Question

Slope Deflection Method - Frames Problem 1. Given: 8 k 2 k/ft 15 ft E constant Support A is a pin Support C is fixed Support D is a roller 12 ft (A) HAND CALCULATIONS Analyze the nonsway frame by the Slope-Deflection Method. Evaluate all the reactions e Consider Special-Case Spans and Statically Determinate (CE 304) Spans where appropriate Note: selected answers are given on p. 687 of your text for Problem 11-13 (B) SAP 2000 CALCULATION Perform a SAP2000 analysis for this problem and submit the following screenshots directly from the program (you may take a screenshot and paste the images on a Word document) Structure with loads acting on it (if you can't show all the loads in one image, you can show the loads acting separately) Support Reactions Shear diagram (show values and don't forget to use a negative sign for the scaling factor) Moment diagram (show values and don't forget to plot moment on the compression side) . Deflected shape (C) COMPARISONS Print out this sheet and fill in the values in the table below, after you complete (A) and (B) Table 1, Comparison ofresults for Problem l Engineering Value Ax (kips) Ay (kips) Cx (kips) Cy (kips) Mc (i.e., McB) (kip-ft) Dy (kips) HAND Calc. SAP 2000 Calc.*Percent rror

Analyze the nonsway frame by the Slope-Deflection Method.

•Evaluate all the reactions

•Consider Special-Case Spans and Statically Determinate (CE 304) Spans where appropriate

Answer 1

* In this problem theta_A, theta_B, & theta_C are need to fixed. Thus the required equations to evaluate theta_A, theta_B, & theta_C are obtained by considering the equilibrium of joint 'B' Step 1: calculation of fixed end moments: - (1) For position AB M F_AB = wL^2/12 n = 2 times 15^2/12 = 37.5 k. tt Therefore M_FBA = -wL^2/12 = -37.5 ktt (2) For position BD M F_BD = wL^2/12 = 24k.tt MFDB = -wL^2/12 = -24 k.tt *For position BF (3) M_fBC = M_FCB = 0 \r\n Step 2: Slope Deflection Equations: - (1) For Span AB: - M_AB = M_FAB + 2EI/l(20_A + theta_B) M_SB = 37.5 + 2EI/15(20_A + theta_B) M_AB = 37.5 + 4EI theta_A/15 + 2EI theta_B/15 rightarrow (a) M_BA = M_FBA + 2EI/L(theta_A + 2theta_B) M_BA = -37.5 + 2EI/15(theta_A + 2theta_B) M_BA = -37.52EI theta_A/15 + 4EI theta_B/15 rightarrow (b) For span BC M_BC = M_FBC + 2EI/2(2 theta_B + Theta_C) theta_C = 0 M_BC = 0 + 2EI/12(2 theta_B + 0) M_BC = 4EI theta_B/12 M_BC = EI theta_B/3 rightarrow (c) M_CB = 0 + 2EI/12(theta_B + 2 theta_C)