Research
Activity
Gravamen, Arvin Raymond F.
BSME – 3GN
Working Stress
Working Stress:
Working stress is the safe stress taken within the elastic range of the material. For brittle materials, it is taken
equal to the ultimate strength divided by suitable factor of safety. However, for materials possessing well
defined yield point, it is equal to yield stress divided by a factor of safety. It is the stress which accounts all sorts
of uncertainties.
It is also called allowable stress, permissible stress, actual stress and safe stress.
Factor of Safety:
Factor of safety is a number used to determine the working stress. It is fixed based on the experimental works
on the material. It accounts all uncertainties such as, material defects, unforeseen loads, manufacturing defects,
unskilled workmanship, temperature effects etc. Factor of safety is a dimensionless number. It is fixed based on
experimental works on each material.
It is defined as the ratio of ultimate stress to working stress for brittle materials or yield stress working stress for
ductile materials.
Yield stress and Ultimate stress
Yield Stress:
Yield stress is defined as the lowest stress at which extension of the test piece increases without increase in
load. It is the stress corresponding to the yield point. For ductile material yield point is well defined whereas for
brittle material it is obtained by offset method. It is also called yield strength.
Ultimate Stress:
Ultimate stress or Ultimate strength corresponds to the highest point of the stress-strain curve. It is the ratio of
maximum load to the original area of cross-section. At this ultimate point, lateral strain gets localized resulting
into the formation of neck.
Sample Problem:
1.1. A spherical pressure vessel 400mm in diameter has a uniform thickness of 6mm. The vessel contains gas
under a pressure of 8,000 KPa. If the allowable tensile stress of material is 420 MPa, what is the factor of
safety with respect to tensile failure?
Solution:
Factor of Safety, F.S. =
Ultimate Stress
Actual Stress
For Spherical Shell:
ςt = ⅓∕⅓ pD
4t
8,000 x (1/1000) 400
4(6)
ςt =
= 133.33 MPa
ςult = 420 MP
420 MPa
133.33 MPa
F.S. =
F.S. = 3.15
1.2. A steel wire 5mm long hanging vertically supports a weight of 1200 N. Determine the required wire
diameter if the stress is limited to 140 MPa and the total elongation must not exceed 4mm. Neglect the
weight of the wire and assume E = 200 Gpa.
Solution:
Diameter based on allowable stress:
ς=
140 =
P
A
1200
D = 3.3m
𝜋
( 4 )D2
a
Diameter based on allowable elongation:
PL
AE
=
4=
1200(5 x 103)
𝜋
( 4 )D2(200 x 103)
D = 3.09
Therefore, use D = 3.3m
1.3. In Fig. 1-12, assume that a 20-mm-diameter rivet joins the plates that are each 110 mm wide. The allowable
stresses are 120 MPa for bearing in the plate material and 60 MPa for shearing of rivet. Determine (a) the
minimum thickness of each plate; and (b) the largest average tensile stress in the plates.
Solution:
Part (a):
From shearing of rivet:
P = τArivets
P = 60 [ 14π(202) ]
P = 6000πN
From bearing of plate material:
P = ςbAb
6000π = 120(20t)
t = 7.85mm
Part (b): Largest average tensile stress in the plate:
P = ςA
6000π = ς*7.85(110−20)+
σ = 26.67MPa
1.4. A steel bar with diameter 30 mm functions in tension as part of a truss. We do not want the bar to
yield. An experienced design engineer recommends a safety factor of 2.5 for this application. What
is the allowable load?
Solution:
Deformable in a System of Axially Loaded Member
The deformation that occurs in a body due to axial loading is known as axial deformation. After the axial
deformation, the axis of structure does not change. The cross section area of the plane will remain
unchanged. After the deformation, the cross section of the structure does not rotate about the same
axis.
Axial deformation depends upon the position of cross section, which is normal to longitudinal axis.
In the linear portion of the stress-strain diagram, the stress is proportional to strain and is given by
since
then
so
To use this formula, the load must be axial, the bar must have a uniform cross-sectional area, and the
stress must not exceed the proportional limit.
If however, the cross-sectional area is not uniform, the axial deformation can be determined by
considering a differential length and applying integration.
where A = ty, and y and t if variable, must be expressed in terms of x.
For a rod of unit mass ρ suspended vertically from one end, the total elongation due to its own weight is
where ρ is in kg/m3, L is the length of the rod in mm, M is the total mass of the rod in kg, A is the crosssectional area of the rod in mm2, and g = 9.81 m/s2.
Stiffness, k
Stiffness is the ratio of the steady force acting on an elastic body to the resulting displacement. It has
the unit of N/mm.
Sample Problem:
2.1. A steel rod having a cross-sectional area of 300 mm2 and a length of 150 m is suspended vertically
from one end. It supports a tensile load of 20 kN at the lower end. If the unit mass of steel is 7850
kg/m3 and E = 200 × 103 MN/m2, find the total elongation of the rod.
