stress

2. Stress
Stress, is the intensity of the internal force acting on a specific plane (area)
passing through a point.

3. Normal Stress. The intensity of the force acting normal to DA is defined as the
normal stress, s (sigma). Since DFz is normal to the area then
If the normal force or stress “pulls” on DA, it is referred to as tensile stress,
whereas if it “pushes” on DA it is called compressive stress.
Shear Stress. The intensity of force acting tangent to DA is called the shear
stress, t (tau). Here we have shear stress components,

4. Units.
Since stress represents a force per unit area, in the International Standard or SI
system, the magnitudes of both normal and shear stress are specified in the
basic units of newtons per square meter (N/m2). This unit, called a pascal
(1Pa=1 N/m2) is rather small, and in engineering work prefixes such as kilo-
(103) symbolized by k, mega-(106) symbolized by M, or giga-(109) symbolized
by G, are used to represent larger, more realistic values of stress.

5. Average Normal Stress in an Axially Loaded Bar
If we pass a section through the bar, and separate it into two parts, then
equilibrium requires the resultant normal force at the section to be P,. Due
to the uniform deformation of the material, it is necessary that the cross
section be subjected to a constant normal stress distribution

6. s = average normal stress at any point on the cross-sectional area internal
P = resultant normal force, which acts through the centroid of the cross-
sectional area. P is determined using the method of sections and the equations
of equilibrium
A = cross-sectional area of the bar where is determined
Each small area DA on the cross section is subjected to a force DF=s DA,
and the sum of these forces acting over the entire cross-sectional area
must be equivalent to the internal resultant force P at the section.

7. Maximum Average Normal Stress
Occasionally, however, the bar may be subjected to several external loads
along its axis, or a change in its cross-sectional area may occur. As a result,
the normal stress within the bar could be different from one section to the
next, and, if the maximum average normal stress is to be determined, then it
becomes important to find the location where the ratio PA is a maximum.

8. Procedure for Analysis
The equation s = P/A gives the average normal stress on the cross-sectional
area of a member when the section is subjected to an internal resultant normal
force P. For axially loaded members, application of this equation requires the
following steps.
• Section the member perpendicular to its longitudinal axis at the point where
the normal stress is to be determined and use the necessary free-body
diagram and force equation of equilibrium to obtain the internal axial force
P at the section.
• Determine the member’s cross-sectional area at the section and calculate
the average normal stress s = P/A.

9. EXAMPLE 1.
The bar in the figure below has a constant width of 35 mm and a thickness of
10 mm. Determine the maximum average normal stress in the bar when it is
subjected to the loading shown.

10. Solution:
Since the cross-sectional area of the bar is constant, the largest average normal
stress occurs the region with the internal axial forces (i.e. region BC).
𝜎𝐵𝐶 =
30 × 1000
35 × 10
= 85.7𝑀𝑃𝑎

11. EXAMPLE 2.
The 80-kg lamp is supported by two rods AB and BC as shown in the figure
below. If AB has a diameter of 10 mm and BC has a diameter of 8 mm,
determine the average normal stress in each rod.

12. Solution:
՜
+
Σ𝐹𝑥 = 0 𝐹𝐵𝐶 ×
4
5
− 𝐹𝐵𝐴 × 𝑐𝑜𝑠60 = 0
+ Σ𝐹𝑦 = 0 𝐹𝐵𝐶 ×
3
5
+ 𝐹𝐵𝐴 × 𝑠𝑖𝑛60 − 784.8 = 0
𝐹𝐵𝐴 = 632.4𝑁 𝐹𝐵𝐶 = 395.2𝑁
𝜎𝐵𝐴 =
𝐹𝐵𝐴
𝐴𝐵𝐴
=
632.4
𝜋 × 52
= 8.05𝑀𝑃𝑎
𝜎𝐵𝐶 =
𝐹𝐵𝐶
𝐴𝐵𝐶
=
395.2
𝜋 × 42
= 7.86𝑀𝑃𝑎

13. EXAMPLE 3.
The casting shown in the figure below is made of steel having a specific
weight of gst = 7850 kg/m3. Determine the average compressive stress acting at
points A and B.

