Analysis of Structures PDF

Republic of the Philippines CAMARINES SUR POLYTECHNIC COLLEGES Nabua, Camarines Sur ANALYSIS OF STRUCTURES ERLY O. CELIZ, RCE  OUTLINE OBJECTIVES...

Republic of the Philippines CAMARINES SUR POLYTECHNIC COLLEGES Nabua, Camarines Sur ANALYSIS OF STRUCTURES ERLY O. CELIZ, RCE  OUTLINE OBJECTIVES Upon completion of this chapter, the 1. Types of Structures students will be able to: 2. Trusses – Terminology and Assumptions 3. Zero-Force Members  Determine internal forces acting on trusses 4. Analysis of Trusses  Use various methods to obtain  Method of Joints internal forces acting on trusses  Method of Sections  Analyze forces acting on trusses 5. Analysis of Frames and Machines and on pins of frames 2 Analysis of Structures: The analysis of a structure is a process by which we determine how the loads applied to a structure are distributed all over the structure. The discussion of this topic is limited to structures with pin-connected types. 3 Types of Structures will be considered: 1) Truss - a stationary structure made up of only 2-force members 2) Frame - a stationary structure containing at least one multi-force member (3 or more forces) 3) Machine - a structure that is designed to move or exert forces (such as a hand tool) containing at least one multi-force member (3 or more forces) 3 TRUSSES – Terminology and Assumptions Trusses  is a structure that is made of straight, slender bars that are joined together to form a pattern of triangles.  consist of members and joints and the entire truss is mounted on supports, as illustrated below. The following assumptions will be made for trusses:  A truss is a stationary structure  Trusses should be rigid (holds its shape and will not collapse)  Trusses will be generally treated as 2D structures, although the analysis methods to be introduced can be applied to 3D trusses (space trusses)  A truss is made up of only 2-force members  All joints in the truss are pinned (thus the reaction at the pin has no moment)  All forces (loads) will be applied at the joints of a truss 4 TRUSSES – Terminology and Assumptions The following assumptions will be made for trusses:  Each member of the truss experiences only axial forces (along the axis of the member). This axial force is either one of tension (T) or compression (C). Forces in truss members will use the designations T or C. 5 TRUSSES RIGID TRUSSES It is important that trusses be rigid. A rigid truss will not collapse under application of a load. SIMPLE TRUSSES A simple truss is:  a rigid truss  a planar truss which begins with a triangular element and can be expanded by adding two members and a joint  will satisfy the formula m = 2n – 3, where m=number of members and n=number of joints 6 ZERO-FORCE MEMBERS o