CIE 113 Mechanics of Deformable Bodies ENGR. KARL ANGELO CUELLAR Lesson 1. Introduction to Stress Lesson 2. Working and Allowable Stresses Lesson 7. Deformation of Bodies using Integration Lesson 8. Thermal Stresses Lesson 13. Shear and Moment using the Area Method Lesson 3. ThinWalled Tubes Lesson 4. Bearing Stress in Beams and Shear Stress in Lap Joints Lesson 5. Introduction to Strain Lesson 6. Statically Indeterminate Structures Lesson 9. Introduction to Torsion Lesson 10. Power Transmission Lesson 11. Introduction to Shear and Moment Lesson 12. Shear and Moment Equations Lesson 14. Introduction to Beam Stresses Lesson 15. Allowable Stresses in Beams and Spacing of Rivets Lesson 16. Additional Problems in Bending Stress and Problem set Lesson 17. Stress Distributions Grading system LESSON 1. INTRODUCTION TO STRESSES FUNDAMENTAL AREAS OF ENGINEERING MECHANICS 1. STATICS – deals with the equilibrium of bodies, those that either at rest or move with a constant velocity. 2. DYNAMICS – is concerned with the accelerated motion of bodies. 3. STRENGTH OF MATERIALS – deals with the internal effects and deformations that are caused by the applied loads. STATIC/DYNAMICS STRENGTH OF MATERIALS Body Rigid Real/deformable Force External effects Internal effects Analysis FBD, equilibrium FBD, equilibirum and deformation diagrams Focus Identification of loads that act on the body. Determine the strength and rigidity of the body. Rigidity – inability to be bent or be forced out of shape Deformable Bodies – bodies that will not retain its shape/form when force is applied. Rigid – a type of body that will retain its shape form once a load is applied onto it. ANALYSIS OF INTERNAL FORCES NORMAL FORCE • Is a type of force applied perpendicular to a crosssectional area of the body. SHEAR FORCE TORQUE • Parallel to the cross-sectional area of the body. • Tends to twist, rotate the cross-sectional area of the body being considered. BENDING MOMENT • Tends to bend the object. STRESS (𝝈) – is the internal resistance of a material to the distorting effects of external forces. 𝑁 𝑘𝑁 𝑁 𝑙𝑏 - Force per unit area. ( 2 , 𝑃𝑎, 2 , 2, 2 , 𝑝𝑠𝑖) 𝑚 𝑚 𝑚𝑚 𝑃 𝜎= 𝐴 UNIT CONVERSION: 1 yard = 3 ft 1 lbf = 4.448 N 1 kg = 2.2 lbm 1 kN = 1000 N 1 MN = 1,000,000 N = 1 x 106 N 1 m = 3.28 ft 1 ft = 12 inches 1 N/m2 = 1 Pa 1 N/mm2 = 1 Mpa 1 kips = 1000 lb 1 ksi = 1000 psi 1 inches = 2.54 cm 1 cm = 10 mm 1 m = 1000 mm 𝑖𝑛 Example 1. Given two bars of same lengths but of different materials suspended from a common supports, determine which bar is stronger. SIMPLE STRESS/ NORMAL STRESS/ AXIAL STRESS/ CENTROIDAL STRESS Normal stress will occur to members that are axially loaded. Axial force can be tension or compression and applied at the centroid of cross-section of the member. Normal stress is either tensile stress or compressive stress. Members subject to pure tension (or tensile force) is under tensile stress, while compression members (members subject to compressive force) are under compressive stress. Compressive force will tend to shorten the member. Tension force on the other hand will tend to lengthen the member. SAMPLE PROBLEM 1. A hollow steel tube with an inside diameter of 100 mm must carry a tensile load of 400 kN. Determine the outside diameter of the tube if the stress is limited to 120 MN/m2. SAMPLE PROBLEM 2. Find the normal stress in the material. SAMPLE PROBLEM 3. Find the normal stress in the material. SAMPLE PROBLEM 4. A rod is composed of an aluminum section rigidly attached between steel and bronze sections, as shown in the figure. Axial loads are applied at the positions indicated. If P = 3000 lb and the crosssectional area of the rod is 0.5 in2, determine the stress in each section. SAMPLE PROBLEM 5. Find the normal stress in member BD. All members have cross-sectional areas of 900 mm2. SAMPLE PROBLEM 6. Determine the average normal stress in each bear if the diameter of each bar is 20 mm. Set P = 40 N . LESSON 2. WORKING AND ALLOWABLE STRESSES COMPUTED STRESS/ ACTUAL STRESS/COMPUTED FORCES/ ACTUAL FORCES: are derived from the effects of externally applied loads acting on the body being analyzed. ALLOWABLE STRESS/ WORKING STRESS/ ALLOWABLE FORCES/ WORKING FORCES: are computed from the structural properties of the member in question. These allowable quantities speak of the prescribed capacity of the member. Problem 1. The figure shows a two-member truss supporting a block of weight W. The cross-sectional areas of the members are 800 mm² for AB and 400 mm² for AC. Determine the maximum safe value of W if the working stresses are 110 MPa for AB and 120 MPa for AC. Problem 2. The 1000-kg uniform bar AB is suspended from two cables AC and BD, each with cross-sectional area 400 mm2. Find the magnitude P and location x of the largest additional vertical force that can be applied to the bar. The stresses in AC and BD are limited to 100 Mpa and 50 Mpa, respectively. SHEAR STRESS Forces parallel to the area resisting the force cause shearing stress. It differs to tensile and compressive stresses, which are caused by forces perpendicular to the area on which they act. Shearing stress is also known as tangential stress. BEARING STRESS Bearing stress is the contact pressure between the separate bodies. It differs from compressive stress, as it is an internal stress caused by compressive forces. This is a special type of normal stress. Problem 3. The lap joint shown below is fastened by four rivets of 3/4-in. diameter. Find the maximum load P that can be applied if the working stresses are 14 ksi for shear in the rivet and 18 ksi for bearing in the plate. Assume that the applied load is distributed evenly among the four rivets, and neglect friction between the plates. Problem 4. The bell crank, which is in equilibrium under the forces shown in the figure, is supported by a 20-mm-diameter pin at D. Determine (a) the required diameter of the connecting rod AB, given that its tensile working stress is 100 MPa; and (b) the shear stress in the pin. Problem 5. The figure shows a roof truss and the detail of the connection at joint B. Members BC and BE are angle sections with the thicknesses shown in the figure. The working stresses are 70 MPa for shear in the rivets and 140 MPa for bearing stress due to the rivets. How many 19-mm-diameter rivets are required to fasten the following members to the gusset plate: (a) BC; and (b) BE? Problem 6. In the clevis shown in Fig. 1-11b, find the minimum bolt diameter and the minimum thickness of each yoke that will support a load P = 14 kips without exceeding a shearing stress of 12 ksi and a bearing stress of 20 ksi. Problem 7. What force is required to punch a 20-mm-diameter hole in a plate that is 25 mm thick? The shear strength is 350 MN/m2. Problem 8. As in the figure below, a hole is to be punched out of a plate having a shearing strength of 40 ksi. The compressive stress in the punch is limited to 50 ksi. (a) Compute the maximum thickness of plate in which a hole 2.5 inches in diameter can be punched. (b) If the plate is 0.25 inch thick, determine the diameter of the smallest hole that can be punched.
