1. CHAPTER OBJECTIVES Review important principles of statics
Use the principles to determine internal resultant loadings in a body
Introduce concepts of normal and shear stress
Discuss applications of analysis and design of members subjected to an axial load or direct shear

2. CHAPTER OUTLINE Introduction Equilibrium of a deformable body Stress
Average normal stress in an axially loaded bar
Average shear stress
Allowable stress
Design of simple connections

3. 1.1 INTRODUCTION Solid Mechanics A branch of mechanics
It studies the relationship of
External loads applied to a deformable body, and
The intensity of internal forces acting within the body
It is used to compute deformations of a body
Study body’s stability when external forces are applied to it

4. Historical development
1.1 INTRODUCTION
Historical development
Beginning of 17th century (Galileo)
Early 18th century (Saint-Venant, Poisson, Lamé and Navier)
In recent times, with advanced mathematical and computer techniques, more complex problems can be solved

5. 1.2 EQUILIBRIUM OF A DEFORMABLE BODY
External loads
Surface forces
Area of contact
Concentrated force
Linear distributed force
Centroid C (or geometric center)
Body force (e.g., weight)

6. 1.2 EQUILIBRIUM OF A DEFORMABLE BODY
Support reactions
for 2D problems

7. 1.2 EQUILIBRIUM OF A DEFORMABLE BODY
Equations of equilibrium
balance of forces
balance of moments
∑ F = 0
∑ MO = 0
Two-Dimension:
Σ Fx = 0
Σ Fy = 0
Σ Mx = 0
Σ My = 0
Σ Mo = 0
Three-Dimension:
Σ Fx = 0
Σ Fy = 0
Σ Fz = 0
Σ Mx = 0
Σ My = 0
Σ Mz = 0
Σ Mo = 0

8. 1.2 EQUILIBRIUM OF A DEFORMABLE BODY
Internal resultant loadings
For coplanar loadings (2-D):
Normal force, N
Shear force, V
Bending moment, M

9. 1.2 EQUILIBRIUM OF A DEFORMABLE BODY
For coplanar loadings:
Apply Fx = 0 to solve for N
Apply Fy = 0 to solve for V
Apply MO = 0 to solve for Mo

10. 1.3 STRESS Concept of stress
To obtain distribution of force acting over a sectioned area
Assumptions of material:
It is continuous (uniform distribution of matter)
It is cohesive (all portions are connected together)

11. 1.3 STRESS
Fig.(a) : A body is subjected to 4-forces and it is in equilibrium
Fig.(b) : Method of Section. Internal force distribution acting on
the exposed imaginary section.

12. Normal Stress, s : The intensity of force, or force per
Consider ΔA in figure above
Small finite force, ΔF acts on ΔA
As ΔA → 0, ΔF → 0
But stress (ΔF / ΔA) → finite limit (∞)
Normal Stress, s : The intensity of force, or force per
unit area, acting normal to DA.
σz =
lim
ΔA →0
ΔFz ΔA

13. 1.3 STRESS ΔFz ΔA σz = ΔA →0 Normal stress
Intensity of force, or force per unit area, acting normal to ΔA
Symbol used for normal stress, is σ (sigma)
σz =
lim
ΔA →0
ΔFz ΔA
Tensile stress: normal force “pulls” or “stretches” the area element ΔA
Compressive stress: normal force “pushes” or “compresses” the area element ΔA

14. 1.3 STRESS ΔFx ΔFy ΔA τzx = ΔA→0 ΔA τzy = ΔA →0 Shear stress
Intensity of force, or force per unit area, acting tangent to ΔA
Symbol used for normal stress is  (tau)
τzx =
lim
ΔA→0
ΔFx ΔA
τzy =
lim
ΔA →0
ΔFy ΔA

15. sz , tzx , tzy sy , tyx , tyz sx , txy , txz 1.3 STRESS
Stress Components :
sy , tyx , tyz
sx , txy , txz

