Chapter Objectives To show how to use the method of sections for determining the internal loadings in a member. To introduce the concepts of normal and shear stress. To analyse and design members subject to axial load and direct shear. To define normal and shear strain, and show how they can be determined for various types of problems. 7.1 Introduction (8.1 in 2 ed.) 7.2 Internal Resultant Loadings (7.1 in 2 ed.) 7.3 Stress (8.2 in 2 ed.) 7.4 Average Normal Stress in an Axially Loaded Bar (8.3 in 2 ed.) 7.5 Average Shear Stress (8.4 in 2 ed.) 7.6 Allowable Stress (8.5 in 2 ed.) 7.7 Design of Simple Connections (8.6 in 2 ed.) 7.8 Deformation (8.7 in 2 ed.) 7.9 Strain (8.8 in 2 ed.) 7.1 INTRODUCTION Mechanics of Materials is a branch of mechanics that studies the relationship between the external loads applied to a deformable body and the intensity of internal forces acting within the body. It mainly involves computing the deformation of the body, and study of the bodys stability when the body is subjected to external forces. The design of any structure or machine involves: - Use of principles of statics to determine the forces acting both on and within its various members - Understanding behaviour of material (mechanics of materials) In short, the size of the members, their deflection, and their stability depend both on internal loading (due to external load) and behaviour of materials under loading 7.2 INTERNAL RESULTANT LOADINGS In Mechanics of Materials statics is primarily used to determine the resultant loadings that act within a body. 7.2 INTERNAL RESULTANT LOADINGS Three Dimensions: four different types of loadings 7.2 INTERNAL RESULTANT LOADINGS Three Dimensions: four different types of loadings 7.2 INTERNAL RESULTANT LOADINGS Coplanar Loadings: only normal-force, shear-force, and bending-moment component exist at the section N, acting normal to the cut section V, acting tangent to the section Couple moment Mo is referred as the bending moment Determine 3-unknowns using three equations of equilibrium + Fx = 0; + Fy = 0; Mo = 0; 7.2 INTERNAL RESULTANT LOADINGS Procedure for Analysis Support Reactions Before cut, determine the members support reactions, if needed Equilibrium equations are used to solve for internal loadings during sectioning Free-Body Diagrams Keep all distributed loadings, couple moments and forces acting on the member in their exact locations Draw FBD of the segment having the least loads Indicate the x, y, and z components of the force, couple moments and resultant couple moments on FBD Only N, V and M act at the section Determine the sense by inspection Equations of Equilibrium Moments should be summed at the section If negative result, the sense is opposite EXAMPLE 7.1 (2 ed and 3 ed) Determine the resultant internal loadings acting on the cross-section at C of the cantilevered beam shown in Fig. 7-3a. 7.3 STRESS (NORMAL & SHEAR STRESS) Copyright 2011 Pearson Education South Asia Pte Ltd AFzAzAA= A 0lim oAFAFyAzyxAzxAA=AA= A A00limlimtt7.4 AVERAGE NORMAL STRESS IN AN AXIALLY LOADED MEMBER Copyright 2011 Pearson Education South Asia Pte Ltd AP= o7.4 AVERAGE NORMAL STRESS IN AN AXIALLY LOADED MEMBER Copyright 2011 Pearson Education South Asia Pte Ltd READING QUIZ 1. What is the normal stress in the bar if P=10 kN and 500 mm? a) 0.02 kPa b) 20 Pa c) 20 kPa d) 200 N/mm e) 20 MPa Copyright 2011 Pearson Education South Asia Pte Ltd EXAMPLE 7.5 (in 2 ed Example 8.1) Copyright 2011 Pearson Education South Asia Pte Ltd The bar in Fig. 7-14a 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. EXAMPLE 7.5 (cont) Copyright 2011 Pearson Education South Asia Pte Ltd Graphically, the normal force diagram is as shown. By inspection, different sections have different internal forces. Solution EXAMPLE 7.5 (cont) Copyright 2011 Pearson Education South Asia Pte Ltd By inspection, the largest loading is in region BC, Since the cross-sectional area of the bar is constant, the largest average normal stress is Solution kN 30 =BCP( )( )( )(Ans) MPa 7 . 8501 . 0 035 . 