Analysis of Indeterminate Structures Force Method Lecture 19 Force Method Indeterminate Structure • Stresses in indeterminate are generally lower than determinate structure. • Deflection in case of indeterminate structure is less than those compared to determinate due to greater stiffness. Indeterminate Structure • Statically indeterminate structure have capacity to redistribute the loads. If a part (or member or support) of such a structure fails, the entire structure will not necessarily collapse, and the loads will be redistributed to the adjacent portions of the structure. https://www.quora.com/What-are-the-advantages-ofindeterminate-structure-over-determinate-structure Methods in the Analysis of Statically Indeterminate Structures • Force Method ➢ Virtual Work Method ➢ Castigliano’s Theorem ➢ Three Moment Theorem • Displacement Method ➢ Slope Deflection Method ➢ Moment Distribution Method ➢ Stiffness Matrix Method ➢ Kani’s Method Force Method BASIS • Originally Developed by James Clerk Maxwell in 1864 • Later refined by Otto Mohr and Heinrich Muller-Breslau • Flexibility Method • Compatibility Method • Method of consistent displacement Force Method • Primary unknown are forces • It is used to calculate reactions and internal forces in statically indeterminate structure due to loads and imposed deformation. • Based on transforming a given structure into a statically determinate primary system and calculating the magnitude of the statically redundant forces required to restore the geometric boundaru condition of original structure. General Procedure Redundant 𝐴𝑐𝑡𝑢𝑎𝑙 𝐵𝑒𝑎𝑚 Redundant 𝑅𝐴 Step 1: determine which redundant to be removed, you may choose to removed the roller support or change the fixed support to hinge or pin support. 