14. Solution:
𝑊𝑠𝑡 = 𝜋 × 0.2282 × 0.838 × 7850 × 9.8/1000 = 10.53𝑘𝑁
+ Σ𝐹𝑧 = 0 𝑃 − 𝑊𝑠𝑡 = 0
𝑃 = 10.53𝑘𝑁
𝜎𝐴 = 𝜎𝐵 =
𝑃
𝐴
=
10.53 × 103
𝜋 × 2282
= 0.064𝑀𝑃𝑎

15. EXAMPLE 4.
Member AC shown below is subjected to a vertical force of 3 kN. Determine
the position x of this force so that the average compressive stress at the smooth
support C is equal to the average tensile stress in the tie rod AB. The rod has a
cross-sectional area of 400 mm2 and the contact area at C is 650 mm2.

16. Solution:
There are three unknowns, namely, FAB, FC and x. To solve this problem we
need three equations:
+ Σ𝐹𝑦 = 0 𝐹𝐴𝐵 − 3000 + 𝐹𝑐 = 0 … … … (1)
+ Σ𝑀𝐴 = 0 − 3000𝑥 + 200𝐹𝑐 = 0 … … … (2)
A necessary third equation can be written that requires the tensile stress in the
bar AB and the compressive stress at C to be equivalent.
𝜎 =
𝐹𝐴𝐵
400
=
𝐹𝐶
650
𝐹𝐶 = 1.625𝐹𝐴𝐵 … … … … … … 3
Solving the above three equations for FAB, FC and x
𝐹𝐴𝐵 = 1143𝑁
𝐹𝐶 = 1857𝑁
𝑥 = 124𝑚𝑚

17. Average Shear Stress
tavg = average shear stress at the section, which
is assumed to be the same at each point located
on the section
V = internal resultant shear force on the section
determined from the equations of equilibrium
A = area at the section
Shear stress is the stress component that acts in the plane of the sectioned
area. To show how this stress can develop, consider the effect of applying a
force F to the bar shown below. If the supports are considered rigid, and F is
large enough, it will cause the material of the bar to deform and fail along the
planes identified by AB and CD.A free-body diagram of the unsupported
center segment of the bar, Fig. 1–20b, indicates that the shear force must be
applied at each section to hold the segment in equilibrium

18. Procedure for Analysis
• Section the member at the point where the average shear stress is to be
determined.
• Draw the necessary free-body diagram, and calculate the internal shear
force V acting at the section that is necessary to hold the part in
equilibrium.
• Determine the sectioned area A, and determine the average shear stress tavg
= V/A

19. EXAMPLE 1.
Determine the average shear stress in the 20-mm-diameter pin at A and the 30-
mm-diameter pin at B that support the beam in the figure below.

20. Solution:
+↶ Σ𝑀𝐴 = 0; 𝐹𝐵 ×
4
5
× 6 − 30 × 2 = 0; 𝐹𝐵 = 12.5𝑘𝑁
՜
+
Σ𝐹𝑥 = 0; −𝐴𝑥 +𝐹𝐵 ×
3
5
= 0; 𝐴𝑥 = 7.5𝑘𝑁
+↑ Σ𝐹𝑦 = 0; 𝐴𝑦 − 30 + 𝐹𝐵 ×
4
5
= 0; 𝐴𝑦 = 20𝑘𝑁
The resultant force acting on pin A is:
𝐹𝐴 = 𝐴𝑥
2
+ 𝐴𝑦
2
= 7.52 + 202 = 21.36𝑘𝑁
𝑉𝐴 =
𝐹𝐴
2
= 10.68𝑘𝑁 (Double shear pin)
𝑉𝐵 = 𝐹𝐵 = 12.5𝑘𝑁 (Single shear pin)
(𝜏𝐴)𝐴𝑣𝑔 =
𝑉𝐴
𝐴𝐴
=
10.68 × 103
𝜋 × 102
= 34𝑀𝑃𝑎
(𝜏𝐵)𝐴𝑣𝑔 =
𝑉𝐵
𝐴𝐵
=
12.5 × 103
𝜋 × 152
= 17.7𝑀𝑃𝑎