Certain truss members may be subjected to zero force under certain loading conditions. o Recognizing zero-force members can simplify the analysis of the truss. o Zero-force members are often more slender than main truss members. Why would zero-force members be used? o To provide stability to a truss during construction o To stiffen the truss o To provide support to a truss if loading conditions change (such as due to snow or wind force on a roof, loading on the deck of a bridge, etc.) In other words, the zero-force members may not be zero-force members when the loading changes. 10  ZERO-FORCE MEMBERS Recognizing zero-force members: If only two members form a truss joint and no If three members form a truss joint for which external load or support reaction is applied to two of the members are collinear, the third the joint, the members must be zero-force member is a zero-force member provided no members. external force or support reaction is applied to the joint. Summing forces in the x and y directions with one axis along the collinear members will quickly verify this result 11 ZERO-FORCE MEMBERS Recognizing zero-force members: 12 Example: Recognizing the zero-force members: 13  Example:  Member CB is a zero-force member since  Member BL is a zero-force member since members AB and BD are collinear members AL and KL are collinear  Member FH is a zero-force member since members GH and HI are collinear 14 ANALYSIS OF TRUSSES Two methods will be introduced for analyzing trusses: Method of joints Method of sections Method of Joints  This is a systematic method for analyzing each joint in the truss in order to determine the forces in all members of the truss.  It is the best method if the forces in all members of the truss are to be determined.  Each joint is considered to be in equilibrium, but the joint is pinned and all member forces go through the joint, so no moments are experienced at the joint. Therefore, only 2 equations are applied in analyzing the joint (for a 2D truss): 𝑭𝒙 = 𝟎 𝑭𝒚 = 𝟎 15 ANALYSIS OF TRUSSES Method of Joints - Procedure 1) Analyze the entire truss as a rigid body to find the external reactions (not always necessary) 2) Pick the first joint to analyze A. Since only two equations are available 𝐹𝑥 = 0 𝑎𝑛𝑑 𝐹𝑦 = 0 , look for joints that only have two unknowns B. Draw a FBD at the joint to be analyzed. C. Show each member force in tension. - If the result is +, then the answer agrees with the way the force was drawn, so the force is in tension (attach a T to the answer). - If the result is -, then the answer disagrees with the way the force was drawn, so the force is in compression (attach a C to the answer). - Express all final answers as positive with either T or C attached. Note: you could similarly draw the forces in compression and a + or – answer would again indicate agreement or disagreement. 