𝑅𝐵 General Procedure Step 2. Using superposition, calculate the force that would be required to achived compatibility with the original structure + 𝑃𝑟𝑖𝑚𝑎𝑟𝑦 𝑆𝑡𝑟𝑢𝑐𝑡𝑢𝑟𝑒 𝐶𝑜𝑚𝑝𝑎𝑡𝑖𝑏𝑖𝑙𝑖𝑡𝑦 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛𝑠: 𝑅𝑒𝑑𝑢𝑛𝑑𝑎𝑛𝑡 𝐵𝑦 𝑎𝑝𝑝𝑙𝑖𝑒𝑑 𝒇𝑩𝑩 ∆𝐵 + ∆′𝐵𝐵 = 0 ∆𝐵 + 𝐵𝑦 𝑓𝐵𝐵 = 0 𝒏𝒐𝒕𝒆: 𝒇𝑩𝑩 − 𝒍𝒊𝒏𝒆𝒂𝒓 𝒇𝒍𝒆𝒙𝒊𝒃𝒊𝒍𝒊𝒕𝒚 𝒄𝒐𝒆𝒇𝒇𝒊𝒄𝒊𝒆𝒏𝒕 Measure of the deflection per unit force, m/N, ft/lb General Procedure 𝑀𝐴 + 𝜃′𝐴𝐴 = 𝑀𝐴 𝛼𝐴𝐴 𝐶𝑜𝑚𝑝𝑎𝑡𝑖𝑏𝑖𝑙𝑖𝑡𝑦 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛𝑠: 𝜃𝐴 + 𝑀𝐴 𝛼𝐴𝐴 = 0 𝛼𝐴𝐴 𝑀𝐴 = −𝜃𝐴 /𝛼𝐴𝐴 𝑛𝑜𝑡𝑒: 𝑠𝑖𝑔𝑛 𝑖𝑛𝑑𝑖𝑐𝑎𝑡𝑒 𝑡ℎ𝑎𝑡𝑀𝐴 𝑎𝑐𝑡𝑠 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑡ℎ𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑡𝑜 𝑡ℎ𝑒 𝑢𝑛𝑖𝑡 𝑐𝑜𝑢𝑝𝑙𝑒 𝑚𝑜𝑚𝑒𝑛𝑡 General Procedure = + + = + + 𝐶𝑜𝑚𝑝𝑎𝑡𝑖𝑏𝑖𝑙𝑖𝑡𝑦 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛𝑠: ∆𝐵 + 𝐵𝑦 𝑓𝐵𝐵 + 𝐵𝑦 𝑓𝐶𝐵 = 0 ∆𝐶 + 𝐶𝑦 𝑓𝐵𝐶 + 𝐶𝑦 𝑓𝐶𝐶 = 0 Deflections and Slopes of Beams (Cantilever Beam) Deflections and Slopes of Beams Deflections and Slopes of Beams (Simple Beam) Deflections and Slopes of Beams (Simple Beam) Deflections and Slopes of Beams (Cantilever Beam) Deflections and Slopes of Beams (Cantilever Beam) Deflections and Slopes of Beams (Cantilever Beam) Deflections and Slopes of Beams (Cantilever Beam) Deflections and Slopes of Beams (Cantilever Beam) Deflections and Slopes of Beams (Simple Beam) Deflections and Slopes of Beams (Simple Beam) Deflections and Slopes of Beams (Simple Beam) Deflections