Chapter 4-b Axially Loaded Members

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1 CIVL 222 STRENGTH OF MATERIALS Chapter 4-b Axially Loaded Members

2 AXIAL LOADED MEMBERS Today s Objectives: Students will be able to: a) Determine the elastic deformation of axially loaded member b) Apply the principle of superposition for total effect of different loading cases c) Deal with compatibility conditions d) Use force method of analysis. In-class Activities: Applications Saint-Venant s principle Elastic deformation in axially loaded member Principle of superposition Compatibility conditions Force method of analysis

3 APPLICATIONS Most concrete columns are reinforced with steel rods; and these two materials work together in supporting the applied load. Are both subjected to axial stress?

4 St. Venant s Principle The stresses and strains in a body at points that are sufficiently remote from points of application of load depend only on the static resultant of the loads and not on the distribution of the loads.

7 ζ = ELASTIC DEFORMATION OF AN AXIALLY LOADED MEMBER P (x) A (x) and ε = dδ dx Provided these quantities do not exceed the proportional limit, we can relate them using Hooke s Law, i.e. ζ = E ε P (x) A (x) δ = L 0 = E ( dδ ) dx P (x) dx A (x) E dδ = P (x) dx A (x) E Fig. 4-4 Fig. 4-2

8 EXAMPLE 1

9 EXAMPLE 1 (continued)

10 PRINCIPLE OF SUPERPOSITION It can be used to simply problems having complicated loadings. This is done by dividing the loading into components, then algebraically adding the results. It is applicable provided the material obeys Hooke s Law and the deformation is small. Fig If P = P 1 + P 2 And d d 1 d 2 Then the deflection at location x is sum of two cases, ie δ x = δ x1 + δ x2

11 COMPATIBILITY CONDITIONS When the force equilibrium condition alone cannot determine the solution, the structural member is called statically indeterminate. In this case, compatibility conditions at the constraint locations shall be used to obtain the solution. For example, the stresses and elongations in the 3 steel wires are different, but their displacement at the common joint A must be the same. Fig. 4-51

12 COMPATIBILITY CONDITIONS EXAMPLE The distributed loading w= 46 kn/m is supported by three suspender bars. AB and EF are made from aluminum and CD is made from steel. If each bar has a cross-sectional area of 450 mm 2, determine the forces in each bar when the distributed loading is applied. E st =200GPa, E al = 70 Gpa.

13 FORCE METHOD OF ANALYSIS It is also possible to solve statically indeterminate problem by writing the compatibility equation using the superposition of the forces acting on the free body diagram. Fig. 4-16

14 EXAMPLE

15 EXAMPLE (continued)

Treat the additional support as redundant and apply the principle of superposition.

16 THERMAL STRESS A temperature change results in a change in length or thermal strain. There is no stress associated with the thermal strain unless the elongation is restrained by the supports. Treat the additional support as redundant and apply the principle of superposition. T T thermal T P L PL AE T L 0 expansion coef. 0 P PL AE The thermal deformation and the deformation from the redundant support must be compatible. T P 0 P AE T P ET A

17 Static Indeterminacy Structures for which internal forces and reactions cannot be determined from statics alone are said to be statically indeterminate. A structure will be statically indeterminate whenever it is held by more supports than are required to maintain its equilibrium. Redundant reactions are replaced with unknown loads which along with the other loads must produce compatible deformations. Deformations due to actual loads and redundant reactions are determined separately and then added or superposed. L R 0

SOLUTION: Consider the reaction at B as redundant, release the bar from that support, and solve for the displacement at B due to the applied

18 Example Determine the reactions at A and B for the steel bar and loading shown, assuming a close fit at both supports before the loads are applied. SOLUTION: Consider the reaction at B as redundant, release the bar from that support, and solve for the displacement at B due to the applied loads. Solve for the displacement at B due to the redundant reaction at B. Require that the displacements due to the loads and due to the redundant reaction be compatible, i.e., require that their sum be zero. Solve for the reaction at A due to applied loads and the reaction found at B.

19 Example SOLUTION: Solve for the displacement at B due to the applied loads with the redundant constraint released, P A L L 0 A L 2 i 2 P i 2 L 3 P L i i A E i P L m E 9 3 N A m P 4 A Solve for the displacement at B due to the redundant constraint, P L δ 1 A 1 1 R P L 2 i 2 R P ili A E m i i B m 2 A E R B 6 m 2 N 6 m 2

Example Require that the displacements due to the loads and due to the redundant reaction be compatible, 1.12510 E R B L R 57710 0 9 3 3 1.

