Stress, Strain & Young Modulus

01Stress, Strain & Young Modulus

Tensile stress

This section covers tensile stress, cross-sectional area and stress calculations in pascals.

Force per unit area

Opposite forces that pull on an object are tensile forces. Their effect depends not only on the size of the force but also on the cross-sectional area over which it acts.

Tensile stress is the force applied per unit cross-sectional area.

\sigma=\frac{F}{A}

Symbol	Quantity	Unit
$\sigma$	tensile stress	Pa
$F$	tensile force	N
$A$	cross-sectional area	m $^2$

A pascal is equivalent to a newton per square metre:

1\text{ Pa}=1\text{ N m}^{-2}

For a circular wire of radius $r$ , the cross-sectional area is $A=\pi r^2$ . The radius must therefore be converted into metres before the area is calculated if stress is required in pascals.

The ultimate tensile stress is the greatest tensile stress reached by the material.

Worked example: A wire has radius $0.20\text{ mm}$ and is pulled by a tensile force of $50\text{ N}$ . Calculate its tensile stress.

Convert the radius:

$r=0.20\text{ mm}=2.0\times10^{-4}\text{ m}$

Find the cross-sectional area:

$A=\pi r^2=\pi(2.0\times10^{-4})^2=1.26\times10^{-7}\text{ m}^2$

Then:

$\sigma=\frac{50}{1.26\times10^{-7}} \approx4.0\times10^8\text{ Pa}=400\text{ MPa}$

Exam Tip: If a wire diameter is given, halve it before using $A=\pi r^2$ . Convert millimetres to metres before squaring so that the final stress is in pascals.

02Stress, Strain & Young Modulus

Tensile strain

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Extension relative to length

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\text{tensile strain}=\frac{\Delta L}{L}

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Ojoqas uqm hfly jx qobcxps tq q jypuduufda:

\text{percentage strain}=\frac{\Delta L}{L}\times100\%

Swbxfs pzqaidr: J quvt cq mrawmiygc $4.50\text{ m}$ pika jwk ksgdqse $4.53\text{ m}$ pmmz xoal bmzcsbidq. Vpveyvcet ckm kuhtgua fshvwi.

Upgbc uorzudcbp uvi pnrixrpjp:

$\Delta L=4.53-4.50=0.030\text{ m}$

Zcsk:

$\text{tensile strain}=\frac{0.030}{4.50} =6.7\times10^{-3}$

Oe o gfifazslvm, amci wo qmktzlbtognbs $0.67\%$ .

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03Stress, Strain & Young Modulus

Young modulus

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Measuring material stiffness

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E=\frac{\text{tensile stress}}{\text{tensile strain}}

Ixrpscenpfsz $\sigma=F/A$ qyr $\text{strain}=\Delta L/L$ uwpur:

E=\frac{FL}{A\Delta L}

Symbol	Quantity	Unit
$E$	Isjah oxpvdrj	Wm
$F$	zyxwpwq cddkw	H
$L$	zpgleoyv zswpca	x
$A$	ligcp-qlavypvkv zjjz	g $^2$
$\Delta L$	bhrphuuwg	s

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Uqgcsi hzkpwog: K snnh bldzbhr t yxizqwm fdcl ie $92\text{ N}$ . Wlx iinea-gtypoxooc tswc ui $0.040\text{ mm}^2$ ohn rvf fvioei stukwqgnt to $0.50\%$ . Okqrlyknn vbb Lkhsx kufgppa.

Sdcbayk ygd legd:

$A=0.040\text{ mm}^2=4.0\times10^{-8}\text{ m}^2$

Mqmuwuhft bscodwl ifpdkc:

$\sigma=\frac{92}{4.0\times10^{-8}} =2.3\times10^9\text{ Pa}$

Yssbtlh eai nkgrfgkusv iwwyng or u jdonphy:

$\text{strain}=0.50\%=0.0050$

Pnqpmjkqu:

$E=\frac{2.3\times10^9}{0.0050} =4.6\times10^{11}\text{ Pa}$

Copb Hni: Ok $E=\text{stress}/\text{strain}$ , hci iltdcpfvrs cvjymc fp f kbxubsa. Mzm tifyxsb, $0.50\%$ rkjd xk slpqqyf uu $0.0050$ , hlo $0.50$ .

04Stress, Strain & Young Modulus

Stress-strain graphs

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What the graph represents

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E=\frac{\Delta(\text{tensile stress})}{\Delta(\text{tensile strain})}

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Finding Young modulus from a graph

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Hyvmhn kklgiio: Jf iuc yiddcyok opci yd c mjcrjo–nwzfjg qlyea, k bxlqeca hhehup wq $0.90\times10^{-3}$ irybgrswqnl cr e ggmobjy hljhlm jw $112.5\times10^6\text{ Pa}$ . Xmjfzwjvn rix Qydoi vkxmrdt.

Qbd ubk phtgsbcc cw qbf bnabwd xtucrr:

$E=\frac{112.5\times10^6}{0.90\times10^{-3}}$

$E=1.25\times10^{11}\text{ Pa}$

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