62 http://www.model-engineer.co.uk Model Engineers’ Workshop

Equation 4

where e is half the beam depth, so substituting gives:

As the lowest design stress level in bending is 9.3 N/sq.mm, the
stress in the roof beam is less than half this value, so is acceptable
and the beam design is satisfactory. As the stress is less than half
the maximum allowable stress, it can be taken that the loading does
not take the beam out of the elastic range, satisfying the previous
assumption. However, there is the possibility that the beams might
buckle sideways under load, but there are ways around this. Firstly,
in my case, the beams were constrained to be vertical at both ends
by being embedded into the top two rows of the bricks of the
garage wall, so would be held more rigidly than if they merely sat on
the top of the wall. Secondly, there is a decking fastened to the top
of the beams consisting of stringers which support the steel sheets
and there is the plywood sheeting fastened to the underside of the
beams. Both these will have some restraining eff ect on the twisting
and buckling of the beams and since the beams are loaded well
below their maximum permissible stress, buckling is unlikely when
the lift ing beam is in use.

Centre wall support

Having decided that using half the width of the garage as a

maximum length for the beam, the support in the centre of the

beam has to be designed. I decided that since I was to split the

garage into two rooms, the centre wall could be used to support the

centre of the whole roof beam. As the load on the beam has been

calculated to be 1570 Newtons, the centre wall has to take half of

this, which is 785 N. As the lift ing beam spreads the load over at

least two of the roof beams, the load on a single upright would be

one quarter of the total load of 1570 N. I built the wall as a series

of uprights secured to a batten on the ceiling and one on the fl oor,

photo 4, and then clad both sides with 9 mm plywood including

a damp proof membrane and some glass fi bre insulation. Each

upright was located directly under a corresponding roof beam. Note

the printed grading marks on the uprights in photo 4. The damp

proof membrane and plywood covering being installed is shown in

photo 5.

Having decided that the part load was to be supported by the

wall upright in photo 4, the ability of the wooden upright to take

the load needed investigating. Taking the worst case that the

upright itself takes the full load (i.e. the plywood surface is non-load

bearing) and assuming that the ends are freely rotating (they are

actually fi xed into slots routed into the horizontal battens) being

the worst case, there are two cases to consider. Firstly, there is the

crushing load which would compress the wood to failure. Reference

1 provides a series of data, again for native American soft woods,

with values of between 16 N/sq.mm and 49 N/sq.mm. As the

upright had actual dimensions of 42 mm x 69 mm (a nominal 3 inch

by 2 inch), and the imposed load is half that of the end of each beam

i.e. 785/2 = 392.5 N, say 400 N, the actual crushing stress will be

400/(42 x 69) = 0.138 N/sq.mm, less than 1% of the minimum value,

so crushing as a failure mechanism can be discounted.

The next consideration is lateral instability of the column, which

results in buckling of the upright column. The Wood Handbook

(ref. 1) suggests that Euler’s formula is suitable for slender vertical

columns, which are columns where the column buckles before the

compressive stress exceeds the proportional limit. This is the worst

case for loading. Euler’s formula is quoted as:

where C is the height of the column and r is the radius of gyration.

For a rectangular beam r is equal to the least dimension of the

cross-section divided by the square root of 12, i.e. r = d/√12.

Substituting in the values for the column, d is 42 mm, so r =

42/√12 = 12.12 mm; the cross-sectional area of the column is 42 x

69 = 2898 sq.mm; and the length is 2100 mm; so the maximum

permissible end load for the column is:

which exceeds the design imposed load of 400 N per upright

by a factor of three. Consequently, the upright is adequate for the

task. In fact, the upright will be able to take a greater load since the

column alone would tend to buckle along the line of the wall as the

column is thinner in that direction. As the plywood wall has been

screwed to the upright, it will be more constrained in that direction,

so would in all probability buckle in a direction at right angles to

the wall if seriously overloaded. However, the eff ect of the plywood

cladding on increasing the buckling potential of the wall cannot be

reliably predicted – only where the ply is glued onto the upright

could any practical increase in load be calculated.

Securing the lift ing beam

The lift ing beam now needs to be secured to the roof beams.

The timber uprights for the wall

(^) M×e
s =
I
(^981250) ×87.5
s = = 4.07Nmm-2
20.1×10.6
2898 ×π× 4146
F = = 1257 N
(
2100
)
²
(^) 12.12
4
F πE

A
(
C
)
2
r