Tiêu đề 11 Statically Indeterminate Members.pdf ✅

PSAD 11: Statically Indeterminate Members Statically Indeterminate Members are members whose support conditions produce more unknowns (reactions) than available equations of equilibrium. For problems involving such members, additional equations can be acquired by analyzing its compatibility conditions. Compatibility conditions are equations that aid in relating a member’s displacement with forces, such as elongation or contraction due to axial loads, and twisting due to torque. 1. Axially Loaded Members Solving statically indeterminate axially loaded members requires the axial deformation equation (Eq. 11-1). Thus, information on the material property and dimensions is necessary. δ = PL AE (Eq. 11 − 1) Below are common problems that demonstrate the use of compatibility equations. Example 1. Consider a given load FB, applied at point B. Bar AB is fully restrained at its ends, and assume no settlement occurs at the supports. Find the reactions. Unknowns (2): Reactions RA and RC Equilibrium Equation (1): ∑FH = 0 | RA + RC = FB Note: initial assumptions on the direction of the reactions are made. By observing that the total length of the bar does not change, the assumed elongation of AB must be cancelled by the assumed contraction of BC. Compatibility Equation (1): δTotal = δAB − δBC = 0 ↓ RALAB AABEAB − RCLBC ABCEBC = 0 Example 2. A bronze bar is rigidly fitted inside an aluminum tube. It resists a load P on its top, while it is fixed on its bottom. Find the force exerted by the bar and tube. Unknowns (2): Forces FBr (bronze bar) and FAl (aluminum tube) Equilibrium Equation (1): ∑Fv = 0 | FBr + FAl = P By assuming that the load P is equally distributed over the entire area, the deformation of the bar and tube must be equal. Compatibility Equation (1): δBr = δAl ↓ FBrLBr ABrEBr = FAlLAl AAlEAl

2. Torque Loaded Members Solving statically indeterminate members with applied torque has the exact principles with its axial counterpart; the main difference lies in the manner of deformation. Torque causes twisting on members, which can be quantified using the angle of twist equation (Eq. 11 – 2). By relating the angle of twists of different parts of the member, compatibility equations can be formed. φ = TL JG (Eq. 11 − 1) Below are common problems that demonstrate the use of compatibility equations. Example 1. Consider a steel shaft, reinforced with a bronze core, subjected to torque P at its free end. The member is fully restrained on its other end. Find the resisting torque of the shaft and core. Unknowns (2): Torque TBr (bronze core) and TSt (steel shaft) Equilibrium Equation (1): ∑M = 0 | TBr + TSt = P Since the two materials are rigidly connected, an applied torque causes them to have the same angle of twist. Compatibility Equation (1): φBr = φSt ↓ TBrLBr JBrGBr = TStLSt JStGSt Example 2. A bar fixed at both ends is subjected to a torque at B. Find the reactions at the support. Unknowns (2): Reactions TA and TC Equilibrium Equation (1): ∑M = 0 | TA + TC = TB Note: initial assumptions on the direction of the reactions are made. Since the bar is fully restrained, the angle of twist of A with respect to C must be zero. This can be seen in the image below. Compatibility Equation (1): φAB − φBC = φA/B = 0 ↓ TABLAB JABGAB − TBCLBC JBCGBC = 0