Giải các bài tập liên quan cơ kết cấu: hệ giàn sử dụng phương pháp tách nút và mặt cắt, giải dầm 2 khớp, hệ siêu tĩnh theo phương pháp lực và phương pháp chuyển vị bằng tiếng anh và hướng dẫn khá chi tiết.
VIETNAM NATIONAL UNIVERSITY HO CHI MINH CITY HO CHI MINH CITY UNIVERSITY OF TECHNOLOGY FACULTY OF CIVIL ENGINEERING MECHANICS OF STRUCTURES (CI3233) Semester 231 - Class P01 ASSIGNMENT Lecturer: NGUYỄN HỒNG ÂN Student: Phạm Thị Thanh Tâm ID 2014439 DOB : 25/10/2002 HO CHI MINH CITY, DECEMBER 2023 Ho Chi Minh City University of Technology Faculty of Civil Engineering Contents Problem 1.1 Problem 1.1 1.2 Problem 1.5 3 Problem 13 2.1 Problem 2.1 13 2.2 Problem 2.5 16 Problem 19 3.1 Problem 3.4 19 3.2 Problem 3.6 24 Problem 29 4.1 Problem 4.2 29 4.2 Problem 4.5 32 Ho Chi Minh City University of Technology Faculty of Civil Engineering Problem 1.1 Problem 1.1 Data: Data 19 λ1 λ2 λ3 Requirements: a Determine the normal forces of bars b Draw the normal force diagram Solution Give the nodes some indices λ4 λ5 Ho Chi Minh City University of Technology Faculty of Civil Engineering Calculation of Reaction Forces As the structure is statically determinate, the reaction forces can be calculated with the Equilibrium equations In our case, we are calculating the support forces HA , VA , and VC X H = : HA = −4P √ a a4P + 4P + a2P M/A = : 2a × VC = a3P + 2 √ ⇒ VC = (4 + 3)P X V = : VA + VC = 4P + 2P + 3P √ √ ⇒ VA = 9P − (4 + 3)P = (5 − 3)P X Calculation of the internal forces Node A Ho Chi Minh City University of Technology Faculty of Civil Engineering Vertical Equilibrium: X √ V = : N4 × sin(60◦ ) + (5 − 3) × P = √ − 10 ⇒ N4 = P (compression) Horizontal Equilibrium: X H = : N1 + N4 × cos 60◦ = 4P √ − 10 ⇒ N1 = 4P − × P × cos 60◦ 3√ 9+5 ⇒ N1 = 4P − × P (tension) Node C Vertical Equilibrium: X √ V = : N7 × sin(60◦ ) + (4 + 3) × P √ √ −(4 + 3) 6+8 ⇒ N7 = P =− P (compression) sin 60◦ Ho Chi Minh City University of Technology Faculty of Civil Engineering Horizontal Equilibrium: X H = : N2 + N7 × cos 60◦ = √ 6+8 ⇒ N2 = P × cos 60◦ 3√ 3+4 ⇒ N2 = P (tension) Section 1-1 (On the left side) 4P E 4P √3 2a N3 ° 60 A 4P N5 N1 (5-√3)P Vertical Equilibrium: X √ V = : N5 × sin(60◦ ) + 4P = (5 − 3)P √ √ −(5 − 3)P − 4P −6 + ⇒ N5 = P = P (compression) sin 60◦ Horizontal Equilibrium: X √ 9+5 H = : N3 + N5 × cos 60 + 4P − 4P + P =0 √ 9+5 ◦ ⇒ N3 = −N5 × cos 60 − P √ ⇒ N3 = (−2 − 3)P (compression) ◦ Ho Chi Minh City University of Technology Faculty of Civil Engineering Node D D 60 ° 60 N3 ° 2P N6 N7 Horizontal Equilibrium: X H = : N3 + N6 × cos 60◦ = N7 × cos 60◦ √ 6+4 ⇒ N6 = P (tension) a The normal force of bars N1 N2 N3 N4 N5 N6 N7 √ 9+5 = 4P − P (tension) √ 3+4 P (tension) = √ = (−2 − 3)P (compression) √ − 10 = P (compression) √ −6 + = P (compression) 3√ 6+4 = P (tension) √ 6+8 =− P (compression) Ho Chi Minh City University of Technology Faculty of Civil Engineering b Plot the normal force diagram Normal force diagram (2+2√3)P (-6+10√3)P (6+4√3)P (6+8√3)P (-6+10√3)P (3+4√3)P (9+5√3)P 1.2 Problem 1.5 Ho Chi Minh City University of Technology Faculty of Civil Engineering Data: Data 18 λ1 λ2 λ3 Requirements: a Determine the normal forces of bars b Draw the normal force diagram Solution Give the nodes and the bars some indices Calculation of the internal forces Node E λ4 λ5 -3 Ho Chi Minh City University of Technology Faculty of Civil Engineering Vertical Equilibrium: X V = : N4 × sin(45◦ ) + 5P = √ −5P ⇒ N4 = = −5 2P (compression) sin(45◦ ) Horizontal