3) Continue analyzing additional joints in the truss until all member forces have been determined. 16 Example: 1.) Determine the force in each member of the truss. Indicate whether the members are in tension or compression. 17 Solution: First, solve for the reaction at the support: From 𝛴𝑀𝐵 = 0 ↻ 𝛴𝑀𝐵 = 𝐴𝑦 25 + 𝐴𝑥 0 + 100 12 − 200 9 = 0 𝑨𝒚 = 𝟐𝟒 𝒌 (𝒖𝒑𝒘𝒂𝒓𝒅) From 𝛴𝑀𝐴 = 0 ↺ 𝛴𝑀𝐵 = 𝐵𝑦 25 + 100 12 + 200 16 = 0 𝑩𝒚 = 𝟏𝟕𝟔 𝒌 (𝒖𝒑𝒘𝒂𝒓𝒅) From 𝛴𝐹𝑥 = 0 ← 𝛴𝐹𝑥 = 𝐴𝑥 − 100 = 0 𝑨𝒙 = 𝟏𝟎𝟎 𝒌 (𝒕𝒐 𝒕𝒉𝒆 𝒍𝒆𝒇𝒕) Isolate joint A Solve for members AC and AB From 𝛴𝐹𝑦 = 0 ↓ 𝛴𝐹𝑦 = 𝐴𝐶𝑦 − 24 = 0 3 0 = 5 𝐴𝐶 − 24 𝑨𝑪 = 𝟒𝟎 𝒌 (𝑪) From 𝛴𝐹𝑥 = 0 → 𝛴𝐹𝑥 = 𝐴𝐵 − 𝐴𝐶𝑥 − 100 = 0 4 0 = 𝐴𝐵 − 5 𝐴𝐶 − 100 4 0 = 𝐴𝐵 − 5 40 − 100 𝑨𝑩 = 𝟏𝟑𝟐 𝒌 (𝑻) Solution (continuation): Isolate Joint C Solve for member BC 3 From 𝛴𝐹𝑥 = 0 ← 𝛴𝐹𝑥 = 5 𝐵𝐶 − 𝐴𝐶𝑥 − 100 = 0 3 4 0 = 5 𝐵𝐶 − 5 (40) − 100 𝑩𝑪 = 𝟐𝟐𝟎 𝒌 (𝑪) 4 From 𝛴𝐹𝑦 = 0 ↑ 𝛴𝐹𝑦 = 5 𝐵𝐶 + 𝐴𝐶𝑦 − 200 = 0 4 3 0 = 𝐵𝐶 + (40) − 200 5 5 𝑩𝑪 = 𝟐𝟐𝟎 𝒌 (𝑪) Check equilibrium at the last Joint B 3 From 𝛴𝐹𝑥 = 0 → 𝛴𝐹𝑥 = 5 220 − 132 = 0 0 = 132 − 132 𝟎 = 𝟎(𝒄𝒉𝒆𝒄𝒌) 4 From 𝛴𝐹𝑦 = 0 ↑ 𝛴𝐹𝑦 = 5 220 − 176 = 0 0 = 176 − 176 𝟎 = 𝟎(𝒄𝒉𝒆𝒄𝒌) Example: 2.) Determine the force in each member of the truss. Indicate whether the members are in tension or compression. 20  Solution: First, solve for the reaction at the support: From 𝛴𝑀𝐷 = 0 ↺ 𝛴𝑀𝐷 = 𝐴𝑥 2.5 − 40 3 − 40(6) = 0 𝑨𝒙 = 𝟏𝟒𝟒 𝒌𝑵 (𝒕𝒐 𝒕𝒉𝒆 𝒓𝒊𝒈𝒉𝒕) From 𝛴𝑀𝐴 = 0 ↺ 𝛴𝑀𝐴 = 𝐷𝑥 2.5 − 40 3 − 40(3) = 0 𝑫𝒙 = 𝟏𝟒𝟒 𝒌𝑵 (𝒕𝒐 𝒕𝒉𝒆 𝒍𝒆𝒇𝒕) From 𝛴𝐹𝑦 = 0 ↑ 𝛴𝐹𝑦 = 𝐷𝑦 − 20 − 40 − 40 = 0 𝑫𝒚 = 𝟏𝟎𝟎 𝒌𝑵 (𝒖𝒑𝒘𝒂𝒓𝒅) Isolate joint D Solve for members AD and DE From 𝛴𝐹𝑥 = 0 → 𝛴𝐹𝑥 = 𝐷𝐸𝑥 − 144 3 0 = 3.25 𝐷𝐸 − 144 𝑫𝑬 = 𝟏𝟓𝟔 𝒌𝑵 (𝑻) 1.25 From 𝛴𝐹𝑦 = 0 ↑ 𝛴𝐹𝑦 = 𝐴𝐷 + 100 − (𝐷𝐸) 3.25 1.25 0 = 𝐴𝐷 + 100 − 3.25 (156) 𝐴𝐷 = −40 𝑘𝑁 𝑨𝑫 = 𝟒𝟎 𝒌𝑵 (𝑻)  Solution (continuation): Isolate Joint A Solve for members AE and AB From 𝛴𝐹𝑦 = 0 ↓ 𝛴𝐹𝑦 = 𝐴𝐸𝑦 + 20 − 40 = 0 1.25 0 = 3.25 𝐴𝐸 − 20 𝑨𝑬 = 𝟓𝟐 𝒌𝑵 (𝑪) From 𝛴𝐹𝑥 = 0 ← 𝛴𝐹𝑥 = 𝐴𝐵 + 𝐴𝐸𝑥 − 144 = 0 3 0 = 𝐴𝐵 + 3.25 52 − 144 𝑨𝑩 = 𝟗𝟔 𝒌𝑵 (𝑪) Isolate Joint E Solve for members CE and BE From 𝛴𝐹𝑥 = 0 → 𝛴𝐹𝑥 = 𝐶𝐸𝑥 + 𝐴𝐸𝑥 − 𝐷𝐸𝑥 = 0 3 3 3 0= 𝐶𝐸 + 52 − 156 3.25 3.25 3.25 𝑪𝑬 = 𝟏𝟎𝟒 𝒌𝑵 (𝑻) From 𝛴𝐹𝑦 = 0 ↓ 𝛴𝐹𝑦 = 𝐵𝐸 + 𝐶𝐸𝑦 − 𝐴𝐸𝑦 − 𝐷𝐸𝑦 = 0 1.25 1.25 1.25 0 = 𝐵𝐸 + 3.25 104 − 3.25 52 − 3.25 156 𝑩𝑬 = 𝟒𝟎 𝒌𝑵 (𝑻)  Solution (continuation): Isolate Joint B Solve for members BC From 𝛴𝐹𝑥 = 0 ← 𝛴𝐹𝑥 = 𝐵𝐶 − 𝐴𝐵 0 = 𝐵𝐶 − 96 𝑩𝑪 = 𝟗𝟔 𝒌𝑵 (𝑪) Check equilibrium at the last Joint C 3 From 𝛴𝐹𝑥 = 0 → 𝛴𝐹𝑥 = 3.25 104 − 96 0 = 96 − 96 𝟎 = 𝟎(𝒄𝒉𝒆𝒄𝒌) 1.25 From 𝛴𝐹𝑦 = 0 ↑ 𝛴𝐹𝑦 = 3.25 104 − 40 0 = 40 − 40 𝟎 = 𝟎(𝒄𝒉𝒆𝒄𝒌) Example: 3.) A Fink truss is loaded as shown. Determine the force in each member of the truss assuming them to be pin-connected. 24  Solution: First, solve for the reaction at the support: Since the truss and the applied forces is symmetrical, thus the reaction need be found in only one-half of it. 