometimes you will need to consider extra weight for wet external walls. From Seward[2] the brick work load is approximately 22 kN/m^3 and let's say the length of the Lintel is 2.7m. For this particular problem we'll assume a typical 9 inch (or a 215 mm modern equivalent) thick wall so the density laid in the x direction is rho. We'll try 2 universal beams (178x102x19) bolted together with spacers and assume the beams act as one box section.

$rho = 22[(kN)/m^3] * 0.215[m] = 4.73 [(kN)/m^2]$

The area of a triangle:

$A = 1/2 * L * h$ where

$tan(theta) = h/(L/2)$ or $ h = (L*tan(60)) /2 $

$A = 1/4 * L^2 * tan(60) $ where

Q is the total triangular load of brick work to be considered on the lintel.

$Q = A*rho = 1/4 * 2.7[m]^2 * tan(60)*4.73 [(kN)/m^2] = 15 [kN] $

but the load is distributed as a triangle so using the equation of a straight line the first half of the beam:

$q(x) = (2*q_0)/L * x $ where $ 0 < x < L/2 $
$Q = 1/2 * q_0 * L$ and transposing for $q_0$

$q_0 = (2*Q)/L = 11.2 [(kN)/m] $

Find the maximum deflection using an analytical method and check your answer using any FEA software.

Please note this is just a student exercise and not an approved or recognized method of specifying a lintel. More information on current building practice is available for designers in the British Standards literature and standard construction methods will need to be employed to satisfy the building inspector. You can't take risks so always consult a structural engineer.

Answers

Working in Newtons and mm

We need the second moment of area,I about the neutral axis:

From tables Ixx for 178x102x19 is 1357cm^4 therefore Ixx for the 2 beams is:

$I = 2* {1357*[cm^4] * (1^4*[mm^4])/(0.1^4 [cm^4])} = 27.14*10^6 mm^4$

Figure 2 - Cross section where x -x is the neutral axis

$E = 200 000 (MN)/m^2 = 200 000 N/(mm^2)$

$q_0 = 11200 [N/m] * 1/(1000 [(mm)/m]) = 11.2 [N/(mm)] $

The theoretical end deflection is given by:

$y(L/2) = (7*q_0 * L^4)/(3840*E*I) $
$y(L/2) = (7*11.2*2700^4) / (3840*200000*27.14*10^6) = 0.2 mm $

More information on the derivation can be found at : Triangular beam load

I've chosen to use Prof. Rieg's Z88 Finite Elements Program because it's free and unlimited and ideal for student use. The theory behind FEM is beyond the scope of this tutorial but you can learn a great deal from the Z88 manual.

Using a Beam Element No. 13 we create a General Structure Data File, Z88i1.txt which basically consists of : 2D Analysis, 12 Nodes, 11 Elements, 3DOF per node 36 in total. Nodal co-ordinates, elements defined and adjacent elements. All elements 1 to 11 have the same properties Young's Modulus but this time working in millimeters = 2e5 N/mm^2. Poisson's Ratio = 0.3 from tables, Area = 4.84e3 mm^2, I = 27.14e6 mm^4

2 12 11 36 1 0 1 0 Z88I1.TXT,typed in by Ty
1 3 +0.00000E+000 +0.00000E+000 +0.00000E+000 node #1
2 3 +1.35000E+002 +0.00000E+000 +0.00000E+000 node #2
3 3 +4.05000E+002 +0.00000E+000 +0.00000E+000 node #3
4 3 +6.75000E+002 +0.00000E+000 +0.00000E+000 node #4
5 3 +9.45000E+002 +0.00000E+000 +0.00000E+000 node #5
6 3 +1.21500E+003 +0.00000E+000 +0.00000E+000 node #6
7 3 +1.48850E+003 +0.00000E+000 +0.00000E+000 node #7
8 3 +1.7550E+003 +0.00000E+000 +0.00000E+000 node #8
9 3 +2.0250E+003 +0.00000E+000 +0.00000E+000 node #9
10 3 +2.2950E+003 +0.00000E+000 +0.00000E+000 node #10
11 3 +2.5650E+003 +0.00000E+000 +0.00000E+000 node #11
12 3 +2.70000E+003 +0.00000E+000 +0.00000E+000 node #12
1 13 element #1
1 2
2 13 element #2
2 3
3 13 element #3
3 4
4 13 element #4
4 5
5 13 element #5
5 6
6 13 element #6
6 7
7 13 element #7
7 8
8 13 element #8
8 9
9 13 element #9
9 10
10 13 element #10
10 11
11 13 element #11
11 12
1 11 +2.00000E+005 +3.00000E-001 1 +4.84000E+003 0 0 +2.71400E+007 +8.90000E+001 0 0

The load on each node is more complicated than for a constant load as used in beam problem 2 Figure 3 shows the beam divided into 10 rectangles where q(midpoint) times h will be the approximate load.

Figure 3 - Applying the loads to the discrete nodal points

$h = L/10 = (2700 [mm])/10 = 270[mm]$

So for the rectangle 4 shown shaded the load on node 5 is q5:

$q_5 = q(x)*h $ when x = 3.5*h$

$q_5 = q(945)*270 = 2116.8 [N] $

and hence the load file Z88i2.txt is shown below.

The Boundary Condition and Load file: Z88i2.txt

17 , typed in by ty
1 1 2 0.00000E+000
1 2 2 0.00000E+000
1 3 2 0.00000E+000
2 2 1 -3.02400E+002
3 2 1 -9.07200E+002
4 2 1 -1.51200E+003
5 2 1 -2.11680E+003
6 2 1 -2.72160E+003
7 2 1 -2.72160E+003
8 2 1 -2.11680E+003
9 2 1 -1.51200E+003
10 2 1 -9.07200E+002
11 2 1 -3.02400E+002
12 1 2 0.00000E+000
12 2 2 0.00000E+000
12 3 2 0.00000E+000

output file Z88O2.TXT : displacements, computed by Z88F V10
*************

Knoten U(1) U(2) U(3) U(4) U(5) U(6)

1 +0.0000000E+000 +0.0000000E+000 +0.0000000E+000
2 +0.0000000E+000 -6.5351048E-003 -9.2588735E-005
3 +0.0000000E+000 -4.8725556E-002 -2.0369949E-004
4 +0.0000000E+000 -1.0831494E-001 -2.2350041E-004
5 +0.0000000E+000 -1.6265997E-001 -1.6823686E-004
6 +0.0000000E+000 -1.9460017E-001 -6.2276888E-005
7 +0.0000000E+000 -1.9443155E-001 +6.3497866E-005
8 +0.0000000E+000 -1.6276826E-001 +1.6817881E-004
9 +0.0000000E+000 -1.0840976E-001 +2.2363480E-004
10 +0.0000000E+000 -4.8773936E-002 +2.0388564E-004
11 +0.0000000E+000 -6.5420631E-003 +9.2685958E-005
12 +0.0000000E+000 +0.0000000E+000 +0.0000000E+000

Figure 4 - Using the Z88 Plot program

Looking at the U(2) displacement between nodes 6 and 7 which equals 0.195 mm which is close to the beam bending theory.

References

(1) Newbold, H.,B., (Revised Lucas, E.,), Modern Practical Building, Caxton, 2nd Ed., London: (1946), .

(2) Seward, S.,, Understanding Structures, Palgrave Macmillan, 3rd Ed., UK: (2003), .

Tested in Firefox 2.04: FireFox homepage ASCII to MathML used in this page: ASCIItoMathML homepage