Equilibrium: X H = : N1 + N4 × cos 45◦ + 4P = √ ⇒ N1 = −4P − (−5 2)P × cos 45◦ = P (tension) Node C Vertical Equilibrium: X V = : N5 = N4 × cos 45◦ + 5P √ ⇒ N5 = 2P × cos 45◦ + 5P = 10 (tension) Horizontal Equilibrium: X H = : N3 + N4 × sin 45◦ = √ ⇒ N3 = −5 2P × sin 45◦ = −5P (compression) 10 Ho Chi Minh City University of Technology Faculty of Civil Engineering a Calculation of Reaction Forces: 25 qL − 2qL = qL 8 X H = : HB = X M/B = : 2L × VA = 3ql2 + 5qL2 + 2qL × L − X V = : VB = VA + 5qL = 25 15 qL × 2L ⇒ VA = qL 8 15 55 qL + 5qL = qL 8 b Plot the M, V, N of the frame below: X1= 25qL A P=5qL M=3qL2 D EI EI C 15qL 2L VA = 15qL q EI 5qL 15qL HB= qL 55 VB= qL L8 qL 8qL 15qL2 25qL 55qL 25qL 25qL V B 2L 5qL qL 17qL2 N M 55qL 22 Ho Chi Minh City University of Technology Faculty of Civil Engineering c Calculation of the vertical displacement of the point where the force P is applied: Calculation of Reaction forces due to PK = as follows: X H = : HB = X V = : VB = VA + = + = 2 X M/B = : 2L × VA = L ⇒ VA = Plot the bending moment due to PK = and external load is shown in below: 8qL D EI EI 15 VA = qL q "m" EI qL C 17qL2 2L X1= 25qL A 15qL2 P=5qL M=3qL2 Mm B HB= qL 55 VB= qL L8 2L PK=1 D A L C 2L VA = "k" B 2L HB=0 VB= L2 23 Mk Ho Chi Minh City University of Technology Faculty of Civil Engineering Using the graph multiplication method: ∆P = (Mm )(Mk ) 1 15 1 = ( × 2L × qL2 × L + 3qL2 × L × L + × L × (8qL2 − 3qL2 ) × L) EI 2 17qL = 3EI 3.2 Problem 3.6 Data: Data 18 λ1 λ2 λ3 λ4 λ5 λ6 λ7 λ8 Requirements: A frame is subjected to loads as shown in above figure Using the force method: a Determine the reaction forces of supports b Draw the M, V, N of the frame c Determine the vertical displacement of the point where the force P is applied Solution Degree of indetermeinacy: DI = The normal equation is : δ11 + X1 + ∆1P = 24 Ho Chi Minh City University of Technology Faculty of Civil Engineering Select the released structure : A EI EI EI q B 2L P=qL M=2qL2 X1 L q A C EI EI RS EI C L P=qL M=2qL2 B 2L 2L 2L Calculation of primary beam subjected to external load: X V = : VA = qL 11 M/A = : L × HB = 2qL2 + 4L × qL − qL × L ⇒ HB = qL 2 X 11 13 H = : HA = HB + qL = qL + qL = qL 2 X Calculation of primary beam subjected to X1 = : X V = : VA = X M/A = : L × HB = ⇒ HB = X H = : HA = HB = L L The bending moment diagram for primary beam due to external load and X1 = is shown in below: 25 Ho Chi Minh City University of Technology Faculty of Civil Engineering qL HA = 13qL P=qL M=2qL2 A EI EI q B 2L HA = L A C EI 2 qL L VA = qL 2 qL qL MoP qL 11 HB= qL 2L X1 EI EI B 2L 1 L EI VA = C M1 L 2L HB= Using the graph multiplication method, the flexibility coeficients are com- puted as follows: 1 7L (1 × 2L × + × × L × ) = EI 3EI 1 2 ql2 ∆1 P = (MPo )(M1 ) = ( × 2qL2 × 2L × + × 6qL2 × L × − × ×L× ) EI 2 3 95qL → ∆1 P = 24EI δ1 = (M1 )(M1 ) = From the normal equation : 95qL3 7L X1 + =0 3EI 24EI 95 → X1 = − qL2 56 a Calculation of Reaction Forces: X V = : VA = qL 95 213 M/A = : L × HB + qL × L + qL2 = 2qL2 + 4L × qL ⇒ BB = qL 56 56 X 269 H = : HA = HB + qL → HA = qL 56 X 26 Ho Chi Minh City University of Technology Faculty of Civil Engineering b Plot the M, V, N of the frame below: 95qL2 56 HA = 269qL 56 P=qL M=2qL2 A EI EI q EI 269qL 56 C L VA= qL qL qL V 213 qL 56 2L B HB= 2L 213 qL 56 qL 269qL 56 95qL2 56 269qL 56 N 2 qL 241 qL2 56 M qL 17qL2 56 c Calculation of the vertical displacement of the point where the force P is applied: Calculation of Reaction