010 303= = =APBCBCo7.5 AVERAGE SHEAR STRESS Copyright 2011 Pearson Education South Asia Pte Ltd section the at aream equilibriu of equations the from determined section the on force shear resultant internalsection the on located point each at same the be to assumed is which section, the at stress shear average====AVAVavgavgtt READING QUIZ (cont) 2. What is the average shear stress in the internal vertical surface AB (or CD), if F=20 kN, and AAB=ACD=1000 mm? a) 20 N/mm b) 10 N/mm c) 10 kPa d) 200 kN/m e) 20 MPa Copyright 2011 Pearson Education South Asia Pte Ltd EXAMPLE 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 following figure. 7.6 ALLOWABLE STRESS Copyright 2011 Pearson Education South Asia Pte Ltd allowfailallowfailallowfailS FS FFFS Fttoo===. .. .. .7.7 DESIGN OF SIMPLE CONNECTION Copyright 2011 Pearson Education South Asia Pte Ltd For normal force requirement For shear force requirement allowPAo=allowVAo= EXAMPLE 7.10 (in 2 ed Example 8.9) The control arm is subjected to the loading as shown. Determine to the nearest 5 mm the required diameter of the steel pin at C if the allowable shear stress for the steel is 55 MPa. 7.8 DEFORMATION Copyright 2011 Pearson Education South Asia Pte Ltd 7.9 NORMAL STRAIN Copyright 2011 Pearson Education South Asia Pte Ltd ss savgA A A='c( ) s s A + ~ A c 1 'ss sn A BA A A= 'lim alongc READING QUIZ (cont) 4. What is the unit of normal strain? a) mm b) mm/m c) Micron d) no unit Copyright 2011 Pearson Education South Asia Pte Ltd 7.9 SHEAR STRAIN Copyright 2011 Pearson Education South Asia Pte Ltd ' lim2 along alongutt A Cn A Bnt =7.9 CARTESIAN STRAIN COMPONENTS The approximate lengths of the sides of the parallelepiped are The approximate angles between sides, again originally defined by the sides x, y and z are Notice that the normal strains cause a change in volume of rectangular element, whereas the shear strain cause a change in shape ( ) ( ) ( ) z y x z y x A + A + A + c c c 1 1 1xz yz xy ttt 2 2 27.9 CARTESIAN STRAIN COMPONENTS (cont) Copyright 2011 Pearson Education South Asia Pte Ltd EXAMPLE 7.14 (in 2 ed Example 8.15) Copyright 2011 Pearson Education South Asia Pte Ltd Due to a loading, the plate is deformed into the dashed shape shown in Fig. 26a. Determine (a) the average normal strain along the side AB, and (b) the average shear strain in the plate at A relative to the x and y axes. EXAMPLE 4 (cont) Copyright 2011 Pearson Education South Asia Pte Ltd Part (a) Line AB, coincident with the y axis, becomes line AB after deformation, thus the length of this line is The average normal strain for AB is therefore The negative sign indicates the strain causes a contraction of AB. Solution ( ) mm 018 . 248 3 2 250 '22= + = AB( ) ( ) (Ans) mm/mm 10 93 . 7240250 018 . 248 '3 ===ABAB ABavg ABc EXAMPLE 4 (cont) Copyright 2011 Pearson Education South Asia Pte Ltd Part (b) As noted, the once 90 angle BAC between the sides of the plate, referenced from the x, y axes, changes to due to the displacement of B to B. Since then is the angle shown in the figure. Thus, Solution '2 u t =xy xy(Ans) rad 121 . 02 2503tan1=|.|

\|= xy CONCEPT QUIZ 1) The thrust bearing is subjected to the loads as shown. Determine the order of average normal stress developed on cross section through B, C and D. a) C > B > D b) C > D > B c) B > C > D d) D > B > C Copyright 2011 Pearson Education South Asia Pte Ltd CONCEPT QUIZ (cont) 2) The rectangular membrane has an unstretched length L1 and width L2. If the sides are increased by small amounts L1 and L2, determine the normal strain along the diagonal AB. Copyright 2011 Pearson Education South Asia Pte Ltd ( )( )222122212 12 1222122212211 D) C) B) A)L LL LL LL LL LL LLLLL+ A + A+ A + A+ A + A A+A CONCEPT QUIZ (cont) 3) The rectangular plate is subjected to the deformation shown by the dashed line. Determine the average shear strain xy of the plate. Copyright 2011 Pearson Education South Asia Pte Ltd ( )2 2200 150320015020031503+ D) C) B) A)Copyright 2011 Pearson Education South Asia Pte Ltd All solved examples All Fundamental Problems related to each section And Selected Problems (3-ed) 7.3,7.8,7.9,7.10,7.19,7.20,7.24,7.25,7.26,7.35,7.38, 7.39,7.50,7.51,7.54,7.55,7.61,7.62,7.83,7.84,7.87 7.88,7.91,7.92,7.95,7.97,7.98,7.99 Problems to be practiced (ch.7 of R.C. Hibbeler, 3ed)