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Bài giảng môn Sức bền vật liệu 1 tóm tắt nhưng đầy đủ các nội dung cơ bản và ví dụ trọng tâm. Sức bền vật liêu 1 là môn cơ sơ ngành quan trọng của khối nghành xây dựng. Vì vậy hy vọng với bài giảng này sẽ giúp các bạn sinh viên hệ thống lại được nhưng kiến thức cần nắm của môn học này
Chương 6 CHUYEÅN VÒ DAÀM CHÒU UOÁN 6.1. Chuyển vị của dầm chịu uốn (1) 1. Khái niệm chung Đường đàn hồi: Đường cong của trục dầm sau khi chịu uốn Trọng tâm mặt cắt ngang của dầm K - trước biến dạng K’ – sau biến dạng KK’ – chuyển vị của trọng tâm mặt cắt ngang Biến dạng bé: u(z)<<v(z) v(z) => độ võng: y(z)=> B F L K K’ K K’ z v(z) u(z) KK’ v(z) - chuyển vị đứng u(z) - chuyển vị ngang Độ võng của dầm chịu uốn là chuyển vị theo phương thẳng đứng của trọng tâm mặt cắt ngang University of Architechture [...]... xoay trên dầm thực University of Architechture Bài tập - Ví dụ 6. 1.1 (1) P Ví dụ 6. 2: Cho dầm có liên kết và chịu tải trọng như hình vẽ Xác định độ võng tạitiết diện đặt lực P A B L/2 L/2 Giải: Bước 1: Vẽ biểu đồ mô men uốn nội lực Bước 2: Xác định liên kết trên dầm giả tạo, tải trọng giả tạo, M>0 nên tải trọng giả tạo hướng xuống M PL qgt EI x 4 EI x Bước 3: Xác định nội lực trên dầm giả tạo.. .6. 1 Chuyển vị của dầm chịu uốn (10) Các bước thực hiện: Vẽ biểu đồ mô men uốn trên dầm thực Chia tung độ biểu đồ cho độ cứng EI để có trị số của tải trọng giả tạo Nếu Mx>0 thì qgt z = 3a => y1(z=a) = 0 y3(z=3a) = 0 Từ hai phương trình độ võng y1(z) và y3(z), áp dụng điều kiện biên: 5qa 4 yo 24EI x Từ đó tính được: qa 3 o 6EI x 7qa 4 y C y 2 (z 2a) 24EI x qa 3 D y'3 (z 3a) 6EI x University of Architechture 6. 2 Bài toán siêu tĩnh (1) July 2009 University of Architechture 6. 2 Bài toán siêu tĩnh (2) July 2009... z ) y0 0 z EI ' a ' i 1 ' i z2 z3 z4 z5 ' M 0 2! Q0 3! q0 4! q0 5! University of Architechture 6. 1 Chuyển vị của dầm chịu uốn (13) ' y0 ,0 , M 0 , Q0 , q0 , q0 , gọi là các thông Các thông số số ban đầu và được xác định từ điều kiện biên Chú : Chiều dương của mô men tập trung, lực tập trung, tải trọng phân bố như hình vẽ Nếu liên kết giữa hai đoạn thứ (i) và (i+1)... tải phân bố, ta nhận được công thức truy hồi của hàm độ võng (hàm độ võng trên đoạn thứ i+1 được xác định khi biết hàm độ võng trên đoạn thứ i) University of Architechture 6. 1 Chuyển vị của dầm chịu uốn (12) Khi độ cứng của dầm EI=const trên cả chiều dài yi 1 ( z ) yi ( z ) ya a ( z a) 1 EI Với ( z a) 2 ( z a )3 ( z a) 4 ( z a)5 M a Qa qa q 'a 2! 3!... 0 Ví dụ University of Architechture Bài tập – Ví dụ 6. 4.2 (1) q Ví dụ 6. 4. 2: Dùng phương pháp thông A số ban đầu, xác định độ võng tại C và góc xoay tại D của dầm chịu tải trọng như hình vẽ 1 a 2 Lập bảng thông số ban đầu Tìm yC => hàm độ võng y2 Tìm D => hàm góc xoay y3’ D C B 2 3 VB a a VD 3a 1 Xác định phản lực 9 VD qa 4 2 M=qa 2a Bài giải: 11 VB qa 4 P=4qa z=0 z=a z = 2a y0 0 0 0 M0... M a 0 Qa P qa 0 , qa 0 University of Architechture Ví dụ 6. 1.2 (2) Công thức truy hồi: yi 1 ( z ) yi ( z ) ya a ( z a) 1 EI 5 ( z a)2 ( z a )3 ( z a)4 ' ( z a) M a 2! Qa 3! qa 4! q a 5! z=0 Xét đoạn 1(AB ): 0 ≤ z ≤ a qz 4 y1 (z) y o o z 24EI x Xét đoạn 2 (BC ): a ≤ z ≤ 2a z=a z = 2a y0 0 0 0 M0 0 Q0 0 q0 q , q0 0 ya . Chương 6 CHUYEÅN VÒ DAÀM CHÒU UOÁN 6. 1. Chuyển vị của dầm chịu uốn (1) 1. Khái niệm chung Đường đàn hồi: Đường cong của trục dầm sau khi chịu uốn Trọng tâm mặt cắt ngang của dầm K -. 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– chuyển vị của trọng tâm mặt cắt ngang Biến dạng b : u(z)<<v(z) v(z) => độ võng: y(z)=> B F L K K’ K K’ z v(z) u(z) KK’ v(z) - chuyển vị đứng u(z) - chuyển vị ngang Độ võng của dầm