Problem 5.26

Welded bimetallic strip subjected to constant temperature increase T.  Properties as shown. Determine

a) the radius of curvature of the strip caused by T.

b) the maximum stress occurring at the interface.

c) the temperature increase that would result in the simultaneous yielding of both elements.

Two solution procedures are presented below:

Procedure 1: The two component strips are isolated, a free thermal deformation is allowed to take place, then forces are applied to reestablish geometric compatibility between the two strips.  Governing equations are formulated, and two solution methods are presented: The force method and the displacement method.

Procedure 2: The strip is fixed against axial deformation, while the temperature increase is applied, and the stresses and end  restraining forces are determined.  The end restraining forces are then applied in reverse sense, and the stresses are determined by treating the strip as a beam of two materials.  The final stresses are obtained by superposition.

I. Procedure 1 - Superposition to the State of Free Thermal Deformation

1) Separate the two strips and allow free thermal deformation

2) Apply self-equilibrating forces and moments at the ends of the strips to re-establish continuity of deformation at the interface

The resulting deformation is as shown below

The preceding formulation assumes that relaxed boundary conditions are acceptable at the strip ends.

Governing Equations

Equilibrium:  Resultant force and resultant moment on a cross-section are zero:

N₁ - N₂ = 0                                     (1) 

M₁ + M₂ - ( h/2)(N₁ + N₂) =  0       (2)

Geometric Continuity: The two strips must have the same curvature k, and the same interface strain e.

Constitutive Equations for strip 1

k = M₁/E₁I₁= 12M₁/E₁h³                (3)

At the interface,

s₁ = N₁/A₁ + M₁h/2I₁ = N₁/h + 6M₁/h²                     (4)

e  =  a₁T  + s₁/E₁ =  a₁T + N₁/E₁h + 6M₁/E₁h²        (5)

Constitutive Equations for strip 2

k  =  M₂/E₂I₂  =  12M₂/E₂h³                                       (6)

At the interface,

s₂ = - N₂/A₂ - M₂h/2I₂ = - N₂/h - 6M₂/h²                 (7)

e  =  a₂T  + s₂/E₂ =  a₂T - N₂/E₂h - 6M₂/E₂h²         (8)

Eq. (1), (2), (3), (5), (6), and (8) form a system of six equations in the six unknowns: N₁, M₁, N₂, M₂, e, and k, and Eq. (4) and (7) serve to compute s₁ and s₂.

Solution by the Force Method

To solve the system of equations (1) to (8) by the force method, we need to eliminate e and k.  To do so, equate the two expressions of k in Eq. (3) and (6), and the two expressions of e in Eq. (5) and (8).  Obtain, after simplification,

M₁/E₁ - M₂/E₂ = 0                                                                          (9)

(a₁ - a₂)T + (N₁/E₁h + N₂/E₂h) + 6(M₁/E₁h²  + M₂/E₂h²)  = 0      (10)

Eq. (1), (2), (9), (10) are now solved for N₁, N₂, M₁, and M₂.  Using Eq. (1) and (9),  Eq. (2) and (10) may be expressed in terms of the two unknowns, N = N₁ = N₂, and M₁.  Obtain, after simplification,

(1 +E₂/E₁)M₁ - hN  =  0                                                  (11)

12(E₂/E₁)M₁ + (1 + E₂/E₁)hN   + (a₁ - a₂)E₂h²T = 0      (12)

Eliminating N,

(1 + E₂/E₁)(Eq. 11) + (Eq. 12) = [(1 + E₂/E₁)²  + 12(E₂/E₁)]M₁   + (a₁ - a₂)E₂h²T = 0

Thus

Eq. (9) yields

Eq.(11) yields

Eq. (3) and (6) yield

and Eq. (4) and (7) yield

Comparing the two expression above, it is seen that the maximum stress in magnitude occurs in the material having the larger modulus of elasticity.

Part (c):  temperature increase that would result in the simultaneous yielding of both elements.

Let s_1Y  =  yield stress of material (1)

Let s_2Y  =  yield stress of material (2)

If the two materials yield simultaneously, the temperature T_Y at which this occurs must satisfy Eq. (17) and (18).  Taking the ratio of s₁/s₂ at yield, we obtain

s_1Y /s_2Y =  (7E₁ + E₂)/(7E₂ + E₁)                              (19)

Assuming Eq.(19) is satisfied by the material properties, T_Y is obtained from either Eq. (17) or (18).

