Explanation Given information: The cross-sectional area ( A ) of the column is 10 in . 2 . The moment of inertia ( I z ) of cross-sectional of the column is 273 in . 4 The strains at A and B are ε A = − 400 × 10 − 6 in ./in . and ε B = − 300 × 10 − 6 in ./in . . The modulus of elasticity ( E ) of the beam is 29 × 10 6 psi . Calculation: Calculate the stress σ A from the strain guage at A using the relation: σ A = E ε A (1) Substitute 29 × 10 6 psi for E and − 400 × 10 − 6 in ./in . for ε A in Equation (1). σ A = 29 × 10 6 psi × ( − 400 ) × 10 − 6 in ./in . = − 11 , 600 psi Calculate the stress σ B from the strain gage at A using the relation: σ B = E ε B (2) Substitute 29 × 10 6 psi for E and − 300 × 10 − 6 in ./in . for ε B in Equation (2). σ B = 29 × 10 6 psi × ( − 300 ) × 10 − 6 in ./in . = − 8 , 700 psi The distance ( c ) between the neutral axis and the outermost fiber is 5 in . Show the concentric force on the cross-section of the column as follows: F = P + Q Show the bending couple on the column as follows: M = 6 P − 6 Q Show the expression for the stress at point A as follows: σ A = − F A + M c I (3) Substitute ( P + Q ) for F , 10 in . 2 for A , ( 6 P − 6 Q ) for M , 5 in . for c , and 273 in . 4 for I in Equation (3). σ A = − P + Q 10 + ( 6 P − 6 Q ) × 5 273 = − P + Q 10 + 30 P − 30 Q 273 (4) Show the expression for the stress at point B as follows: σ B = F A − M c I (5) Substitute ( P + Q ) for F , 10 in . 2 for A , ( 6 P − 6 Q ) for M , 5 in

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Author: BEER

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Axial Load

When a bar of the uniform cross-section is acted by a load, such that the load acts along the longitudinal axis of the bar, such kinds of loadings are known as axial loadings. In general terms, a load is an external force acting on a component. An axial …

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Chapter 4.7, Problem 124P

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The magnitude of the load P and Q.

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Chapter 4.7, Problem 123P

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The magnitude of the load P and Q .

Solution Summary: The author explains the magnitude of the load P and Q, the moment of inertia, and the modulus of elasticity.

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