A criterion that identifies the failure mode as well as the load is valuable because
Answer: A
Matrix cracking may be tolerable while fibre failure is not, and each calls for a different change to the layup. Mode-based criteria such as Hashin's are preferred for that reason despite being more involved than a single interactive expression.
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Free-edge stresses in a laminate are significant because they
Answer: B
Mismatch between adjacent plies produces interlaminar stresses that classical laminate theory cannot represent, since it assumes plane stress. They peak at a free edge and are a common site for delamination initiation.
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Ply drop-offs, where plies terminate to taper a laminate, are fatigue-sensitive because
Answer: C
Ending a ply forces its load into the neighbours over a short distance, producing out-of-plane stress at the drop. Staggering the drops and tapering gradually are the standard measures, and the detail is a classic delamination origin.
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A laminate has A-matrix terms giving an in-plane modulus of \( 52 \) GPa over \( 2 \) mm thickness and \( 100 \) mm width. Its axial stiffness is
Answer: D
Correct answer: 10.4 MN
Axial stiffness is the modulus times the cross-sectional area: \[ EA = 52 \times 10^9 \times (0.002 \times 0.1) \] \[ = 52 \times 10^9 \times 2 \times 10^{-4} = 1.04 \times 10^7\ \mathrm{N} = 10.4\ \mathrm{MN}. \] The commonest slip is to work the area in square millimetres while leaving the modulus in gigapascals, which puts the answer out by a factor of a million.
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A quadratic failure criterion returns a value of \( 0.64 \) for a given stress state. The margin of safety on load, for a linear criterion in stress, is
Answer: A
Correct answer: 0.25
A quadratic criterion is quadratic in stress, so scaling the load by a factor \( R \) scales the criterion value by \( R^2 \). Failure occurs at unity: \[ R^2 (0.64) = 1 \quad\Rightarrow\quad R = \frac{1}{0.8} = 1.25, \] and the margin of safety is \( R - 1 = 0.25 \). Taking \( 1 - 0.64 = 0.36 \) as the margin is the trap, and it would overstate the strength remaining.
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Assertion (A): Free-edge stresses cannot be predicted by classical laminate theory.
Reason (R): The theory assumes a state of plane stress in each ply.
Select the correct answer from the options below.
Answer: B
Correct answer: Both A and R are true, and R is the correct explanation of A
Both are true and the reason is exactly why. Classical laminate theory assumes plane stress in every ply, so the through-thickness stress components are zero by construction and cannot be computed. At a free edge those very components are large, arising from the mismatch in Poisson’s ratio and shear coupling between adjacent plies - which is why edge delamination needs a three-dimensional analysis.
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Assertion (A): Ply drop-offs are staggered rather than terminated together.
Reason (R): Terminating several plies at one station concentrates the interlaminar stress there.
Select the correct answer from the options below.
Answer: C
Correct answer: Both A and R are true, and R is the correct explanation of A
Both are true and the reason is the design rule. Where a ply terminates, the load it carried must be shed into its neighbours through interlaminar shear, and ending several plies at the same station concentrates that shear and creates a resin pocket. Staggering the drop-offs spreads the transfer over a length and keeps the interlaminar stress within what the matrix can carry.
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Consider the following statements about classical laminate theory:
1. It assembles ply stiffnesses transformed to the laminate axes.
2. It produces the A, B and D matrices.
3. It predicts interlaminar stresses at a free edge.
Which of the statements given above is or are correct?
Answer: D
Correct answer: 1 and 2 only
Statements 1 and 2 describe what the theory does: transform each ply stiffness to the laminate axes and assemble them into A, B and D. Statement 3 is beyond it, for the reason the assertion question makes explicit - the plane stress assumption sets the interlaminar components to zero, so free-edge stresses need a three-dimensional model.
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Consider the following statements about composite failure criteria:
1. Maximum stress criteria check each direction independently.
2. Interactive criteria such as Tsai-Wu use a single quadratic expression.
3. Interactive criteria identify which failure mode occurred.
Which of the statements given above is or are correct?
Answer: A
Correct answer: 1 and 2 only
Statements 1 and 2 are correct. Statement 3 is the trade-off between the two families: a maximum stress criterion checks each component separately, so it tells you which one was exceeded, while an interactive criterion such as Tsai-Wu combines them into one quadratic expression that predicts failure more accurately but says nothing about the mode.
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Match each analysis concern in List I with the appropriate treatment in List II.
List I
A. In-plane laminate stiffness
B. Free-edge delamination
C. Ply drop-off
D. First-ply failure
List II
1. A staggered, gradual taper
2. Classical laminate theory and the A matrix
3. A ply-level failure criterion
4. Interlaminar stress analysis beyond plane stress
Select the correct match.
Answer: B
Only the first is handled by the classical theory alone; the rest need either a detailed three-dimensional treatment or a design rule. That gap between the convenient theory and the real failure modes is why composite design leans so heavily on test.
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