𝐹 = 1000 3 + 2000 2 = 7000 𝑙𝑏 𝑅𝐴 𝑅𝐺 7000 The next pin at which no more than = = = 3500 𝑙𝑏 two unmarked members appear is 2 2 2 The next pin at which no more seen to be joint C, repeat the After determining the reactions technique used at joint A and B. from symmetry, consider joint A than two unmarked members Since the truss and the appear is seen to be joint B, 𝐵𝐶 = 866.03 𝑙𝑏 which has only two unmarked 𝐶𝐷 applied forces is members (AB and AC) acting repeat the technique used at symmetrical, thus member upon it. 𝐴𝐵 joint A. AB=GF, AC=EG, BD=DF 𝑦 1000 𝑙𝑏 𝑥 60° 60° and CD=DE. 30° 𝐶𝐸 𝐵𝐷 𝐴𝐶 = 6062.18 𝑙𝑏 30° 𝐴𝐶 Summary: 2000 𝑙𝑏 𝑨𝑩 = 𝑭𝑮 = 𝟕𝟎𝟎𝟎 𝒍𝒃 𝑪 𝐹𝑦 = 0 𝑨𝑪 = 𝑬𝑮 = 𝟔𝟎𝟔𝟐. 𝟏𝟖 𝒍𝒃 𝑻 𝐶𝐷 sin 60° = 𝐵𝐶 sin 60° + 2000 𝑩𝑫 = 𝑫𝑭 = 𝟔𝟓𝟎𝟎 𝒍𝒃 𝑪 𝑅𝐴 = 3500 𝑙𝑏 𝐴𝐵 = 7000 𝑙𝑏 𝐵𝐶 𝐶𝐷 sin 60° = 866.03 sin 60° + 2000 𝑩𝑪 = 𝑬𝑭 = 𝟖𝟔𝟔. 𝟎𝟑 𝒍𝒃 𝑪 𝐹𝑦 = 0 𝐹𝑥 = 0 𝑪𝑫 = 𝟑𝟏𝟕𝟓. 𝟒𝟑 𝒍𝒃 𝑻 𝑪𝑫 = 𝑫𝑬 = 𝟑𝟏𝟕𝟓. 𝟒𝟑 𝒍𝒃 𝑻 3500 = 𝐴𝐵 sin 30° 𝐴𝐵 = 𝐵𝐷 + 1000 sin 30° 𝑨𝑩 = 𝟕𝟎𝟎𝟎 𝒍𝒃 𝑪 7000 = 𝐵𝐷 + 1000 sin 30° 𝐹𝑥 = 0 𝑩𝑫 = 𝟔𝟓𝟎𝟎 𝒍𝒃 𝑪 𝐶𝐸 + 𝐶𝐷 cos 60° + 𝐵𝐶 cos 60° = 𝐴𝐶 𝐹𝑥 = 0 𝐶𝐸 + 3175.43 cos 60° + 866.03 cos 60° = 6062.18 𝐴𝐶 = 𝐴𝐵 cos 30° 𝐹𝑦 = 0 𝑪𝑬 = 𝟒𝟎𝟒𝟏. 𝟒𝟓 𝒍𝒃 𝑻 𝐴𝐶 = 7000 cos 30° 𝐵𝐶 = 1000 cos 30° 𝑨𝑪 = 𝟔𝟎𝟔𝟐. 𝟏𝟖 𝒍𝒃 𝑻 𝑩𝑪 = 𝟖𝟔𝟔. 𝟎𝟑 𝒍𝒃 𝑪  Republic of the Philippines CAMARINES SUR POLYTECHNIC COLLEGES Nabua, Camarines Sur ANALYSIS OF STRUCTURES ERLY O. CELIZ, RCE  OUTLINE OBJECTIVES Upon completion of this chapter, the 1. Types of Structures students will be able to: 2. Trusses – Terminology and Assumptions 3. Zero-Force Members  Determine internal forces acting on trusses 4. Analysis of Trusses  Use various methods to obtain  Method of Joints internal forces acting on trusses  Method of Sections  Analyze forces acting on trusses 5. Analysis of Frames and Machines and on pins of frames 2 ANALYSIS OF TRUSSES Method of Sections  In the method of sections, a truss is divided into two parts by taking an imaginary “cut” (shown here as a-a) through the truss.  Since truss members are subjected to only tensile or compressive forces along their length, the internal forces at the cut members will also be either tensile or compressive with the same magnitude. 3 ANALYSIS OF TRUSSES Method of Sections - Procedure 1) Decide how you need to “cut” the truss. This is based on: a) where you need to determine forces, and, b) where the total number of unknowns does not exceed three (in general). 2) Decide which side of the cut truss will be easier to work with (minimize the number of reactions you have to find). 3) If required, determine any necessary support reactions by drawing the FBD of the entire truss and applying the equations of equilibrium. 