Take note of the coverage of the f irst long exam. Meeting No.
Topic
Lecture Objectives (After the Lecture, the student must be able to do the following)
Suggested Problems
Class Policies 1
Review of Statics Equations of Equilibrium & Shear and Bending Moment Diagram Introduction to the Concept of Stress Normal Stress and Centric Loading
2 Shear Stress
1. Identify the relationship of force, area of application and stress developed from the previous quantities 2. Identify & calculate the different stresses developed in a member/part of a structure
1.54, 1.98 (Hibbeler) 1.8; 1.55; 1.60 (Beer)
Bearing and Punching Stress Strain Concepts Normal Strain Shear Strain 3
1. Determine the strains caused by stresses 2. Identify how materials react when stressed within the Elastic Limit 3. Determine the limitations of the Hooke's Law 4. Understand the Stress-Strain Diagram for Axial Loading
2.3, 2.19 (Hibbeler) 2.26, 2.81-82 (Beer)
Material Properties The Stress-Strain Diagram and Hooke's Law Strain Energy
4
Poisson's Ratio & Generalized Hooke's Law Shear Stress-Strain Diagram & Hooke's Law for Shear
1. Determine how strains affect one another 2. Describe the General Form of the Hooke's law 3. Understand the stress-strain diagram for Shear stress 4. Determine the allowable stresses for a given Factor of Safety and vice versa
3.16, 3.22 (Hibbeler) 2.63 (Beer)
Allowable Stresses and Factors of Safety Stresses and Deformations Arising from Axial Loading Axial Deformation Formulae from Statics 5&6
Analysis of Statically Determinate Axially Loaded Members Temperature Effects, Thermal Stress
7
Statically Indeterminate Axially Loaded Members
Axial Loading: 4.45, 4.59 (Hibbeler) 1. Derive the Axial Deformation Formulae 2. Apply the Axial Deformation Formula Correctly 3. Determine the stresses and/or strains caused by change in temperature 4. Design/Analyze statically determinate/indeterminate, axially loaded members (with the use of suitable compatibility equations)
2.40, 2.41 (Beer) Thermal: 4.86, 4.93 (Hibbeler) 2.49, 2.60 (Beer)
1st Long Exam
ES 13 Mechanics of Materials
1 of 5
UP College of Engineering
Member A of the timber step joint for a truss is subjected to a compressive force (kN) as shown in Figure 2. The ultimate stress of the steel material used in the hanger rod C is 470MPa while the maximum shear stress in wood B is 30MPa.
Determine the maximum allowable force
if the Factor of
Safety against failure by shearing of B is 3 and the Factor of Safety against failure by fracture of C is 2.5. The steel rod C has a diameter of 10mm and the wood B has a width of 30mm. Note: Neglect bearing of rod C on wood B.
At the proportional limit, a 200 mm gage length of a 15 mm diameter alloy bar has elongated 0.90 mm when a 58.4 kN axial load was applied. At a load of 70 kN the same material ruptures with a total elongation of 2.0 mm. Determine the following properties of this material: [Assume the plastic region being linear] a. The proportional limit. b. The modulus of elasticity. c. Modulus of resilience d. Approximate modulus of toughness. e. Assuming the material is initially unloaded. Determine the permanent deformation acquired by the material if a load equivalent to 40 kN is applied then unloaded. f. Assuming the material is initially unloaded. Determine the permanent deformation acquired by the material if a load equivalent to 60 kN is applied then unloaded. Load P is applied at end A of rigid bar . This causes normal strains of 0.008 and 0.003 for wire BC and bar DE respectively. The connection at D has a 0.5 mm clearance. The angles at F and G are both 45 o. (a) What is the displacement of point B? (b) What is the new length, in mm, of bar DE? (c) What is the shear strain of point E on tip of plate EFG? (d) What is the normal strain on side EF of the triangle?
ES 13 Mechanics of Materials
2 of 5
UP College of Engineering
The data shown in the accompanying table were obtained from a tensile test of highstrength steel. The test specimen had a dimater of 0.505 in and a gage length of 2.00 in. At fracture, the elongation between the gage marks was 0.12 in and the minimum diameter was 0.42 in. Plot the conventional stress-strain curve for the steel and determine the proportional limit, modulus of elasticity (i.e., the slope of the initial part of the stress-strain curve for the steel and determine the proportional limit, modulus of elasticity (i.e., the slope of the initial part of the stressstrain curve), yield stress at 0.1 % offset, ultimate stress, percent elongation in 2.00 in, and percent reduction in area.
1,000 2,000 6,000 10,000 12,000 12,900 13,400 13,600 13,800 14,000 14,400 15,200 16,800 18,400 20,000 22,400 22,600
ES 13 Mechanics of Materials
3 of 5
0.0002 0.0006 0.0019 0.0033 0.0039 0.0043 0.0047 0.0054 0.0063 0.0090 0.0102 0.0130 0.0230 0.0336 0.0507 0.1108 Fracture
UP College of Engineering
The main cables of a suspension bridge [see part (a) of the figure] follow a curve that is nearly parabolic because the primary load on the cables is the weight of the bridge deck, which is uniform in intensity along the horizontal. Therefore, let us represent the central region AOB of one of the main cables [see part (b) of the figure] as a parabolic cable supported at points Aand Band carrying a uniform load of intensity q along the horizontal. The span of the cable is L, the sag is h, the axial rigidity is EA, and the origin of coordinates is at midspan.
(a) Derive the following formula for the elongation of cable AOB shown in part (b) of the figure:
2 16ℎ = 8ℎ (1 + 32 ) (b) Calculate the elongation δ of the central span of one of the main cables of the Golden Gate Bridge, for which the dimensions and properties are L = 4200 ft , h = 470 ft , q = 12,700 lb/ft , and E = 28,800,000 psi . The cable consists of 27,572 parallel wires o f diameter of 0.196 in.
Hint:Determine the tensile force T at any point in the cable from a free-body diagram of part of the cable; then determine the elongation of an element of the cable of length ds; finally, integrate along the curve of the cable to obtain an equation for the elongation δ.
ES 13 Mechanics of Materials
4 of 5
UP College of Engineering
Three steel cables jointy support a load of 12 k (see figure). The diameter of the middle cable is ¾ in. And the diameter of each outer cable is ½ in. The tensions in the cables are adjusted so that each cable carries one-third of the load (i.e., 4 k). Later, the load is increased by 9 k to a total load of 21 k. (a) What percent of the total load is now carried by the middle cable? (b) What are the stresses σ Mand σ O in the middle and outer cables, respectively?
The structure shown is subjected to a concentrated force P as shown and an increase in temperature of 90 ° C . The material and section properties are provided in Table below. The steel and bronze rods are fixed at one end while the aluminum bar and the other end of the steel bar is attached to a rigid plate. The aluminum and bronze bar is separated by a 0.2 mm gap. Determine the maximum allowable force P such that the allowable stresses will not be exceeded.
Area (mm2)
300
1000
3500
E (GPa)
200
70
83
9x10-6
25x10-6
21x10-6
90
50
66
α
(1/°C)
all (MPa)
σ
ES 13 Mechanics of Materials
5 of 5
UP College of Engineering