Solution:
Elongation due to its own weight:
δ1 =
PL
AE
Where:
P = W = 7850(1/1000)3(9.81)[300(150)(1000)]
P = 3465.3825 N
L = 75(1000) = 75 000 mm
A = 300 mm2
E = 200 000 MPa
Thus,
δ1 =
3465.3825(75000)
300(200000)
δ1 = 4.33 mm
Elongation due to applied load:
δ2 =
PL
AE
Where:
P = 20 kN = 20 000 N
L = 150 m = 150 000 mm
A = 300 mm2
E = 200 000 MPa
Thus,
20000(150000)
300(200000)
δ2 = 50 mm
δ2 =
Total elongation:
δ = δ1 + δ2
δ = 4.33 + 50
δ =54.3mm
2.2. A steel tire, 10 mm thick, 80 mm wide, and 1500.0 mm inside diameter, is heated and shrunk onto a
steel wheel 1500.5 mm in diameter. If the coefficient of static friction is 0.30, what torque is required to
twist the tire relative to the wheel? Neglect the deformation of the wheel. Use E = 200 GPa.
Solution:
δ=
PL
AE
Where:
δ = π (1500.5 - 1500) = 0.5π mm
P=T
L = 1500π mm
A = 10(80) = 800 mm2
E = 200 000 MPa
Thus,
T(1500π)
800(200000)
T = 53333.33N
0.5π =
F = 2T
p(1500)(80) = 2(53333.33)
p = 0.8889MPa→ internal pressure
Total normal force, N:
N = p × contact area between tire and wheel
N = 0.8889 × π(1500.5)(80)
N = 335 214.92 N
Friction resistance, f:
f = μN = 0.30(335 214.92)
f = 100 564.48 N = 100.56 kN
Torque = f × ½(diameter of wheel)
Torque = 100.56 × 0.75025
Torque = 75.44 kN · m
2.3. The rigid bar AB, attached to two vertical rods as shown in Fig. P-213, is horizontal before the load P
is applied. Determine the vertical movement of P if its magnitude is 50 kN.
Solution:
Free body diagram:
For aluminum:
ΣMB = 0
6Pal = 2.5(50)
Pal = 20.83kN
δ=
PL
AE
20.83(3)1000
2500(70000)
δal = 1.78mm
δal =
For steel:
ΣMA=0
6Pst=3.5(50)
Pst=29.17kN
δ=
PL
AE
29.17(4)(10002)
300(200000)
δst = 1.94mm
δst =
Movement diagram:
y3.5 = 1.94 − 1.786
y = 0.09mm
δB = vertical movement of P
δB = 1.78 + y
=1.78 + 0.09
δB = 1.87mm
Temperature Effects on Axially Loaded Members
Temperature changes cause the body to expand or contract. The amount δT, is given by
ΔT = αL(Tf−Ti) = αLΔT
where α is the coefficient of thermal expansion in m/m°C, L is the length in meter, Ti and Tf are the initial
and final temperatures, respectively in °C. For steel, α = 11.25 × 10-6 m/m°C.
If temperature deformation is permitted to occur freely, no load or stress will be induced in the
structure. In some cases where temperature deformation is not permitted, an internal stress is created.
The internal stress created is termed as thermal stress.
For a homogeneous rod mounted between unyielding supports as shown, the thermal stress is
computed as:
deformation due to temperature changes;
δT=αLΔT
deformation due to equivalent axial stress;
δ = PL = ςL
P
AE
δT = δP
E
αLΔT=
σ = EαΔT
where ς is the thermal stress in MPa, E is the modulus of elasticity of the rod in MPa.
If the wall yields a distance of x as shown, the following calculations will be made:
δT = x + δP
αLΔT = x +
where ς represents the thermal stress.
σL
E
Take note that as the temperature rises above the normal, the rod will be in compression, and if the
temperature drops below the normal, the rod is in tension.
Sample Problem
3.1. A steel rod is stretched between two rigid walls and carries a tensile load of 5000 N at 20°C. If the
allowable stress is not to exceed 130 MPa at -20°C, what is the minimum diameter of the rod?
Assume α = 11.7 µm/(m·°C) and E = 200 GPa.
Solution:
δ = δT + δst
ςL
E
= αL(ΔT) +
ς = αE(ΔT) +
PL
AE
P
A
130 = (11.7×10-6)(200000)(40) +
A=
5000
A
5000
36.4
= 137.36 mm2
14πd2=137.36
d = 13.22 mm
3.2. A steel rod 3 feet long with a cross-sectional area of 0.25 in.2 is stretched between two fixed points.
The tensile force is 1200 lb at 40°F. Using E = 29 × 106 psi and α = 6.5 × 10-6 in./(in.·°F), calculate (a) the
temperature at which the stress in the bar will be 10 ksi; and (b) the temperature at which the stress will
be zero.
