Transcript Document

Loads and beams
Loads and beams
Starter exercise
Load Bearing beam
3m
3m
1m
S2
S1
Work
and S2
1m
10N
out the loads on supports S1
The UDL of the beam is 2N/m
15 N
Loaded Beam
F
F/2
F/2
The reactions at the end supports
with the load at the centre are
F/2
Shear force diagram
Maximum shear
force = reaction
force
Bending moment diagram
The Maximum bending moment due
to the load if found by calculating the
area under the Shear Force Diagram
between zeros
The area is (L/2) x (F/2) (area of
triangle ½ base x height)
This could be
written as
FL/4
UDL (Uniformly Distributed Load)
The UDL (Uniformly Distributed Load) of a
beam is found by the mass per metre by
the length and g(gravitational field
strength)
UDL = (kg/m) x g x L (which is weight per
metre x length)
mid span moment (Due to weight of
beam)
The equation for mid span
moment is (FL/4) x (L/2)
FL2/8 or wt x L2/8
(L/2) = Length form
end of beam to
midpoint
Maximum design loading
Maximum design loading (M design) is the
Maximum bending moment due to the
load
plus
the mid span (bending) moment (Due to
weight of beam)
Elastic section modulus of a beam
The elastic section modulus of a beam is the
ratio of the maximum bending moment, M to
the maximum allowable stress, .
That is, elastic section modulus
= maximum bending moment / maximum
allowable stress (sigma, σ)
z=M/σ (m3)
Second Moment of Area
 Area
Moment of Inertia, also known
as Second Moment of area - I, is a
property of shape that is used to
predict deflection, bending and stress
in beams.
The second moment of area

The second moment of area


I = ∫ y2dA
where dA is a thin section of the area taken as a
slice parallel to the neutral axis, and y is the
distance separating that slice from the neutral axis.
The second moment of area

I (second moment of area) can also be found using
the equation Z (elastic section modulus ) x Y
(depth to neutral axis)

Equation forms

I=Zxy

Y = I/Z

Z = I/Y
Table 1
Square hollow section 30 x 30 x 3.5
Size DxD Thickness Mass per
mm
t
metre kg
mm
30 x30
3.5
1.74
Area of
section
cm2
3.71
Second
Elastic
moment section
of area Ixx modulus
cm4
Zxxcm3
2.22
1.78
Table 2
Universal beam 470 x 205 x 54
Serial size
mm
Mass per
metre kg
Depth of
section
mm
Width of
section
mm
Web
thickness
mm
427 x 205
54
422
205
8.28
Flange Second
thickness moment
mm
of area Ixx
cm4
11.64
22500
Elastic
section
modulus
Zxxcm3
1200
Area of
section
Cm2
82.3
Worked example Q1
An overhead gantry in a workshop is of the
form shown in Figure 1 below. It is required to
lift a light goods vehicle of mass 6000kg.
Fig 1
60o (all)
A
3m
B
3m
C
Worked example Q1

a) Convert the loading to kN and calculate the
reactions at the end supports. (g = 9.81m/s2)

6000kg x 9.81 = 58.86 kN

Load is at the centre so the reactions at the end
supports are half the load = 29.43kN
Worked example Q1

Show that the maximum force (tensile or
compressive) in any member is 33.98 kN.
29.43kN
60o
Max Force in 60o member
= 29.43/sin 60 = 33.98kN
Max force in horizontal member
force = 33.98 x cos60 =16.99kN
Worked example Q1
All of the members in the truss are 30 x 30 x 3.5
square hollow sections.
Using the data in Table 1 find the cross-sectional
area and calculate the maximum axial stress in
the section.
Worked example Q1

Area of section = 3.71cm2

Max load is 33.98kN

Max stress = 33.98/3.71 = 9.16kN/cm2
Worked example Q1


Safety factor = maximum allowable
stress/working stress
The maximum allowable stress in this example is
27.5 kN/cm2

The working stress is 9.16 kN/cm2
 Safety
factor = 27.5/9.16 = 3
Worked example Q1
 It
is decided that a factor of safety of 3.5 is
required for the gantry in Question 1. The
truss is to be replaced with a universal
beam of section 470 x 205 x 54

(see Table 2) and of the same length as the
span of the truss in Question 1.
Worked example Q2
 Draw
shear force and bending moment
diagrams for the 6000kg loading case.
Worked example Q2
Shear Force Diagram
29.43kN
- 29.43kN
Worked example Q2
29.43kN
6m
Bending
moment
diagram
Bending
moment due
to imposed
load
The area is (L/2) x (F/2) (area of triangle
½ base x height)
=29.43 x 3 = 88.29kNm
Worked example Q2
 Convert
the mass per metre of the beam into
a distributed load and calculate the
maximum mid-span moment in the beam
due to this load. Add this to the maximum
moment due to the imposed load to find the
maximum design loading.
Worked example Q2
 UDL
= 54 x9.81 N/m
 529.74N/m
Mid span moment FL2/8 or wt x L2/8
 529.74x 36/8
 2.38
kNm
Worked example Q2
Maximum design loading (M design) is the
Maximum bending moment due to the
load
plus
the mid span (bending) moment (Due to
weight of beam)
Worked example Q2
=
88.29kNm + 4.74kNm
= 90.67kNm
Worked example Q2
 Find
values for the second moment of
area Ixx and the elastic section
modulus zxx from table 2 and hence
calculate the depth to the neutral
axis, y.
Worked example Q2
Y
= I/Z
 22500cm4/1200cm3
 18.75cm
 Or
187.5mm
Worked example Q2
 Using
the formula Mdesign = σ calculate the
I
Y
maximum bending stress in the beam
Worked example Q2
Stress
(σ) =(Mdesign x Y)/I
=90.67 x106Nmm x 187.5mm
225 x 106mm4
=
75.56N/mm2
Worked example Q2
 Calculate
the factor of safety for this
arrangement assuming the maximum
allowable
stress in 275N/mm2.
Worked example Q2
Safety
factor
275/75.56
= 3.64