21. EXAMPLE 2.
If the wood joint in the figure below has a width of 150 mm, determine the
average shear stress developed along shear planes a–a and b–b. For each plane,
represent the state of stress on an element of the material.

22. Solution:
՜
+
Σ𝐹𝑥 = 0; 6 − 𝐹 − 𝐹 = 0; 𝐹 = 3𝑘𝑁
Now consider the equilibrium of segments cut across shear planes a–a and b–b
՜
+
Σ𝐹𝑥 = 0; 𝑉
𝑎 − 3 = 0; 𝑉
𝑎 = 3𝑘𝑁
՜
+
Σ𝐹𝑥 = 0; 𝑉𝑏 − 3 = 0; 𝑉𝑏 = 3𝑘𝑁
(𝜏𝐴)𝐴𝑣𝑔 =
𝑉
𝑎
𝐴𝑎
=
3 × 103
100 × 150
= 0.2𝑀𝑃𝑎
(𝜏𝐴)𝐴𝑣𝑔 =
𝑉𝑏
𝐴𝑏
=
3 × 103
125 × 150
= 0.16𝑀𝑃𝑎

23. EXAMPLE 3. The inclined member in the figure below is subjected to a
compressive force of 3 kN. Determine the average compressive stress along
the smooth areas of contact defined by AB and BC, and the average shear
stress along the horizontal plane defined by DB.

24. Solution:
՜
+
Σ𝐹𝑥 = 0; 𝐹𝐴𝐵 − 3 ×
3
5
= 0; 𝐹𝐴𝐵 = 1.8𝑘𝑁
+↑ Σ𝐹𝑦 = 0; 𝐹𝐵𝐶 − 3 ×
4
5
= 0; 𝐹𝐵𝐶 = 2.4𝑘𝑁
From the free-body diagram of the top segment ABD
of the bottom member, the shear force acting on the
sectioned horizontal plane DB is:
՜
+
Σ𝐹𝑥 = 0; 𝑉 = 1.8𝑘𝑁
𝜎𝐴𝐵 =
𝐹𝐴𝐵
𝐴𝐴𝐵
=
1.8 × 103
25 × 38
= 1.89𝑀𝑃𝑎
𝜎𝐵𝐶 =
𝐹𝐵𝐶
𝐴𝐵𝐶
=
2.4 × 103
50 × 38
= 1.26𝑀𝑃𝑎
(𝜏𝐷𝐵)𝐴𝑣𝑔 =
𝑉
𝐴𝐷𝐵
=
1.8 × 103
76 × 38
= 0.62𝑀𝑃𝑎
3kN
1.8kN

25. Allowable Stress
To properly design a structural member or mechanical element it is necessary to
restrict the stress in the material to a level that will be safe. To ensure this safety, it
is therefore necessary to choose an allowable stress that restricts the applied load to
one that is less than the load the member can fully support. There are many reasons
for doing this. For example, the load for which the member is designed may be
different from actual loadings placed on it. The intended measurements of a
structure or machine may not be exact, due to errors in fabrication or in the
assembly of its component parts. Unknown vibrations, impact, or accidental
loadings can occur that may not be accounted for in the design. Atmospheric
corrosion, decay, or weathering tend to cause materials to deteriorate during
service. And lastly, some materials, such as wood, concrete, or fiber-reinforced
composites, can show high variability in mechanical properties.