4) Draw the FBD of the selected part of the cut truss. We need to indicate the unknown forces at the cut members. Initially we may assume all the members are in tension, as we did when using the method of joints. Upon solving, if the answer is positive, the member is in tension as per our assumption. If the answer is negative, the member must be in compression. (Please note that you can also assume forces to be either tension or compression by inspection as was done in the figures above.) 4 ANALYSIS OF TRUSSES Method of Sections - Procedure 5) Apply the scalar equations of equilibrium to the selected cut section of the truss to solve for the unknown member forces. Please note, in most cases it is possible to write one equation to solve for one unknown directly. So look for it and take advantage of such a shortcut! Recall that there are several choices for 2D equations of equilibrium as listed below: 𝑭𝒙 = 𝟎 𝑭𝒚 = 𝟎 𝑴𝑶 = 𝟎 (for any point) 5 Example: 1.) Use the method of sections to determine the forces in members DE, AB and AE of the truss described in the figure and indicate whether the member is in tension or compression. 6  Solution: First, solve for the reaction at the support: From 𝛴𝑀𝐷 = 0 ↺ 𝛴𝑀𝐷 = 𝐴𝑥 2.5 − 40 3 − 40(6) = 0 𝑨𝒙 = 𝟏𝟒𝟒 𝒌𝑵 (𝒕𝒐 𝒕𝒉𝒆 𝒓𝒊𝒈𝒉𝒕) From 𝛴𝑀𝐴 = 0 ↺ 𝛴𝑀𝐴 = 𝐷𝑥 2.5 − 40 3 − 40(3) = 0 𝑫𝒙 = 𝟏𝟒𝟒 𝒌𝑵 (𝒕𝒐 𝒕𝒉𝒆 𝒍𝒆𝒇𝒕) From 𝛴𝐹𝑦 = 0 ↑ 𝛴𝐹𝑦 = 𝐷𝑦 − 20 − 40 − 40 = 0 𝑫𝒚 = 𝟏𝟎𝟎 𝒌𝑵 (𝒖𝒑𝒘𝒂𝒓𝒅) Pass a cutting plane separating the truss into two parts and expose the forces cut by the plane Solution (continuation): Example: 2.) The Warren truss loaded as shown in the figure is supported by a roller at C and a hinge at G. By method of sections, compute the force in the members BC, DF, CE and DE. 𝑨𝒏𝒔𝒘𝒆𝒓: 𝑩𝑪 = 𝟒𝟒. 𝟕𝟐 𝒌𝑵, 𝑪 𝑫𝑭 = 𝟖𝟎 𝒌𝑵, 𝑪 𝑪𝑬 = 𝟏𝟎 𝒌𝑵, 𝑻 𝑫𝑬 = 𝟐𝟐. 𝟑𝟔 𝒌𝑵, 𝑻 9  Example: 3.) Determine the force in the bar CD of the truss shown in the figure. 𝑨𝒏𝒔𝒘𝒆𝒓: 𝑪𝑫 = 𝟓𝟎 𝒌𝑵, 𝑻 10  Example: 4.) For the truss shown, determine the force in member BF by the method of section. 𝑨𝒏𝒔𝒘𝒆𝒓: 𝑩𝑭 = 𝟐. 𝟓 𝒌𝑵, 𝑪 Practice Problems: 1.) Determine the force in members DF, DG, and EG of the Howe truss shown. 12 Practice Problems: 2.) Using the method of sections, compute the force in bars FH, GH, and EK. 13 Practice Problems: 3.) For the cantilever truss shown, determine the forces in members DF, FH, FI, GI, and FG. 14  Assignment: 1.) Remove the 1200 lb forces and determine the 2.) Determine the force in each member of the truss greatest force P that can be applied to the truss so shown. State if the members are in tension or that none of the members are subjected to a force compression. exceeding either 2000 lb in tension and 1500 lb in compression. 15  Assignment: 3.) Using method of sections, determine the force in 4.) the Howe bridge truss is subjected to the loading members DC, HI and JI of the truss. State if the shown. determine the force in members HI, HB and members are in tension or compression. BC, and state if the members are in tension or compression. 