Chapter 12

7.1 (a) Equations 7.4a and 7.4b are expressions for normal (σ′) and shear (τ′) stresses, respectively, as a function of the applied tensile stress (σ) and the inclination angle of the plane on which these stresses are taken (θ of .4). Make a plot on which is presented the orientation parameters of these expressions (i.e., cos 2 θ and sin θ cos θ) versus θ.

tz, 1992

When designing a structure and its components, the designer must decide on the appropriate structural model. The choice of the model effects: • the analysis of the structure, which is aimed at the determination of the stress (internal forces and moments), and • the calculation of cross section resistance Thus a model implies the use of a method of analysis combined with a method of cross section resistance calculation. There are several possible combinations of methods of analysis and methods of cross-section calculation, for the ultimate limit state, involving either an elastic or plastic design approach; the possible combinations are: a) Plastic-plastic model: This is related to plastic design of structures. Full plasticity may be developed within cross-sections, so that plastic hinges can form. These have suitable moment rotation characteristics giving sufficient rotation capacity for the formation of a plastic mechanism, as the result of moment redistribution in the structure. b) Elastic-plastic model: For structures composed of sections which can achieve their plastic resistance, but have not sufficient rotation capacity to allow for a plastic mechanism in the structure. The stresses from the elastic analysis are compared with the plastic section capacity. c) Elastic-elastic model: When the cross section of a structure cannot achieve their plastic capacity both analysis and verification of cross section conducted elastically. Elastic analysis of reinforced concrete beams gives reasonable results up to working loads. Beyond working loads the elastic analysis is not applicable because of the non linearity in the stress-strain curves for the materials and the cracks which develop in concrete. When beam is loaded beyond working loads, plastic hinges form at certain locations. On further loading of the beam, bending moments do not increase beyond the ultimate moment capacities of the these sections, however, rotations at the plastic hinges keep on increasing. A redistribution of moments takes place, the moment now being received by the less stressed sections. The rotation at a plastic hinges keeps on increasing with out any increase in the moment until the ultimate rotation capacity is reached beyond which the section collapses 4.2 Non-Linear Analysis of Indeterminate Structures The linear elastic analysis of structures is based on the assumption that there is a linear relationship between the stress and strain in a member, i.e. Where: E is the elastic modulus (young's modulus) F 0 6 5 is strain

13-1. When a mass of 500 g is hung from a spring, the spring stretches 3 cm. What is the spring constant? [ m = 0.500 kg; x = 0.03 m, F = W = mg ] F =-kx; 2 (0.50 kg)(9.8 m/s) 0.03 m F k x = = ; k = 163 N/m 13-2. What will be the increase in stretch for the spring of Problem 13-1 if an additional 500-g mass is hung blow the first? [ F = W = mg ] 2 (0.500 kg)(9.8 m/s) 163 N/m F x k ∆ ∆ = = ; ∆x = 3.00 cm 13-3. The spring constant for a certain spring is found to be 3000 N/m. What force is required to compress the spring for a distance of 5 cm? F = kx = (3000 N/m)(0.05 m); F = 150 N 13-4. A 6-in. spring has a 4-lb weight hung from one end, causing the new length to be 6.5 in. What is the spring constant? What is the strain? [ ∆x = 6.5 in. – 6.0 in. = 0.50 in. ] (4 lb) ; 0.5 in. F k x = = k = 8.00 lb/in. 0.50 in. 6.00 in. L Strain L ∆ = = ; Strain = 0.0833 13-5. A coil spring 12 cm long is used to support a 1.8-kg mass producing a strain of 0.10. How far did the spring stretch? What is the spring constant? ; () (12.0 cm)(0.10) o L Strain L L strain L ∆ = ∆ = = ; ∆L = 1.20 cm 2 (1.8 kg)(9.8 m/s) ; 0.0120 m F k L ∆ = = ∆ k = 1470 N/m 174