and Slopes of Beams (Simple Beam) Deflections and Slopes of Beams (Simple Beam) Deflections and Slopes of Beams (Simple Beam) Problem #1 Determine the reactions and draw the shear and moment diagram of the given beam Compatibility Equations ∆𝐴 + 𝐴𝑦 𝑓𝐴𝐴 = 0 ∆𝐴 𝑃𝑟𝑖𝑚𝑎𝑟𝑦 𝐵𝑒𝑎𝑚 𝐴𝑦𝑓𝐴𝐴 Ay = 1 𝑅𝑒𝑑𝑢𝑛𝑑𝑎𝑛𝑡 𝐴𝑦 Primary Beam 𝑃𝑎2 𝑀𝑜 𝐿2 ∆𝐴 = − 3𝐿 − 𝑎 + 6𝐸𝐼 2𝐸𝐼 𝑃𝑎2 ∆= (3𝐿 − 𝑎) 6𝐸𝐼 120 62 ∆𝐴 = − 6𝐸𝐼 ∆𝐴 = − 𝑀0𝐿2 ∆= 2𝐸𝐼 75 122 3 12 − 6 + 2𝐸𝐼 16200 𝑘𝑁𝑚3 𝐸𝐼 Redundant Ay 𝑓𝐴𝐴 = (1)123 3𝐸𝐼 1 Compatibility Equations ∆𝐴 𝑃𝑟𝑖𝑚𝑎𝑟𝑦 𝐵𝑒𝑎𝑚 ∆𝐴 = − ∆𝐴 + 𝐴𝑦 𝑓𝐴𝐴 = 0 16200 𝑘𝑁𝑚3 𝐸𝐼 𝐴𝑦𝑓𝐴𝐴 Ay = 1 𝑅𝑒𝑑𝑢𝑛𝑑𝑎𝑛𝑡 𝐴𝑦 𝐴𝑦 = 28.125 𝑘𝑁 ↑ Ex.1:Determine the reactions and draw the shear and moment diagram of the given beam 𝑀𝐶 = 307.5 𝑘𝑁𝑚 𝐵𝑦 = 91.875 𝑘𝑁 𝐴𝑦 = 28.125 𝑘𝑁 28.125 𝑘𝑁 75 𝑘𝑁𝑚 243.75 𝑘𝑁𝑚 −91.875 𝑘𝑁 −307.5 𝑘𝑁𝑚 Problem#2 Determine the reactions and draw the shear and moment diagram of the given beam ∆𝐶 𝑃𝑟𝑖𝑚𝑎𝑟𝑦 𝐵𝑒𝑎𝑚 𝐶𝑦𝑓𝐶𝐶 Compatibility Equations ∆𝐶 + 𝐶𝑦 𝑓𝑐𝑐 = 0 𝐶𝑦 = 1 𝑅𝑒𝑑𝑢𝑛𝑑𝑎𝑛𝑡 𝐶𝑦 Primary 𝐴𝑦 = 356.25 𝑘𝑁 𝐷𝑦 = 318.75 𝑘𝑁 𝐴𝑦 = 0.5 𝑘𝑁 𝐷𝑦 = 0.5 𝑘𝑁 Redundant 𝑹𝒆𝒂𝒍 𝑽𝒊𝒓𝒕𝒖𝒂𝒍 𝐶 𝐴𝑦 = 0.5 𝑘𝑁 𝐶 𝐷𝑦 = 0.5 𝑘𝑁 𝐴𝑦 = 0.5 𝑘𝑁 𝐷𝑦 = 0.5 𝑘𝑁 Primary Segment origin Limit M m AB A 0-5 356.25𝑥 0.5𝑥 BC A 5-10 356.25𝑥 − 375 𝑥 − 5 0.5𝑥 DC D 0-10 318.75𝑥 − 15𝑥 2 0.5𝑥 1 𝐿 න 𝑀𝑚 𝑑𝑥 𝐸𝐼 0 1 5 2473.96 න 356.25𝑥 0.5𝑥 𝑑𝑥 = 3𝐸𝐼 0 𝐸𝐼 10 1 න 356.25𝑥 − 375 𝑥 − 5 0.5𝑥 𝑑𝑥 3𝐸𝐼 5 10807.29 = 𝐸𝐼 10 1 34375 න 318.75𝑥 − 15𝑥 2 0.5𝑥 𝑑𝑥 = 𝐸𝐼 0 𝐸𝐼 47656.25 ∆𝐶 = 𝑘𝑁𝑚3 ↓ 𝐸𝐼 Redundant Segment origin Limit M m 1 𝐿 න 𝑀𝑚 𝑑𝑥 𝐸𝐼 0 AC A 0-10 −0.5𝑥 −0.5𝑥 1 10 න −0.5𝑥 −0.5𝑥 𝑑𝑥 = 27.78 3𝐸𝐼 0 DC D 0-10 −0.5𝑥 −0.5𝑥 1 10 න −0.5𝑥 −0.5𝑥 𝑑𝑥 = 83.33 𝐸𝐼 0 𝑓𝐶𝐶 111.11 = 𝑘𝑁𝑚3 ↑ 𝐸𝐼 𝐶𝑜𝑚𝑝𝑎𝑡𝑖𝑏𝑖𝑙𝑖𝑡𝑦 