Solution:
(a) Without temperature change:
ς=
P
A
= 12000.25 = 4800 psi
ς = 4.8ksi < 10ksi
A drop of temperature is needed to increase the stress to 10 ksi. See figure below.
δ = δT + δst
ςL = αL(ΔT) + PL
AE
E
ς = αE(ΔT) +
P
A-6
10000 = 6.5×10 )(29×106)(ΔT) + 1200
0.25
ΔT=27.59°F
Required temperature: (temperature must drop from 40°F)
T= 40−27.59 = 12.41°F
T = 12.41°F
(b) Temperature at which stress will be zero:
From the figure below:
δ = δT + δst
PL
AE
= αL(ΔT)
P = αAE(Tf−Ti)
1200 = (6.5×10-6)(0.25)(29×106)(Tf − 40)
Tf = 65.46°F
3.3. Calculate the increase in stress for each segment of the compound bar shown in Fig. P-266 if the
temperature increases by 100°F. Assume that the supports are unyielding and that the bar is suitably
braced against buckling.
Solution:
δT = αLΔT
δT(st) = 6.5×10−6)(15)(100)
= 0.00975
δT(al) = 12.8×10−6)(10)(100)
= 0.0128 in
δst + δal = δT(st) + δT(al)
(PL/AE)st+(PL/AE)al = 0.00975 + 0.0128
Where Pst = Pal = P. Thus,
P(15)
+
1.5(29×106)
P(10)
2(29×106)
= 0.02255
P = 26691.84 lb
ς=P/A
ςst = 26691.84/1.5
ςst = 17794.56 psi
ςal = 26691.84/2.0
ςal = 13345.92 psi
Statistical Intermediate Members
When the reactive forces or the internal resisting forces over a cross section exceed the number of
independent equations of equilibrium, the structure is called statically indeterminate. These cases
require the use of additional relations that depend on the elastic deformations in the members.
Sample Problem
4.1. A steel bar 50 mm in diameter and 2 m long is surrounded by a shell of a cast iron 5 mm thick.
Compute the load that will compress the combined bar a total of 0.8 mm in the length of 2 m. For steel,
E = 200 GPa, and for cast iron, E = 100 GPa.
Solution:
δ = PL/AE
δ = δcastiron = δsteel = 0.8mm
δcastiron =
Psteel (2000)
= 0.8
(1/4)π(602−502)](100000)
Pcastiron = 11000πN
δsteel=
Psteel (2000)
(1/4)π(502)]( 200000)
= 0.8
Psteel=50000πN
ΣFv = 0
P = Pcastiron + Psteel
P = 11000π + 50000π
P = 61000πN
P = 191.64kN
4.2. A timber column, 8 in. × 8 in. in cross section, is reinforced on each side by a steel plate 8 in. wide
and t in. thick. Determine the thickness t so that the column will support an axial load of 300 kips
without exceeding a maximum timber stress of 1200 psi or a maximum steel stress of 20 ksi. The moduli
of elasticity are 1.5 × 106 psi for timber, and 29 × 106 psi for steel.
Solution:
δsteel = δtimber
(ςL/E)steel = (ςL/E)timber
ςbronzeL
ςsteelL
6
29×106 = 1.5×10
1.5ςsteel=29ςtimber
When ςtimber = 1200 psi
1.5ςsteel = 29(1200)1.5ςsteel = 29(1200)
ςsteel=23200 psi =23.2 ksi >20 ksi (not ok!)
When ςsteel = 20 ksi
1.5(20×1000) = 29ςtimber
ςtimber = 1034.48 psi < 1200 psi (ok!)
Use ςsteel = 20 ksi and ςtimber = 1.03 ksi
ΣFV=0
Fsteel + Ftimber = 300
(ςA)steel+(ςA)timber=300
20[4(8t)]+1.03(82)=300
t=0.365 in
4.3. The rigid platform in Fig. P-239 has negligible mass and rests on two steel bars, each 250.00 mm
long. The center bar is aluminum and 249.90 mm long. Compute the stress in the aluminum bar after
the center load P = 400 kN has been applied. For each steel bar, the area is 1200 mm2 and E = 200 GPa.
For the aluminum bar, the area is 2400 mm2 and E = 70 GPa.
Solution:
δst = δal + 0.10
(ςL/E)st = (ςL/E)al +0.10
ςst(250) ςal(249.90)
=
+ 0.10
200000
70000
0.00125ςst = 0.00357ςal + 0.10
ςst = 2.856ςal + 80
ΣFV = 0
2Pst + Pal = 400000
2ςstAst+ςalAal = 400000
2(2.856ςal + 80)1200 + ςal(2400) = 400000
9254.4ςal + 192000 = 400000
ςal = 22.48 MPa