Question 9.2: Built to Last GOAL Calculate a compression due to tensile st...

Built to Last

GOAL Calculate a compression due to tensile stress, and maximum load.

PROBLEM A vertical steel beam in a building supports a load of 6.0 × 10^4 N. (a) If the length of the beam is 4.0 m and its cross-sectional area is 8.0 × 10^{-3} m², find the distance the beam is compressed along its length. (b) What maximum load in newtons could the steel beam support before failing?

STRATEGY Equation 9.3 pertains to compressive stress and strain and can be solved for ΔL, followed by substitution of known values. For part (b), set the compressive stress equal to the ultimate strength of steel from Table 9.3. Solve for the magnitude of the force, which is the total weight the structure can support.

stress = elastic modulus × strain            [9.3]

Table 9.3 Ultimate Strength of Materials
Material Tensile Strength

(N/m²) 

Compressive Strength

(N/m²)

Iron 1.7 × 10^8 5.5 × 10^8
Steel 5.0 × 10^8 5.0 × 10^8
Aluminum 2.0 × 10^8 2.0 × 10^8
Bone 1.2 × 10^8 1.5 × 10^8
Marble 8.0 × 10^7
Brick 1 × 10^6 3.5 × 10^7
Concrete 2 × 10^6 2 × 10^7
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(a) Find the amount of compression in the beam.
Solve Equation 9.5 for ΔL and substitute, using the value of Young’s modulus from Table 9.2:

\frac{F}{A}=Y \frac{\Delta L}{L_0}

\Delta L=\frac{F L_0}{Y A}=\frac{\left(6.0  \times  10^4  N \right)(4.0  m )}{\left(2.0  \times  10^{11} Pa \right)\left(8.0  \times  10^{-3}  m ^2\right)}

=1.5 \times 10^{-4}  m

(b) Find the maximum load that the beam can support.
Set the compressive stress equal to the ultimate compressive strength from Table 9.3, and solve for F:

\frac{F}{A}=\frac{F}{8.0  \times  10^{-3}  m ^2}=5.0 \times 10^8  Pa

F=4.0 \times 10^6  N

REMARKS In designing load-bearing structures of any kind, it’s always necessary to build in a safety factor. No one would drive a car over a bridge that had been designed to supply the minimum necessary strength to keep it from collapsing.

Table 9.2 Typical Values for the Elastic Modulus
Substance Young’s Modulus (Pa) Shear Modulus (Pa) Bulk Modulus (Pa)
Aluminum 7.0 × 10^{10} 2.5 × 10^{10} 7.0 × 10^{10}
Bone 1.8 × 10^{10} 8.0 × 10^{10}
Brass 9.1 × 10^{10} 3.5 × 10^{10} 6.1 × 10^{10}
Copper 11 × 10^{10} 4.2 × 10^{10} 14 × 10^{10}
Steel 20 × 10^{10} 8.4 × 10^{10} 16 × 10^{10}
Tungsten 35 × 10^{10} 14 × 10^{10} 20 × 10^{10}
Glass 6.5–7.8 × 10^{10} 2.6–3.2 × 10^{10} 5.0–5.5 × 10^{10}
Quartz 5.6 × 10^{10} 2.6 × 10^{10} 2.7 × 10^{10}
Rib Cartilage 1.2 × 10^7
Rubber 0.1 × 10^7
Tendon 2 × 10^7
Water 0.21 × 10^{10}
Mercury 2.8 × 10^{10}

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