26. One method of specifying the allowable load for a member is to use a number
called the factor of safety. The factor of safety (F.S.) is a ratio of the failure
load to the allowable load Here is found from experimental testing of the
material, and the factor of safety is selected based on experience so that the
above mentioned uncertainties are accounted for when the member is used
under similar conditions of loading and geometry. Stated mathematically,
If the load applied to the member is linearly related to the stress developed
within the member, as in the case of using s = P/A and tavg = V/A then we can
also express

27. Design of Simple Connections
EXAMPLE 1.
The control arm is subjected to the loading shown below. Determine the
required diameter of the steel pin at C if the allowable shear stress for the steel
is tallow = 55MPa.

28. Solution:
+↶ Σ𝑀𝐶 = 0; 𝐹𝐵𝐴 × 0.2 − 13 × 0.075 − 22 ×
3
5
× 0.125 = 0; 𝐹𝐵𝐴 = 13.1𝑘𝑁
՜
+
Σ𝐹𝑥 = 0; 𝑅𝐶𝑥 + 22 ×
4
5
− 13.1 = 0; 𝑅𝐶𝑥 = −4.5𝑘𝑁
+↑ Σ𝐹𝑦 = 0; 𝑅𝐶𝑦 − 13 − 22 ×
3
5
= 0; 𝑅𝐶𝑦 = 26.2𝑘𝑁
𝐹𝐶 = 𝑅𝐶𝑥
2
+ 𝑅𝐶𝑦
2
= (−4.5)2+26.22 = 26.6𝑘𝑁
Since the pin is subjected to double shear, a shear force of (26.6/2=13.3kN)
acts over its cross-sectional area between the arm and each supporting leaf for
the pin.
𝜏𝑎𝑙𝑙𝑜𝑤 =
𝑉
𝐴
; 𝐴 =
𝑉
𝜏
=
13.3 × 103
55
= 242𝑚𝑚2
𝐴 =
𝜋
4
𝑑2; 242 =
𝜋
4
𝑑2; 𝑑 = 17.6𝑚𝑚
Use a pin having a diameter of at least 17.6mm or higher

29. EXAMPLE 2.
The suspender rod is supported at its end by a fixed-connected circular disk as
shown below. If the rod passes through a 40-mm-diameter hole, determine the
minimum required diameter of the rod and the minimum thickness of the disk
needed to support the 20-kN load. The allowable normal stress for the rod is
sallow = 60 MPa and the allowable shear stress for the disk is tallow = 35 MPa.

30. Solution:
The required cross-sectional area of the rod:
𝜎𝑎𝑙𝑙𝑜𝑤 =
𝑃
𝐴
; 𝐴 =
𝑃
𝜎𝑎𝑙𝑙𝑜𝑤
;
𝜋
4
𝑑2 =
20 × 103
60
; 𝑑 = 20.6𝑚𝑚
Thickness of Disk:
𝜏𝑎𝑙𝑙𝑜𝑤 =
𝑉
𝐴
; 𝐴 =
𝑉
𝜏
=
20 × 103
35
= 571𝑚𝑚2
𝐴 = 2𝜋. 𝑟. 𝑡; 𝑡 =
𝐴
2𝜋. 𝑟
=
571
2𝜋 × 20
= 4.55𝑚𝑚

31. EXAMPLE 3.
The shaft shown below is supported by the collar at C, which is attached to the
shaft and located on the right side of the bearing at B. Determine the largest
value of P for the axial forces at E and F so that the bearing stress on the collar
does not exceed an allowable stress of (sb)allow = 75 MPa and the average
normal stress in the shaft does not exceed an allowable stress of (st)allow = 55
MPa.

