Using Double Integration Method for the deflection of beams analyze the givenstructure below.12KN/m25KNAВ-4m5m2m+ 2mSOLVE for the SLOPE and DEFLECTION under the 25KN concentrated load

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Answered: structure below. 12KN/m 25KN A -4m 5m…

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Part A Learning Goal: To be able to soive throe-dimensona oquibnum problems using the equations of equilibium The Jshaped member shown in the fgure(Eigua1)s supponed ty a cabie DE and a single journal beaing with a square shalt at A Determine the reaction foroes A, and A. atsupport A ropared to keep the system in eguiltrium. The cylinder has a weight Wa5.00 Ib, and F-1.40 Ibis a verical force applied to the member at C The dimensions of the member are 1.50 n 600 t, and =200 . As with two-dimensional problens, a free-body diagram Is the est step in solving thre-dimensional equabnum problems For the tree-body diagram itis important to identity the appropriate reaction foroes and coupie moments that act in three dimensions At a support, a force arises when translation of the attached member is restricled and a couple momet arises when rotation i prevented For a rigid body to be in equilbrium when sutjected to a force system, both the resultant force and the resultart couple moment actng…
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# 2-14. Normal Strain in rods constrained by walls with misfit At its initial temperature, the bar is too long by a distance of 8 to fit into the gap between the rigid walls, which are separated by a distance L. In order to make it fit tightly, the bar is cooled to a very low temperature so that it shrinks enough to fit easily into the gap. After it is fit in the gap, the temperature of the bar is raised to a temperature AT below the initial temperature and the bar fits tight between the walls. What is the stress in the bar at this temperature AT? The bar has elastic modulus E = 35. 10³ ksi, thermal expansion coefficient a = 6.10-6 /°F and cross-sectional area A = 1 in². Use L = 10 in, 8 = 3 . 10-² in and AT = -200 °F. 0 = ksi L Initial AT=0 L Final AT
B 300 mm ↑ 15 mm 70 KN 70 KN 20 mm 500 mm Steel Aluminum AMIN ATTIV 70 kN 70 KN + Rc A steel rod (20 mm diameter) 500 mm long is connected to two parallel aluminum rods (15 mm diameter) each 300 mm long via a rigid plate as shown. Two 70 kN loads at applied at the end of each aluminum rod. The structure is fixed at both ends resulting in a statically indeterminate system that can be treated as the superposition of two determinate systems as presented above. Note that Rc is the reaction at C. Assume Esteel = 200 GPa and Ealuminum = 70 GPa. a) Determine the reaction force at C, Rc b) Determine the reaction force at A and B and the normal stress in the steel and aluminum rods c) Which rods are in tension and which are in compression?
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structure below. 12KN/m 25KN A -4m 5m 2m-- 2m- SOLV E tor the CLODE SEELEOTION,