20 Example Require that the displacements due to the loads and due to the redundant reaction be compatible, E R B L R E N 577kN R B 0 Find the reaction at A due to the loads and the reaction at B F R A y 0 R 323kN A 300kN 600kN 577kN R R A B 323kN 577kN

Question # 1: Coaxial tube and core An aluminum tube (1) encases a brass core (2). The two components are bonded together to comprise an axial member that is subjected to a downward force of 30 kn.

21 Question # 1: Coaxial tube and core An aluminum tube (1) encases a brass core (2). The two components are bonded together to comprise an axial member that is subjected to a downward force of 30 kn. Tube (1) has an outer diameter of D= 30 mm and an inner diameter of d= 22 mm. The elastic modulus of the aluminum is 70 GPa. The brass core (2) has a diameter of D= 22 mm and an elastic modulus of 105 GPa. Compute the normal stresses in tube (1) and core (2).

22 Question # 2: Coaxial tube and core

23 Question # 3: Compound bars with temperature change The compound bar, composed of three segments shown; bronze segment [A= 2000 mm 2, E= 83 GPA: =19 x 10-6 mm/mm/c], aluminum segment [A= 1400 mm 2, E= 70 GPA: =23 x 10-6 mm/mm/c], and steel segment [A= 800 mm 2, E= 200 GPA: =11.7 x 10-6 mm/mm/c], is initially stress-free. Compute the stress in each material if the temperature drops 25C. Assume that the walls do not yield.

Question # 4: Coaxial bars with temperature change A rectangular bar 30-mm wide and 24-mm thick made of aluminum [E= 70 GPA: =23 x 10-6 mm/mm/c] and two rectangular copper bars 30-mm wide and 12-mm

24 Question # 4: Coaxial bars with temperature change A rectangular bar 30-mm wide and 24-mm thick made of aluminum [E= 70 GPA: =23 x 10-6 mm/mm/c] and two rectangular copper bars 30-mm wide and 12-mm thick [E= 120 GPA: =16 x 10-6 mm/mm/c] are connected (in longitudinal direction) by two smooth 11-mm diameter pins. When the pins are initially inserted into the bars, both the copper and aluminum bars are stress free. After the temperature of the assembly has increased by 65C. (a) Determine the internal axial force in the aluminum bar. (b) Determine the normal strain in the copper bars. (c) Determine the shear stress in the 11-mm diameter pins.

25 READING QUIZ 1) The stress distributions at different cross sections are different (see figure below). Fig. 4-1 However, at locations far enough away from the support and the applied load, the stress distribution becomes uniform. This is due to A)Principle of superposition C) Poisson s effect B) Inelastic property D) Saint Venant s Principle

26 READING QUIZ 2) The principle of superposition is valid provided that a) The loading is linearly related to the stress or displacement b) The loading does not significantly change the original geometry of the member c) The Poisson s ratio v 0.45 d) Young s Modulus is small A)a, b and c C) a and b only B) a, b and d D) All

CONCEPT QUIZ 1) The assembly consists of two posts made from material 1 having modulus of elasticity of E 1 and a cross-sectional area A 1 and a material 2 having modulus of elasticity E 2 and

27 CONCEPT QUIZ 1) The assembly consists of two posts made from material 1 having modulus of elasticity of E 1 and a cross-sectional area A 1 and a material 2 having modulus of elasticity E 2 and crosssectional area A 2. If a central load P is applied to the rigid cap, determine the force in each post. The support is also rigid. Let r = E 1 A 1 E 2 A 2 Prob r A) P 1 = ( ) P 2r P 2 = ( ) P 2r B) P 1 = ( ) P 2r + 1 r P 2 = ( ) P 2r + 1 C) P 1 = r P P 2 = (2r-1) P D) P 1 = r (r+1) P P 2 = (r+1) P

MECHANICS OF MATERIALS

MECHANICS OF MATERIALS Third CHTR Stress MCHNICS OF MTRIS Ferdinand. Beer. Russell Johnston, Jr. John T. DeWolf ecture Notes: J. Walt Oler Texas Tech University and Strain xial oading Contents Stress & Strain: xial oading Normal