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

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

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

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

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)

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

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

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

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

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)

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.

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

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

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

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

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)

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

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

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

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

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

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

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.

EXAMPLE 1-2 y Equation of equilibrium FAC FAB W 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 = 173.2 N

EXAMPLE 1-2 Average normal stress AAB = ¼  (12)2 mm2 = 113.1 mm2 AAC = ¼  (10)2 mm2 = 78.54 mm2 60o W FAC FAB x y Substituting the appropriate values, we have sAB = 0.884 N/mm2 = 0.884 MPa sAC = 2.205 N/mm2 = 2.204 MPa The greater normal stress is in rod AC.

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

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

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

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

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).

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

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).

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

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

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

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

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

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.

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

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

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

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 = 31.56  10−6 m2 = AA = (dA2/4) dA = 6.3 mm Choose a size larger to nearest millimeter: dA = 7 mm

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 = = 74.11  10−6 m2 AB = (dB2/4) dB = 9.7 mm Choose a size larger to nearest millimeter: dB = 10 mm

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