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 ∆𝐶 𝑃𝑟𝑖𝑚𝑎𝑟𝑦 𝐵𝑒𝑎𝑚 𝐶𝑦 𝑓𝐶𝐶 𝐶𝑦 = 1 ∆𝐶 = 47656.25 𝑘𝑁𝑚3 ↓ 𝐸𝐼 𝑓𝐶𝐶 = ∆𝐶 + 𝐶𝑦 𝑓𝑐𝑐 = 0 111.11 𝑘𝑁𝑚3 ↑ 𝐸𝐼 𝑅𝑒𝑑𝑢𝑛𝑑𝑎𝑛𝑡 𝐶𝑦 𝐶𝑦 = 428.911 𝑘𝑁 428.91 𝑘𝑁 141.795 𝑘𝑁 104.295 𝑘𝑁 195.705 𝑘𝑁 141.795 𝑘𝑁 10 − 𝑥 = 3.476𝑚 x = 6.524m −104.295 𝑘𝑁 −233.205 𝑘𝑁 708.975 kNm 181.34 𝑘𝑁𝑚 −457.05 𝑘𝑁𝑚 Ex.3:Determine the reactions and draw the shear and moment diagram of the given Frame 𝐷𝑦𝑓𝐷𝐷 ∆𝐷 𝑃𝑟𝑖𝑚𝑎𝑟𝑦 𝐹𝑟𝑎𝑚𝑒 𝐶𝑜𝑚𝑝𝑎𝑡𝑖𝑏𝑖𝑙𝑖𝑡𝑦 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 ∆𝐷 + 𝐷𝑦 𝑓𝐷𝐷 = 0 𝑅𝑒𝑑𝑢𝑛𝑑𝑎𝑛𝑡 𝐷𝑦 Primary frame 𝐷 200𝑘𝑁 200 kN 1750 + 10P segme origin nt limit DC D 0-5 CB D 5-10 AB A 0-10 M 𝑴 𝑷 M (P=0) 𝟏 𝑳 𝑴 න 𝑴 𝒅𝒙 𝑬𝑰 𝒐 𝑷 Primary frame 𝐷 200𝑘𝑁 200 kN 1750 + 10P segme origin nt limit M 𝑴 𝑷 M (P=0) 𝟏 𝑳 𝑴 න 𝑴 𝒅𝒙 𝑬𝑰 𝒐 𝑷 DC D 0-5 −𝑃𝑥 −𝑥 0 0 CB D 5-10 −𝑃𝑥 − 150(𝑥 −5) −𝑥 −150(𝑥 −5) 15625/𝐸𝐼 AB A 0-10 − 1750 + 10𝑃 − 10𝑥 2 + 200𝑥 −10 −1750 − 10𝑥 2 + 200𝑥 108333.333/𝐸𝐼 ∆𝑫 = 𝟏𝟐𝟑𝟗𝟓𝟖. 𝟑𝟑𝟑/𝑬𝑰 𝐷 Redundant 10𝑚 10𝑚 10P seg orig lim me in it nt DB D 0-5 AB A 0-10 M M (P=1) 𝐷 Redundant 10𝑚 10𝑚 10P segme origin nt limit M 𝑴 𝑷 M (P=1) 𝟏 𝑳 𝑴 න 𝑴 𝒅𝒙 𝑬𝑰 𝒐 𝑷 DB D 0-10 𝑃𝑥 𝑥 𝑥 333.333/𝐸𝐼 AB A 0-10 10𝑃 10 10 1000/𝐸𝐼 𝒇𝑫𝑫 = 𝟏𝟑𝟑𝟑. 𝟑𝟑𝟑/𝑬𝑰 𝐷𝑦𝑓𝐷𝐷 ∆𝐷 𝑃𝑟𝑖𝑚𝑎𝑟𝑦 𝐹𝑟𝑎𝑚𝑒 ∆𝑫 = 𝟏𝟐𝟑𝟗𝟓𝟖. 𝟑𝟑𝟑/𝑬𝑰 𝐶𝑜𝑚𝑝𝑎𝑡𝑖𝑏𝑖𝑙𝑖𝑡𝑦 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 ∆𝐷 + 𝐷𝑦 𝑓𝐷𝐷 = 0 𝑅𝑒𝑑𝑢𝑛𝑑𝑎𝑛𝑡 𝐷𝑦 𝒇𝑫𝑫 = 𝟏𝟑𝟑𝟑. 𝟑𝟑𝟑/𝑬𝑰 𝟏𝟐𝟑𝟗𝟓𝟖. 𝟑𝟑𝟑 𝐷𝑦 = 1333.333 𝑫𝒚 = 𝟗𝟐. 𝟗𝟔𝟗 𝒌𝑵 𝑫𝒚 = 𝟗𝟐. 𝟗𝟕 𝒌𝑵 𝑨𝒙 = 𝟐𝟎𝟎 𝒌𝑵 𝑴𝑨 = 𝟖𝟐𝟎. 𝟑 𝒌𝑵𝒎 𝑨𝒚 = 𝟓𝟕. 𝟎𝟑 𝒌𝑵 𝑴𝑩 = 𝟏𝟕𝟗. 𝟕𝒌𝑵𝒎 𝑩𝒚 = 𝟓𝟕. 𝟎𝟑 𝒌𝑵 179.7 𝑘𝑁𝑚 𝑩𝒙 =0 𝑨𝒙 = 𝟐𝟎𝟎 𝒌𝑵 𝑴𝑨 = 𝟖𝟐𝟎. 𝟑 𝒌𝑵𝒎 𝑨𝒚 = 𝟓𝟕. 𝟎𝟑 𝒌𝑵 200 𝑘𝑁 −820.3 𝑘𝑁 𝑴𝑩 = 𝟏𝟕𝟗. 𝟕𝒌𝑵𝒎 𝑩𝒙 =0 𝑩𝒚 = 𝟓𝟕. 𝟎𝟑 𝒌𝑵 𝑫𝒚 = 𝟗𝟐. 𝟗𝟕 𝒌𝑵 𝟓𝟕. 𝟎𝟑 𝒌𝑵 𝟒𝟔𝟒. 𝟖𝟓𝒌𝑵𝒎 𝟏𝟕𝟗. 𝟕𝒌𝑵𝒎 −𝟗𝟐. 𝟗𝟕 𝒌𝑵