The intent of the question is probably to find the temperature increase at which both strips have yielded.  To do so, assume that yield starts in one of the strips, say in strip (1), at a temperature increase T₁.  For increasing  T,  a plastic zone develops in strip (1) near the interface until the stress at the interface in strip (2) reaches the yield stress for material (2). 

Solution by the Displacement Method

To solve the system of equations (1) to (8) by the displacement method, we solve Eq. (3), (5), (6), and (8) for N₁, M₁, N₂, M₂, in terms of  e and k, and substitute into the equilibrium equations, Eq. (1) and (2).  Equivalently, if we had in mind to use the displacement method, we could have written the four equations mentioned above directly in the desired form, as shown below:

M₁ = E₁h³k/12              (3')

M₂ = E₂h³k/12              (6')

N₁  = E₁h(e - a₁T) -  E₁h²k/2       (5')

N₂  = - E₂h(e - a₂T) -  E₂h²k/2       (8')

Substituting into Eq. (1) and (2) yields, after simplification,

(E₁+ E₂)e  + (E₂- E₁)hk/2 - (E₁a₁ + E₂a₂)T   = 0      (20)

(E₂ - E₁)e  + 2(E₁+ E₂)hk/3 + (E₁a₁ - E₂a₂)T = 0       (21)

It may be verified that eliminating e from Eq. (20) and (21) yields for k the solution found above in Eq. (16).  

Instead of solving for e, and then computing the stresses at the interface, we can solve Eq. (20) and (21) directly for s₁ = E₁(e - a₁T) and s₂ = E₂(e - a₂T), as shown next.

Eq.(20) - Eq.(21) = 2E₁(e - a₁T) - (hk/6)( 7E₁+ E₂) = 0

Eq.(20) + Eq.(21) = 2E₂(e - a₂T) + (hk/6)(7E₂ + E₁) = 0

The solution of these two equations for s₁ = E₁(e - a₁T) and s₂ = E₂(e - a₂T) coincides with that found in Eq. (17) and (18).

II. Procedure 2 - Superposition to the Fixed State

1) Restrain the axial deformation of the strips, while the temperature increase is applied, and determine the stresses and the external restraining forces.

2) Apply in reverse sense the restraining forces, and determine the stresses treating the strips as a beam of two materials.

3) Superimpose states (1) and (2).

1) In the fixed state, the strips have the stresses,

s_1F = -E₁a₁T              (22-a)

s_2F = -E₂a₂T              (22-b)

The axial forces at the ends are

N_1F = s_1F h  = -E₁a₁Th               (23-a)

N_2F = s_2F h  = -E₂a₂Th               (23-b)

2) To treat the strips as a beam of two materials, we first locate the weighted centroid C of the section.  With origin at the interface, the y coordinate of C is given by (E₁h + E₂h)y_c = E₁h²/2 - E₂h²/2, or

y_c = h(E₁ - E₂)/2(E₁ + E₂)            (24)

The axial and bending rigidities of the strip are

(EA) = E₁h + E₂h = (E₁ + E₂)h     (25)

(EI) = E₁I₁ + E₂I₂ = E₁(h - y_c)³/3 + E₂(h + y_c)³/3 - E₂y_c3/3

This simplifies to

EI = h³*(E₁²+14E₁E₂+E₂²)/12(E₁+E₂)       (26)

The strip is subjected at each of its ends to -N_1F and -N_2F.  The resultant force and moment at C, at each of the ends, are

N =  -N_1F -N_2F = (E₁a₁+ E₂a₂)Th         (27)

M = (h/2 - y_c)N_1F - (h/2 + y_c)N_2F           

This simplifies to

M = E₁E₂Th²(a₂-a₁)/(E1+E2)                 (28)

Using stress formulas for a beam of two materials, we have for the stresses at the interface,

s₁₍₂₎ = E₁N/EA + E₁My_c/EI                     (29-a)

s₂₍₂₎ = E₂N/EA + E₂My_c/EI                     (29-b)

3) Superposition of states (1) and (2) yields

s₁ =  s_1F + s₁₍₂₎ = -E₁a₁T + E₁N/EA + E₁My_c/EI         (30-a)

s₂ =  s_2F + s₂₍₂₎ = -E₂a₂T + E₂N/EA + E₂My_c/EI         (30-b)

Substituting for N, M, EA, EI and y_c yields after simplification the same results as found by the previous method.