32. Solution:
To solve the problem we will determine P for each possible failure condition.
Then we will choose the smallest value.
Normal Stress.
(𝜎𝑡)𝑎𝑙𝑙𝑜𝑤 =
𝐹
𝐴
; 𝐹 = (𝜎𝑡)𝑎𝑙𝑙𝑜𝑤𝐴; 3𝑃 = 55 ×
𝜋
4
× 602/1000
P = 51.8kN 1
Bearing Stress.
Bearing area, 𝐴𝑏 =
𝜋
4
802
−
𝜋
4
602
= 2199𝑚𝑚2
(𝜎𝑏)𝑎𝑙𝑙𝑜𝑤 =
𝐹
𝐴𝑏
; 𝐹 = (𝜎𝑏)𝑎𝑙𝑙𝑜𝑤𝐴𝑏; 3𝑃 = 75 × 2199/1000
P = 55kN 2
Therefore, the largest load that can be applied to the shaft is 51.8kN

33. EXAMPLE 4.
The rigid bar AB shown below is supported by a steel rod AC having a
diameter of 20 mm and an aluminum block having a cross-sectional area of
1800mm2. The 18-mm-diameter pins at A and C are subjected to single shear.
If the failure stress for the steel and aluminum is (sst)fail = 680 MPa and (sal)fail
= 70 MPa respectively, and the failure shear stress for each pin is tfail = 900
MPa, determine the largest load (P) that can be applied to the bar. Apply a
factor of safety of F.S. = 2.

34. (𝜎𝑠𝑡)𝑎𝑙𝑙𝑜𝑤 =
(𝜎𝑠𝑡)𝐹𝑎𝑖𝑙
𝐹. 𝑆.
=
680
2
= 340𝑀𝑃𝑎
(𝜎𝑎𝑙)𝑎𝑙𝑙𝑜𝑤 =
(𝜎𝑎𝑙)𝐹𝑎𝑖𝑙
𝐹. 𝑆.
=
70
2
= 35𝑀𝑃𝑎
𝜏𝑎𝑙𝑙𝑜𝑤 =
𝜏𝐹𝑎𝑖𝑙
𝐹. 𝑆.
=
900
2
= 450𝑀𝑃𝑎
+↶ Σ𝑀𝐵 = 0; −𝐹𝐴𝐶 × 2 + 𝑃 × 1.25 = 0; 𝑃 = 1.6𝐹𝐴𝐶 … … . 1
+↶ Σ𝑀𝐴 = 0; 𝐹𝐵 × 2 − 𝑃 × 0.75 = 0; 𝑃 = 2.67𝐹𝐵 … … . 2
Allowable stress in the rod AC:
𝐹𝐴𝐶 = (𝜎𝑠𝑡)𝑎𝑙𝑙𝑜𝑤× 𝐴𝐴𝐶 = 340 × 𝜋 × 102 × 10−3 = 106.8𝑘𝑁
Using Eq. 1, 𝑃 = 171𝑘𝑁
Allowable stress in the aluminum block:
𝐹𝐵 = (𝜎𝑎𝑙)𝑎𝑙𝑙𝑜𝑤× 𝐴𝐵 = 35 × 1800 × 10−3
= 63𝑘𝑁
Using Eq. 2, 𝑃 = 168𝑘𝑁
Single shear stress in the pin
𝐹𝐴𝐶 = 𝑉 = 𝜏𝑎𝑙𝑙𝑜𝑤 × 𝐴 = 450 × 𝜋 × 92 × 10−3 = 114.5𝑘𝑁
Using Eq. 1, 𝑃 = 183𝑘𝑁
Therefore, the maximum value of P that can be applied is 𝑃 = 168𝑘𝑁

35. EXAMPLE 5.
The assembly consists of three disks A, B, and C that are used to support the
load of 140 kN. Determine the smallest diameter d1 of the top disk, the diameter
d2 within the support space, and the diameter d3 of the hole in the bottom disk.
The allowable bearing stress for the material is (sallow)b = 350MPa and
allowable shear stress is tallow = 125 MPa.

37. Problem:
If the allowable bearing stress for the material under the supports at A and B is
(sb )allow = 1.5 MPa, determine the maximum load P that can be applied to the
beam. The bearing plates A’ and B’ have square cross sections of
150mmG150mm and 250mmG250mm, respectively.