16  Republic of the Philippines CAMARINES SUR POLYTECHNIC COLLEGES Nabua, Camarines Sur CENTROIDS AND CENTERS OF GRAVITY ERLY O. CELIZ, RCE  OUTLINE OBJECTIVES Upon completion of this 1. Centroids and Center of Gravity chapter, the students will be able to: 2. Axis of Reference Familiarize the concepts of 3. Centroids of Composite Bodies centroids, center of gravity Locate centroid of plane and composite figures 2 CENTROIDS AND CENTER OF GRAVITY Representation of Centroids and Center of Gravity CENTROIDS  It is defined as a point about which the entire line, area or volume is assumed to be concentrated.  It is related to distribution of Length, Area or Volume. CENTER OF GRAVITY  It is defined as a point about which the entire weight of the body is assumed to be concentrated.  Center of mass.  It is related to distribution of mass. 3 AXIS OF REFERENCE The center of the gravity of a body is always calculated with a reference to some assumed axis known as axis of reference (or sometimes with reference to some point of reference). The axis of reference, of the plane figures, Is generally taken as the lowest line of the figure for calculating 𝑦 and the left line of the figure for calculating 𝑥. Consider the figure shown, 4 CENTROIDS OF COMPOSITE BODIES Centroids determined by integration 𝑏 𝐴𝑥 = 𝑥 𝑑𝐴 𝑎 𝑐 𝑏 𝐴𝑦 = 𝑦 𝑑𝐴 𝑎 𝑐 Centroids of areas 𝐴𝑥 = 𝑎𝑥 𝐴𝑦 = 𝑎𝑦 Centroids of lines 𝐿𝑥 = 𝑙𝑥 𝐿𝑦 = 𝑙𝑦 5 CENTROIDS OF THE COMMON GEOMETRIC SHAPES 9 CENTROIDS OF THE COMMON GEOMETRIC SHAPES 10 CENTROIDS OF THE COMMON GEOMETRIC SHAPES 11 Problem 1: The dimensions of the T-section of a cast-iron beam are shown. How far is the centroid of the area above the base? 12 Solution 𝐴1 = 6 1 = 6 𝑖𝑛2 𝑦1 = 0.5 𝑖𝑛 𝐴2 = 8 1 = 8 𝑖𝑛2 𝑦2 = 5 𝑖𝑛 𝐴 = 𝐴1 + 𝐴2 = 6 + 8 = 14 𝑖𝑛2 𝐴𝑦 = 𝑎𝑦 14𝑦 = 6 0.5 + 8(5) 𝒚 = 𝟑. 𝟎𝟕 𝒊𝒏 𝒂𝒃𝒐𝒗𝒆 𝒕𝒉𝒆 𝒃𝒂𝒔𝒆 13 Problem 2: 3” Determine the coordinates of the centroid of the area shown with respect to the given axes. 9” 14 Solution 1 𝐴1 = (6) 9 = 27 𝑖𝑛2 𝐴𝑦 = 𝑎𝑦 2 1 41.14𝑦 = 27 6 + 14.14(10.27) 𝑥1 = 6 = 2 𝑖𝑛 3 𝒚 = 𝟕. 𝟒𝟕 𝒊𝒏 2 𝑦1 = 9 = 6 𝑖𝑛 3 Coordinates of the centroid is at (𝟐. 𝟑𝟒, 𝟕. 𝟒𝟕) 1 𝐴2 = 𝜋(3)2 = 14.14 𝑖𝑛2 2 𝑥2 = 𝑟 = 3 𝑖𝑛 4(3) 𝑦2 = 9 + = 10.27 𝑖𝑛 3𝜋 𝐴 = 𝐴1 + 𝐴2 = 27 + 14.14 = 41.14 𝑖𝑛2 𝐴𝑥 = 𝑎𝑥 41.14𝑦 = 27 2 + 14.14(3) 𝒙 = 𝟐. 𝟑𝟒 𝒊𝒏 15 Problem 3: Locate the centroid of the shaded area created by cutting a semicircle of diameter r from a quarter circle of radius r. 16 Solution 𝐴𝑥 = 𝑎𝑥 1 2 1 2 4𝑟 1 2 2𝑟 𝐹𝑜𝑟 𝑞𝑢𝑎𝑟𝑡𝑒𝑟 𝑐𝑖𝑟𝑐𝑙𝑒: 𝜋𝑟 𝑥 = 𝜋𝑟 − 𝜋𝑟 1 8 4 3𝜋 8 3𝜋 𝐴1 = 𝜋𝑟 2 1 2 1 1 4 𝜋𝑟 𝑥 = 𝑟 3 − 𝑟 3 4𝑟 8 3 12 𝑥1 = 3𝜋 1 2 1 4𝑟 𝜋𝑟 𝑥 = 𝑟 3 𝑦1 = 8 4 3𝜋 𝟐𝒓 𝒙= = 𝟎. 