Transcript Document

External effects of
forces on structures
External effects of forces on structures
Force
Cross sectional Stress = Force/cross sectional
area
area
σ = F/A
Strain = extension/ original
length
External effects of forces on structures
Young’s modulus of elasticity (E)
= stress/strain
= (F/A) ÷(e/L)
Or
FL/Ae
External effects of forces on structures
Force
A
B
compound bars
of 2 materials
with different
values of Young’s
modulus in
parallel
External effects of forces on structures
Force
A
B
Young’s modulus
for A = E1 and
young’s modulus
for E2
Stress for A = σ1
and stress for B
= σ2
External effects of forces on structures
Force
A
B
The extension
for both is the
same which
means that
strain is the
same
External effects of forces on structures
Force
E1 = σ1/strain
strain = σ1/E1
E2 = σ2/strain
Strain = σ2/E2
Strain is equal for both bars
Therefore
AA
BB
σ1/E1 = σ2/E2
External effects of forces on structures
A reinforced concrete column 300mm square
contains 4, 16 mm diameter steel bars and
carries an axial load of 400kN.
Calculate the stresses in the steel and the
concrete.
Esteel = 200kN/mm2 and Econcrete = 20kN/mm2
External effects of forces on structures
Total area of concrete pillar + steel
bars = 300mm x 300 mm = 9 x
104mm2
Total cross sectional area of one
steel bar = πr2
=π x82
= 64π
201.1mm2
Cross sectional area of 4 bars
= 804.4 mm2 = 8.04 x 102mm2 (A1)
External effects of forces on structures
Total area of concrete pillar
= = 9 x 104mm2 - 8.04 x 102mm2
=8.92 x 104mm2 (A2)
σ1is stress in steel and σ2 is
stress in concrete
F/A =σ
F = σA
System force, F= σ1 A1 + σ2A2
External effects of forces on structures
F= σ1 A1 + σ2A2
(1)400 x103 =σ1 x 8.04 x 102mm2
+ σ2 x 8.92 x 104mm2
σ1/E1 = σ2/E2
σ1/σ2 = E1/E2
σ1/σ2 =
200kN/mm2 ÷ 20kN/ mm2
= 10/1
σ1= 10 x σ2
External effects of forces on structures
(1)400 x103 =σ1 x 8.04 x 102mm2
+ σ2 x 8.92 x 104mm2
Substituting (𝜎1=10 x σ2) in (1)
400 x103 =σ2 x 8.04 x 103mm2
+ σ2 x 8.92 x 104mm2
= σ2 ( 8.04 x 103 + 8.92 x 104)
400 x103 = σ2 x 9.7 x104
σ2 = 400 x103 /9.7 x104
= 4.12N/mm2
σ1 = 41.2N/mm2
x10
Shear stress
A shearing force is applied to the top of the
rectangle while the bottom is held in place.
The resulting shear stress deforms the rectangle into
a parallelogram.
The area involved would be the top of the
parallelogram.
(shear stress) Τ = F/A
F =T x A
Shear stress
 When
two cross sections of the element
takes the shear load it is called double
shear.
For double shear the force is
shared (F/2)
Shear stress, T = F/2A
F =T x 2A
Shear stress
The hitching pin in the assembly
shown is 10mm diameter and in
double shear. If the shear strength of
the steel is 260 N/mm2 and the
factor of safety employed is 2, what
is the maximum force F
that can be applied to
the hitch?
10mm
Shear stress
Τ = F/2A F = T x 2A
260 x 2 x 78.5 = 40.84kN
Applying a safety factor of 2
Max force = 40.84/2kN
=20.42kN
10mm
Resolving forces
Consider a jib crane
situation: we need to
resolve forces in each
member
A
B
40o
C
15kN
Resolving forces
Draw the forces as a
vector triangle
A
A
15kN
B
40o
C
B
15kN
40o
C
Resolving forces
Sin40o = 15kN/AC
AC = 15kN/sin 40o
AC = 23.3kN (tension)
A
15kN
B
40o
C
Cos40o = BC/AC
Cos40o = BC/23.3kN
BC = Cos40o x 23.3kN
BC =17.8kN (compression)
Resolving forces
A
B
40o
C
15kN
In this situation
axial load are
horizontal and
shear loads are
vertical
Resolving forces
A
Max axial load
(horizontal)
BC = 17.8kN
B
40o
C
15kN
Max shear load
(vertical)
=15kN
Resolving forces

The tension member (AC) is a 20mm Diameter steel
rod, which has an ultimate yield stress of 275N/mm2

The actual stress = F/A

= 23.3x103 N/314.2mm2


=74.2N/mm2
The safety factor against failure for this member
275/74.2 = 3.7
Resolving forces
 The
compression member is 25 x 25 x 2 steel
square hollow section ( this means that the
external dimensions are 25mm x 25mm and
the thickness of the steel is 2mm. (the
inside dimensions 21mm x 21mm) The cross
sectional area of steel is (25mm x 25mm) –
(21mm x 21mm) = 184mm2
Resolving forces
 The

Stress in the horizontal member
= F/A = 17.8 x 103N/184mm2
=
 The
96.7N/mm2
safety factor = 275/96.7 = 2.8