16. 1.3 STRESS Units (SI system) General state of stress
Figure shows the state of stress acting around a chosen point in a body
Units (SI system)
Newtons per square meter (N/m2) or a pascal (1 Pa = 1 N/m2)
1 kPa = 103 N/m2 (kilo-pascal)
1 MPa = 106 N/m2 (mega-pascal)
1 GPa = 109 N/m2 (giga-pascal)

17. 1.4 AVERAGE NORMAL STRESS IN AXIALLY LOADED BAR
Examples of axially loaded bar
Usually long and slender structural members
Truss members, hangers, bolts, etc.
Prismatic means all the cross sections are the same

18. 1.4 AVERAGE NORMAL STRESS IN AXIALLY LOADED BAR
Assumptions
Uniform deformation: Bar remains straight before and after load is applied, and cross section remains flat or plane during deformation
In order to get uniform deformation, force P be applied along centroidal axis of cross section

19. 1.4 AVERAGE NORMAL STRESS IN AXIALLY LOADED BAR
Tension
Compression P A
σ = P A
σ =

20. EXAMPLE 1-1 Bar width = 35 mm, thickness = 10 mm
Determine maximum average normal stress in bar when subjected to loading shown.

21. EXAMPLE 1-1 Internal loading Normal force diagram
PCD
PAB = 12 kN
PBC = 30 kN
PCD = 22 kN
Normal force diagram A B C D
By inspection, largest loading segment is BC, where PBC = 30 kN

22. Average maximum normal stress
EXAMPLE 1-1
PBC = 30 kN
Average maximum normal stress
σBC =
PBC A
30(103) N
(0.035 m)(0.010 m) =
= 85.7 MPa

23. EXAMPLE 1-2
The 200 N ( 20 kg) lamp is supported by two steel rods connected by a ring at A. Determine which rod is subjected to the greater average normal stress and compute its value. Take q = 60o. The diameter of each rod is given in the figure.

24. EXAMPLE 1-2 y Equation of equilibrium FAC FABW
FAC
FAB x y
Equation of equilibrium
+Fx = 0, FACcos60o – FABsin60o = 0
+Fy = 0, FACsin60o + FABcos60o – W = 0
The 2 equations yield
FAB = 100 N
FAC = N

25. EXAMPLE 1-2 Average normal stress AAB = ¼  (12)2 mm2 = 113.1 mm2
AAC = ¼  (10)2 mm2 = mm2
60o W
FAC
FAB x y
Substituting the appropriate values,
we have
sAB = N/mm2 = MPa
sAC = N/mm2 = MPa
The greater normal stress is in rod AC.

26. 1.5 AVERAGE SHEAR STRESS
Shear stress is the stress component that act in the plane of the sectioned area.
Consider a force F acting to the bar
For rigid supports, and F is large enough, bar will deform and fail along the planes identified by AB and CD
Free-body diagram indicates that shear force, V = F/2 be applied at both sections to ensure equilibrium

27. 1.5 AVERAGE SHEAR STRESS V τavg = A
Average shear stress over each section is: V A
τavg =
τavg = average shear stress at section, assumed to be the same at each point on the section
V = internal resultant shear force at section determined from equations of equilibrium
A = area of section

28. 1.5 AVERAGE SHEAR STRESS
Case discussed above is example of simple or direct shear
Caused by the direct action of applied load F
Occurs in various types of simple connections, e.g., bolts, pins, welded material

29. 1.5 AVERAGE SHEAR STRESS Single shear
Steel and wood joints shown above are examples of single-shear connections, also known as lap joints.
Since we assume members are thin, there are no moments caused by F

30. 1.5 AVERAGE SHEAR STRESS Single shear
For equilibrium, cross-sectional area of bolt and bonding surface between the two members are subjected to single shear force, V = F
The average shear stress equation can be applied to determine average shear stress acting on colored section in (d).