𝟔𝟑𝟔𝟔𝒓 𝝅 𝐹𝑜𝑟 𝑠𝑒𝑚𝑖𝑐𝑖𝑟𝑐𝑙𝑒: 1 1 2 1 1 2 1 2 4𝑟 1 2 1 𝐴2 = 𝜋𝑟 = 𝜋𝑟 2 𝜋𝑟 𝑦 = 𝜋𝑟 − 𝜋𝑟 𝑟 2 4 8 8 4 3𝜋 8 2 1 1 2 1 1 4 𝑟 2𝑟 𝜋𝑟 𝑦 = 𝑟 3 − 𝜋𝑟 3 2 𝑥2 = = 8 3 16 3𝜋 3𝜋 1 2 1 𝜋𝑟 𝑦 = 0.137𝑟 3 𝑦2 = 𝑟 8 2 𝒚 = 𝟎. 𝟑𝟒𝟖𝟖𝒓 𝐹𝑜𝑟 𝑡ℎ𝑒 𝑠ℎ𝑎𝑑𝑒𝑑 𝑎𝑟𝑒𝑎: 1 1 1 Coordinates of the centroid is at (𝟎. 𝟔𝟑𝟔𝟔𝒓, 𝟎. 𝟑𝟒𝟖𝟖𝒓) 𝐴 = 𝐴1 − 𝐴2 = 𝜋𝑟 2 − 𝜋𝑟 2 = 𝜋𝑟 2 4 8 8 17 Problem 4: Find the coordinates of the centroid of the shaded area shown. 18 Solution 1 𝐴2 = 𝜋 42 = 25.133 𝑖𝑛2 2 𝑥2 = 4′′ 𝑦2 = 12 − 𝑎 = 12 − 1.698 = 10.302′′ 1 𝐴3 = 𝜋 62 = 28. 274 𝑖𝑛2 4 𝑥3 = 18 − 𝑏 = 18 − 2.546 = 15.454′′ 𝑦3 = 12 − 𝑏 = 12 − 2.546 = 9.454′′ 1 𝐴4 = (6)(6) = 18 𝑖𝑛2 2 𝑥4 = 18 − 𝑐 = 18 − 2 = 16′′ 𝑦4 = 𝑐 = 2′′ 4(4) 𝑎= = 1.698′′ 𝐴 = 𝐴1 − 𝐴2 − 𝐴3 − 𝐴4 3𝜋 4(6) 𝐴 = 216 − 25.133 − 28.274 − 18 = 144.593 𝑖𝑛2 𝑏= = 2.546′′ 3𝜋 1 𝐴𝑥 = 𝑎𝑥 𝑐 = 6 = 2′′ 3 144.593𝑥 = 216 9 − 25.133 4 − 28.274 15.454 − 18(6) 𝒙 = 𝟕. 𝟕𝟑𝟔′′ 𝐴1 = 18 12 = 216 𝑖𝑛2 1 𝐴𝑦 = 𝑎𝑦 𝑥1 = 18 = 9′′ 144.593𝑦 = 216 6 − 25.133 10.302 − 28.274 9.454 − 18(2) 2 1 𝒚 = 𝟓. 𝟎𝟕𝟓′′ 19 𝑦1 = 12 = 6′′ 2 Problem 5: Locate the centroid of the shaded area enclosed by the curve y2 = ax and the straight line shown. Hint: Observe that the curve y2 = ax relative to the y-axis is of the form y = kx2 with respect to the x-axis. 20 Solution 𝐹𝑜𝑟 𝑝𝑎𝑟𝑎𝑏𝑜𝑙𝑖𝑐 𝑠𝑒𝑔𝑚𝑒𝑛𝑡: 2 𝐴1 = 6 12 = 48 𝑓𝑡 2 3 3 𝑥1 = 12 = 7.2 𝑓𝑡 5 3 𝑦1 = 6 = 2.25 𝑓𝑡 8 𝐹𝑜𝑟 𝑡𝑟𝑖𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑎𝑟𝑒𝑎: 1 𝐴𝑥 = 𝑎𝑥 𝐴2 = 6 12 = 36 𝑓𝑡 2 2 𝐴𝑥 = 𝐴1 𝑥1 − 𝐴2 𝑥2 2 𝑥2 = 12 = 8 𝑓𝑡 12𝑥 = 48 7.2 − 36(8) 3 𝒙 = 𝟒. 𝟖 𝒇𝒕 1 𝑦2 = 6 = 2 𝑓𝑡 3 𝐴𝑦 = 𝑎𝑦 𝐴𝑦 = 𝐴1 𝑦1 − 𝐴2 𝑦2 𝐹𝑜𝑟 𝑡ℎ𝑒 𝑠ℎ𝑎𝑑𝑒𝑑 𝑎𝑟𝑒𝑎: 12𝑦 = 48(2.25) − 36(2) 𝐴 = 𝐴1 − 𝐴2 = 48 − 36 = 12 𝑓𝑡 2 𝒚 = 𝟑 𝒇𝒕 Coordinates of the centroid is at (𝟒. 𝟖, 𝟑) 21 Problem 6: Locate the centroid (𝑥 , 𝑦) of the cross-sectional area. 22 Solution 𝐴1 = 4 0.5 = 2 𝑖𝑛2 0.5 𝑥1 = = 0.25 𝑖𝑛 2 4 𝑦1 = = 2 𝑖𝑛 2 𝐴2 = (3 − 0.5) 0.5 = 1.25 𝑖𝑛2 3 − 0.5 𝑥2 = 0.5 + = 1.75 𝑖𝑛 2 0.5 𝑦2 = = 0.25 𝑖𝑛 2 𝐴𝑥 = 𝑎𝑥 3.25𝑥 = 2 0.25 + 1.25(1.75) 𝐹𝑜𝑟 𝑡ℎ𝑒 𝑡𝑜𝑡𝑎𝑙 𝑎𝑟𝑒𝑎: 𝒙 = 𝟎. 𝟖𝟐𝟔𝟗 𝒊𝒏. 𝐴 = 𝐴1 + 𝐴2 = 2 + 1.25 = 3.25 𝑖𝑛2 𝐴𝑦 = 𝑎𝑦 3.25𝑦 = 2 2 + 1.25(0.25) 𝒚 = 𝟏. 𝟑𝟐𝟔𝟗 𝒊𝒏. Coordinates of the centroid is at (𝟎. 𝟖𝟐𝟔𝟗, 𝟏. 𝟑𝟐𝟔𝟗) 23 Problem 7: Locate the centroid 𝑦 of the cross-sectional area of the built-up plan. 24 Solution 𝐴1 = 150 20 = 3000 𝑚𝑚2 20 𝑦1 = 450 + = 10 𝑚𝑚 2 𝐴2 = 450 20 (3) = 27000 𝑚𝑚2 450 𝑦2 = = 225 𝑚𝑚 2 𝐴3 = (450 − 200) 20 = 5000 𝑚𝑚2 450 − 200 𝑦3 = = 125 𝑚𝑚 2 𝐹𝑜𝑟 𝑡ℎ𝑒 𝑡𝑜𝑡𝑎𝑙 𝑎𝑟𝑒𝑎: 𝐴 = 𝐴1 + 𝐴2 − 𝐴3 𝐴 = 3000 + 27000 − 5000 = 25000 𝑚𝑚2 𝐴𝑦 = 𝑎𝑦 25000𝑦 = 3000 460 + 27000 225 − 5000 125 𝒚 = 𝟐𝟕𝟑. 𝟐 𝒎𝒎 25  Worksheet 1 (Finals): 1.) Determine the centroid (𝑥 , 𝑦) of the 2.) Locate the centroid (𝑥 , 𝑦) of 3.) Locate the centroid (𝑥 , 𝑦) of area. the composite area. the cross-sectional area of the channel. 