In this question about combined tensile strength, all of the tensile bars are of the same material. But what if the case that they are made of different materials?

Say we have $200$ bars of $A=10 (\rm mm^2)$.

  • $100$ bars of steel with $E=200 \, (\rm {GPa})$, at tensile limit, $\sigma=200\, (\rm {MPa})$, $\eta=20 (\%)$.
  • $100$ bars of aluminum with $E=60 \, (\rm {GPa})$, at tensile limit, $\sigma=80\, (\rm {MPa})$, $\eta=10(\%)$.

What is total strength (the maximum force capacity) of the bars? Assume that one end is fixed and the other is a rigid body moving in longitudinal direction, so that all bars have the same change in length.

Note: We have to assume that the material is of plasticity, and will not hardening after yield. (The tensile limit is the same as the yield strength.) This may be beyond the text book of elasticity theory.


2 Answers 2


My answer of that question covers this a little bit, and the link fully explains it, but I can go into more detail so you don't have to wade through that giant PDF. I'll link it again here at the bottom, because I'm going to reference equations in it for this answer.

As you might imagine, the total strength of a composite such as the example above is a combination of the two materials. The proportions in which each material contribute to the overall strength are related to the volume fractions of the constituent materials. Because in this case, the lengths of the bars is always equal and the bars are of constant area along their length, we can simplify the volumes to areas. This means that we can compute the elastic modulus using a version of equation 3.1 in the book.


Here, St and Al are steel and alumin(i)um, respectively, the c subscript refers to the composite, and A is the area fraction, not the total area. So in the example problem, $A_{St}=0.5$ and $A_{Al}=0.5$.

To calculate the stress in the bars, we simply use the definition of stress, given that we can measure the strain and we know the strain is the same in all bars.

$$\sigma_c=\epsilon E_c=\epsilon E_{St}A_{St}+\epsilon E_{Al}A_{St}=\sigma_{St} A_{St}+\sigma_{Al} A_{Al}$$

At this point, though, it's important to note that we are seeing different stresses in the bars of different material, which is a natural result of them having the same strain but different elastic moduli. Unless the materials break at the same strain, one of them will fail first, at which point, the same load will be applied to the whole system, but only one type of bar will still be intact to carry that load. The total cross sectional area of the system is reduced as well though.

We need to determine which material breaks first. Because we're applying a known strain rate here, we can find the ultimate strain of each material using the stiffness and ultimate strength. Once we know the strain at which the first material breaks, we plug that strain into the second equation above to see the total stress in the composite at this elongation. However, because the same force is being applied over a smaller area, the strength of the composite after the breakage of the first material (in this case, the steel) is given by

$$\sigma_{TS,c}=\sigma_{Al} A_{Al}$$

We can compare this value of tensile strength to the value of tensile strength just before the steel breaks. If the value before steel fracture is greater, then the steel fracture results in the fracture of the entire system. If the value after steel fracture is greater, the system will continue to carry load until it reaches the tensile strength of just the aluminum bars.

Mechanical Properties of Materials, David Roylance, 2008.

  • 1
    $\begingroup$ Correct, if and only if, with the assumption that all the materials follow Hooke's law till break. $\endgroup$
    – PdotWang
    Feb 11, 2015 at 18:17
  • $\begingroup$ I noticed that you have plastic theory in the PDF notes. I need to take some time to study it. Thanks. $\endgroup$
    – PdotWang
    Feb 11, 2015 at 18:20
  • 1
    $\begingroup$ That's true, this is based in linear-elastic fracture mechanics, not elastic-plastic. EPFM is complicated enough for a single material, applying it to a composite is probably beyond the realistic expectations for this forum, and it's definitely beyond my ability. $\endgroup$ Feb 11, 2015 at 18:27

Short answer: the applied stress will be distributed across the area of all of the bars in accordance with the material stiffness, and the tensile strength of the system (assuming no breakage) is simply the combined tensile strength of all of the materials.

The force capacity of each material is as follows:

  • Steel: $$100\text{bars}\frac{10\text{mm}^2}{\text{bar}}200\text{MPa}=200\text{kN}$$
  • Alumninum: $$100\text{bars}\frac{10\text{mm}^2}{\text{bar}}80\text{MPa}=80\text{kN}$$

So the total force capacity $P_t$ of the entire system is 280kN.

You didn't ask about Young's modulus but it was implied: The composite Young's modulus (and the corresponding stress-strain curve) of the combined system is found by simple area-averaging:




The ratio of load carried by each material is determined by the combination relative stiffness and area, and will be consistent no matter the load (that is, up to a certain point; see the additional comment below):

  • Steel: $$\frac{E_sA_s}{E_{comp}A_t}=76.9\%$$
  • Aluminum: $$\frac{E_aA_a}{E_{comp}A_t}=23.1\%$$

One principle to recognize here is that stiffness attracts load. This is sometimes surprising to students, but it makes sense if you think about it. If you have some load attached to a steel cable, and then add a rubber band attached to the same load, the load in the steel cable will be slightly reduced. However, most of the load will remain in the steel cable (because of its higher stiffness).

However, once the strength capacity of one of the materials is reached, 100% of any additional load will be taken up by the other material (regardless of its stiffness). In this case, the steel will reach its full capacity first (since it is more stiff, and the areas are equal), so the portion of load carried by the steel will go down (and the portion carried by aluminum will go up) beyond the point that the steel has reached full load carrying capacity (note that this ignores any additional strain required to mobilize the excess capacity of the aluminum, which would break the steel):

  • Steel (just before composite breakage): $$\frac{\sigma_sA_s}{\sigma_{comp}A_t}=69.2\%$$
  • Aluminum (just before composite breakage): $$\frac{\sigma_aA_a}{\sigma_{comp}A_t}=30.8\%$$

EDIT: Things are more complicated than this, however, because as load is increased, the area decreases. And as Trevor Archibald points out in his answer, things get even more complicated if you allow for one of the materials to break. In the rubber band example, if the steel is about to break, but the rubber band has excess capacity, you might be able to add a few more paper clips!

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    $\begingroup$ Just to clarify, isn't it really stiffness that attracts load, rather than strength? In your example, the steel cable could just as well be a material that's weaker than the rubber band (maybe yarn), as long as it doesn't stretch very well, right? $\endgroup$
    – Ethan48
    Feb 11, 2015 at 16:51
  • $\begingroup$ Correct. Good catch there; I had the wording totally wrong and it's important to keep that straight. Fixed now. $\endgroup$
    – Rick
    Feb 11, 2015 at 17:15
  • $\begingroup$ I think my question runs a little deeper, and it looks like Trevor Archibald's answer might have addressed it. Imagine if the rods were steel and rubber bands. Even though the rubber bands have some ultimate tensile capacity, they don't contribute their whole capacity to the system since the displacement of the steel bars at their ultimate strength is much less than the displacement of the rubber bands at their ultimate strength, right? $\endgroup$
    – Ethan48
    Feb 11, 2015 at 17:35
  • $\begingroup$ That's right. I added a word about that at the end, and it's a good point. His answer is clearly better. $\endgroup$
    – Rick
    Feb 11, 2015 at 17:40
  • $\begingroup$ I am making a simplifying assumption that the materials are arranged in such a way that you are able to fully mobilize the strength of each (so you "prestress" the rubber band by stretching it before attaching it to the load). $\endgroup$
    – Rick
    Feb 11, 2015 at 18:00

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