forces due to PK = as follows: X V = : VA = X M/A = : L × HB = 4L × ⇒ HB = X H = : HA = HB = Plot the bending moment due to PK = and external load is shown in below: 27 Ho Chi Minh City University of Technology Faculty of Civil Engineering qL 95qL2 56 HA = 269qL 56 P=qL M=2qL2 95qL2 56 A EI EI q C EI B 2L 17qL2 56 L VA = qL qL 213 qL HB= 56 2L 2 qL 241 qL2 56 Mm 2L HA =4 PK =1 A EI EI C B 2L 2L L EI VA =1 4L Mk HB=4 2L Using the graph multiplication method: ∆P = (Mm )(Mk ) 17 4 1 95 = (− × qL2 × 2L × L) + ( × qL2 × L × L) + ( × 2qL2 × 2L × L) EI 56 56 3 241 qL + (2qL2 × 2L × L) + ( × qL × L × L) − ( × × L × 2L) 56 3 475qL4 = 42EI 28 Ho Chi Minh City University of Technology Faculty of Civil Engineering Problem 4.1 Problem 4.2 Data: Requirements: Data 14 λ1 λ2 λ3 λ4 λ5 λ6 λ7 λ8 A frame is subjected to loads as shown in below figure Ignore the effect of axial and shear forces on system displacement Using the displacement method: a Draw the M, V of the frame Solution Degree of kinematic indeterminacy (DKI): n = n1 + n2 = + = Normal Equations: r11 Z11 + R1P = 29 Ho Chi Minh City University of Technology Faculty of Civil Engineering Select the Kinematically Determinate Structure: KIS M=4qL2 q P=3qL B A C 4EI 2EI 3L 2L KDS q M=4qL2 P=3qL B A C 4EI 2EI 3L 2L Draw M1 and MPo (using the table of sample elements): KDS q A M=4qL2 B C 4EI 2EI 3L qL2 qL2 P=3qL MoP qL2 2L qL2 Z1=1 A 8EI 3L B 2EI 3L C 4EI qL2 M1 2L 4EI 3L 8EI L 30 4EI L Ho Chi Minh City University of Technology Faculty of Civil Engineering Determination of Coefficients by using equilibrium equations: r11 = 8EI 8EI 32EI + = 3L L 3L 3 R1P = qL2 − qL2 − 4qL2 = −4qL2 4 Solving the normal equations: Z1 = − R1P 4qL2 3qL3 = 32EI = r11 8EI 3L Shear force and bending moment diagram of Kinematically Indeterminate Structure: With M = M1 Z1 + MPo qL2 qL2 M1 Z1 qL2 3qL2 qL2 qL2 qL2 MoP qL2 qL2 31 Ho Chi Minh City University of Technology Faculty of Civil Engineering qL2 qL2 qL2 M qL2 qL2 qL2 qL qL V qL 2qL 4.2 15qL 15qL Problem 4.5 Data: Data 19 λ1 λ2 λ3 λ4 32 λ5 λ6 λ7 λ8 -1 Ho Chi Minh City University of Technology Faculty of Civil Engineering Requirements: A frame is subjected to loads as shown in below figure Ignore the effect of axial and shear forces on system displacement Using the displacement method: a Draw the M, V of the frame Solution Degree of kinematic indeterminacy (DKI): n = n1 + n2 = + = Normal Equations: r11 Z11 + R1P = Select the Kinematically Determinate Structure: M=qL2 P=3qL KIS 2EI 2EI 2L 2EI q 2L L 4L M=qL2 P=3qL KDS 2EI 2EI 2L 2EI q 2L L 4L Draw M1 and MPo (using the table of sample elements): 33 Ho Chi Minh City University of Technology Faculty of Civil Engineering M=qL2 2EI MoP 2EI q qL2 qL2 2L 2EI qL2 P=3qL 15qL2 qL2 2L L 4L 8EI L Z1=1 M1 2EI 4EI 4EI L L 2L 2EI 2EI 2EI L 2L L 3EI 2L 4L Determination of Coefficients by using equilibrium equations: r11 = 3EI 4EI 8EI 27EI + + = 2L L L 2L 43 R1P = − qL2 − qL2 − qL2 = − qL2 12 Solving the normal equations: Z1 = − R1P = r11 34 43 qL2 12 27 qL2 = 43qL3 162EI Ho Chi Minh City University of Technology Faculty of Civil Engineering Shear force and bending moment diagram of Kinematically Indeterminate Structure: With M = M1 Z1 + MPo qL2 MoP qL2 qL2 15qL2 qL2 172qL2 81 86qL2 81 86qL2 81 43qL2 108 M1 Z1 43qL2 81 35 Ho Chi Minh City University of Technology Faculty of Civil Engineering 172qL2 81 M 86qL2 81 4369qL2 5832 50qL2 27 59qL2 81 56qL2 27 70qL2 81 53qL 27 53qL 11qL27 54 V 86qL 27 28qL 27 86qL 27 97qL 54 36 28qL 27