31. 1.5 AVERAGE SHEAR STRESS Bolt F Shear area
The bolt is undergoing shear stress
d = bolt diameter

32. 1.5 AVERAGE SHEAR STRESS Double shear
The joints shown below are examples of double-shear connections, often called double lap joints.
For equilibrium, cross-sectional area of bolt and bonding surface between two members subjected to double shear force, V = F/2
Apply average shear stress equation to determine average shear stress acting on colored section in (d).

33. 1.5 AVERAGE SHEAR STRESS Bolt F F/2 Shear area
The bolt is undergoing shear stress:
d = bolt diameter

34. 1.5 AVERAGE SHEAR STRESS F Bolt Single shear Which one stronger ??
WHY ?? F
Bolt
F/2
double shear
Smaller shear stress !!

35. 1.6 ALLOWABLE STRESS Ffail Fallow F.S. =
When designing a structural member or mechanical element, the stress in it must be restricted to safe level
Choose an allowable load that is less than the load the member can fully support
One method used is the factor of safety (F.S.)
F.S. =
Ffail
Fallow

36. 1.6 ALLOWABLE STRESS F.S. = σfail σallow F.S. = τfail τallow
If load applied is linearly related to stress developed within member, then F.S. can also be expressed as:
F.S. =
σfail
σallow
F.S. =
τfail
τallow
In all the equations, F.S. is chosen to be greater than 1, to avoid potential for failure
Specific values will depend on types of material used and its intended purpose

37. 1.7 DESIGN OF SIMPLE CONNECTIONS
To determine area of section subjected to a normal force, use
A = P
σallow
To determine area of section subjected to a shear force, use
A = V
τallow

38. EXAMPLE 1-3
The two members pinned together at B. If the pins have an allowable shear stress of τallow = 90 MPa, and allowable tensile stress of rod CB is (σt)allow = 115 MPa
6 kN B A C
2 m
1 m 5 4 3
6 kN B A C
2 m
1 m 5 4 3
6 kN B A C
2 m
1 m 5 4 3
Determine to nearest mm the smallest diameter of pins A and B and the diameter of rod CB necessary to support the load.

39. Draw free-body diagram of member AB:
EXAMPLE 1-3
Draw free-body diagram of member AB: Ax Bx By Ay
Reaction forces:
∑ MA = 0; +
By = 4 kN +
∑ Fy = 0;
Ay = 2 kN
∑ Fx = 0; +
Ax = Bx

40. EXAMPLE 1-3 B Components : B By By = 4 kN = B (3/5) Bx Bx = B (4/5)
We get Bx = 5.32 kN
Resultant reaction force: B = 6.67 kN

41. EXAMPLE 1-3 B = 6.67 kN Ay = 2 kN A Ax = 5.32 kN
Resultant reaction force:
= 5.68 kN

42. EXAMPLE 1-3 VA A = 5.68 kN Diameter of pin A (double shear):
Shear force at pin A :
VA = A/2 = 2.84 kN
AA = VA
tallow
2.84 kN
90  103 kPa
=  10−6 m2 =
AA = (dA2/4)
dA = 6.3 mm
Choose a size larger to nearest millimeter: dA = 7 mm

43. Diameter of pin B (single shear):
EXAMPLE 1-3
B = 6.67 N VB
Diameter of pin B (single shear):
Shear force at pin B :
VB = B = 6.67 kN
Diameter of pin B :
AB = VB
tallow
6.67 kN
90  103 kPa =
=  10−6 m2
AB = (dB2/4)
dB = 9.7 mm
Choose a size larger to nearest millimeter: dB = 10 mm

44. EXAMPLE 1-3 Diameter of rod BC: = (dBC2/4)
(The analysis is based on the normal stress)
ABC = B
(σt)allow
6.67 kN
115  103 kPa =
= 58  10−6 m2
= (dBC2/4)
dBC = 8.59 mm
Choose a size larger to nearest millimeter: dBC = 9 mm