26  Assignment: 4.) Locate the centroid 𝑦 of the cross-sectional area 5.) Locate the centroid 𝑦 of the cross-sectional area of the concrete beam. of the built-up plan. 27  Republic of the Philippines CAMARINES SUR POLYTECHNIC COLLEGES Nabua, Camarines Sur MOMENTS OF INERTIA ERLY O. CELIZ, RCE Moment of Inertia Moment of inertia, also called the second moment of area, is the product of area and the square of its moment arm about a reference axis. Moment of inertia about the x-axis: 𝑰𝒙 = 𝒚𝟐 𝒅𝑨 Moment of inertia about the y-axis: 𝑰𝒚 = 𝒙𝟐 𝒅𝑨 2 Moment of Inertia The 𝒂𝒓𝒆𝒂 𝒎𝒐𝒎𝒆𝒏𝒕 𝒐𝒇 𝒊𝒏𝒆𝒓𝒕𝒊𝒂 is used to determine how a solid member performs under a given load. It correlates to the beam's 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑡𝑜 𝑏𝑒𝑛𝑑𝑖𝑛𝑔. The area moment of inertia has units of length to the fourth. 3 Radius of Gyration The radius of gyration is an alternate way of expressing the distribution of area away from an axis which combines the effects of the moments of inertia and cross sectional area. It is usually denoted by symbol 𝑘 (sometimes by 𝑟) and is defined by the relation 𝑰 𝒌= 𝐨𝐫 𝐈 = 𝐀𝒌𝟐 𝑨 Where: 𝐼 = 𝑚𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝐼𝑛𝑒𝑟𝑡𝑖𝑎 𝐴 = 𝑐𝑟𝑜𝑠𝑠 − 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎 4 Transfer Formula for Moment of Inertia 𝑰 = 𝑰 + 𝑨𝒅𝟐 Where: 𝑥 ′ = 𝑐𝑒𝑛𝑡𝑟𝑜𝑖𝑑𝑎𝑙 𝑎𝑥𝑖𝑠 𝑥 = 𝑎𝑛𝑦 𝑎𝑥𝑖𝑠 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑡𝑜 𝑡ℎ𝑒 𝑐𝑒𝑛𝑡𝑟𝑜𝑖𝑑𝑎𝑙 𝑎𝑥𝑖𝑠 𝐼 = 𝑚𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑖𝑛𝑒𝑟𝑡𝑖𝑎 𝑎𝑏𝑜𝑢𝑡 𝑡ℎ𝑒 𝑥 − 𝑎𝑥𝑖𝑠 𝐼 = 𝑐𝑒𝑛𝑡𝑟𝑜𝑖𝑑𝑎𝑙 𝑚𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑖𝑛𝑒𝑟𝑡𝑖𝑎 𝐴 = 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝑑 = 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑥 𝑎𝑛𝑑 𝑥′ 5 Moment of Inertia of Common Shapes 6 Moment of Inertia of Common Shapes 7 Moment of Inertia of Common Shapes 8 Example: 1.) Determine the moment of inertia of the T-section shown with respect to its centroidal Xo axis. 9 Solution: 𝐴1 = 8 2 = 16 𝑖𝑛.2 8(2)3 16 𝐼1 = = 𝑖𝑛.4 12 3 𝑦1 = 1 𝑖𝑛. 𝐴2 = 8 2 = 16 𝑖𝑛.2 2(8)3 256 𝐼2 = = 𝑖𝑛.4 12 3 𝑦2 = 2 + 4 = 6 𝑖𝑛. 𝐴 = 𝐴1 + 𝐴2 = 16 + 16 = 32 𝑖𝑛.2 𝐴𝑦 = 𝐴1 𝑦1 + 𝐴2 𝑦2 32𝑦 = 16 1 + 16(6) 𝑦 = 3.5 𝑖𝑛. 𝐼 = 𝐼1 + 𝐴1 𝑦 − 𝑦1 2 + [𝐼2 + 𝐴2 𝑦2 − 𝑦 2 ] 16 256 𝐼= + 16 3.5 − 1 2 + + 16 6 − 3.5 2 3 3 𝑰 = 𝟐𝟗𝟎. 𝟔𝟕 𝒊𝒏.𝟒 10 Example: 2.) Determine the moment of inertia of the area shown with respect to its centroidal axes. 11  𝐴1 = 12 1 = 12 𝑖𝑛.2 Solution: 12(1)3 𝐼1 = = 1 𝑖𝑛.4 12 𝐴3 = 6 1 = 6 𝑖𝑛.2 𝑦1 = 0.5 𝑖𝑛. 6(1)3 𝐼3 = = 0.5 𝑖𝑛.4 12 𝐴2 = 1 12 = 12 𝑖𝑛.2 𝑦3 = 13.5 𝑖𝑛. 1(12)3 𝐼2 = = 144 𝑖𝑛.4 12 𝑦2 = 7 𝑖𝑛. 𝐴 = 𝐴1 + 𝐴2 + 𝐴3 = 12 + 12 + 6 = 30 𝑖𝑛.2 𝐴𝑦 = 𝐴1 𝑦1 + 𝐴2 𝑦2 + 𝐴3 𝑦3 30𝑦 = 12 0.5 + 12 7 + 6(13.5) 𝑦 = 5.7 𝑖𝑛. 𝐼𝑥 = 𝐼1 + 𝐴1 𝑦 − 0.5 2 + [𝐼2 + 𝐴2 7 − 𝑦 2 ] + [𝐼3 + 𝐴3 13.5 − 𝑦 2 ] 𝐼𝑥 = 1 + 12 5.7 − 0.5 2 + [144 + 12 7 − 5.7 2 ] + [0.5 + 6 13.5 − 5.7 2 ] 𝑰𝒙 = 𝟖𝟓𝟓. 𝟑 𝒊𝒏.𝟒 1 3 1 3 1 3 𝐼𝑦 = 1 12 + 12 1 + 1 6 12 12 12 𝑰𝒚 = 𝟏𝟔𝟑 𝒊𝒏.𝟒 12 Worksheet 2: Determine the moment of inertia 𝐼𝑥 and 𝐼𝑦 of the structural members. 13