**solution manual An Introduction to Biomechanics Solids and Fluids, Analysis and Design Humphrey O’Rourke 2nd Edition**

**Complete Solution Manual With Answers**

**Sample Chapter is Posted Below**

**Chapter 1**

1.9

For synthesis, the *k* in [ ] [ ]

*d C dt k C*= − is equal to . *s*

*k*− For degradation, *k* is . *d*

*k *Thus

we have

*d C d C*

[ ] [ ] [ ] [ ] and , *s d*

*k C k C*

*dt dt*= = −

*s d*

which if

*d C*

*d C d C*

[ ] [ ] [ ]

*s d*

= +

*dt dt dt*

*d C*

⇒ = −

[ ] [ ] [ ]

*k C k C*

*s d*

*dt*

*d C*

⇒ = −

[ ] ( )[ ]

*k k C*

*s d*

*dt*

*d C*

⇒ =

[ ] [ ]

*k C*

*dt*

*o*

where *o*

*k *is the overall reaction rate. Integrating with respect to time, we have

*d C*

∫ ∫

[ ] [ ]

*dt k C dt*

=

*dt*

*o*

⇒ = +

∫

*d C k C t c*

[ ] [ ]

*o*

1

⇒ = +

[ ] [ ]

*C k C t c*

*o*

1

1

⇒ − =

( )[ ]

*k t C c*

*o*

1

*c c*

⇒ = =

*C*

1 1

[ ] ( )

1 1

− − −

*k t k k t*

*o s d*

,

where we could calculate 1 *c *if we were given an initial concentration. Note that if*c*

*k k k k C*

[ ]

, 0, 1

1

> − > ∴ = ↑

*s d s d*

−

*k t*

*o*

*c*

*k k k k C*

[ ]

, 0, 1

1

< − < ∴ = ↓

*s d s d*

−

*k t*

*o*

*c k k k k C c*

= − = ∴ = =

*s d s d*

[ ]

1

, 0, . 1 0

1

−

*t*

1.17

(xx-Insert figure showing set-up)

Let point *p* be a point halfway between the applied forces. For convenience, let the point

be located at the origin of a 2-D Cartesian coordinate system, with the applied forces located at

2

*x d*= and 2,

*x d*= − ˆ

and the forces oriented in the *y* direction such that 1

*F*= ** F j **and

ˆ

*F*= − ** F j **Computing the moments at point

*p*, we have )1 1 2 2 ,

2

.

= × + ×

∑ ** M r F r F **where at

*p*,

*p*

ˆ

ˆ

1 2 *r i **d*= and 2

1 2 . *r i **d*= −

1

∑

*d d F F*

ˆ ˆ ˆ ˆ

⎛ ⎞ = × + − − × ⎜ ⎟ ⎝ ⎠

*M i j i j*

) ( ) ( ) ( )

*p*

2 2

⇒ = × =

∑

*M i j k*

) ( )

ˆ ˆ ˆ

*dF dF*

.

*p*

If we choose an arbitrary point *a* located by the vector = + *r i j*

*r r *α α

*a*

cos sin ,

ˆ ˆ

1 1 ˆ ˆ ˆ ˆ

*d r r d r r *α α α α

⎛ ⎞ ⎛ ⎞ = − + = − + + ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ *r i j r i j*

cos sin , cos sin 2 2

1 2

and thus

1 ˆ ˆ ˆ ˆ

*d r F r F*

cos sin

⎛ ⎞ = − × + × ⎜ ⎟ ⎝ ⎠

∑…

*M i j j j*

) ( ) ( ) ( )

α α

*a*

2

1 ˆ ˆ ˆ ˆ

*d r F r F*

cos sin

⎛ ⎞ + − − − × + − × ⎜ ⎟ ⎝ ⎠

α α

2

( ) ( ) ( ) ( ) ( )

** i j j j**but ˆ ˆ

× = **0**

** j j **leaving us with

,

1 1 ˆ ˆ ˆ ˆ

⎛ ⎞ ⎛ ⎞ = − × + − − − × ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠

∑

*d r F d r F*

*M i j i j*

) ( ) ( ) ( )

cos cos

α α

*a*

2 2

1 1 ˆ ˆ

∑

*dF rF dF rF*

⎛ ⎞ ⇒ = − + + × ⎜ ⎟ ⎝ ⎠

*M i j*

) ( )

cos cos

α α

*a*

2 2

⇒ =

∑

*M k*

)

ˆ

*dF*

.

*a*

1.18

Using the equations found in example A1.2, we have

⎛ ⎞

ˆ ˆ ˆ

⎛ ⎞ − + − + − ⎜ ⎟

*r*

*F*

*BA*

( ) ( ) ( )

*x x y y z z*

*i j k*

*T T*

*A B A B A B*

= = ⎜ ⎟ ⎜ ⎟ ⎜ ⎟

*r*

2 2 2

⎝ ⎠ − + − + − ⎝ ⎠

*BA A B A B A B*

( ) ( ) ( )

*x x y y z z*

ˆ ˆ ˆ

⎛ ⎞ − + − + − ⎜ ⎟

0 0 0

*L L L*

( ) ( ) ( )

*i j k*

*F*

*T*

2 2 2

⇒ = ⎜ ⎟ − + − + − ⎝ ⎠

0 0 0

*L L L*

( ) ( ) ( )

ˆ

⎛ ⎞

*k F k*

*L T T*

ˆ

,

⇒ = = ⎜ ⎟ ⎜ ⎟ ⎝ ⎠

*L*

and therefore,

ˆ

*R F k*

= − = −

*T*

⇒ = = = −

*R R R T*

0, 0, and . *x y z*

Similarly, for the reaction moments, we have

= − = −

0

*M z F y F L L T*

*x A y A z*

( ) ( )

*M x F z F T L*

= − = −

0 0

*y A z A x*

( ) ( )

*M y F x F*

= − = −

0 0 0 0

*z A x A y*

( ) ( )

*M TL M M*

⇒ = − = =

, 0, and 0.

*x y z*

1.19(xx-Insert 3-D diagram showing angles alpha, beta, and gamma from *x*, *y*, and *z* axes,

respectively).

ˆ ˆ ˆ

If

ˆ

cos cos cos , α β γ = + + ** e i j k **then

2 2 2

*e*

ˆ 1 cos cos cos

= = + +

α β γ

2 2 2 2

⇒ = + + =

1 cos cos cos 1

α β γ

−

1 2 2

⇒ = − −

cos 1 cos cos

α β γ

( )

−

1 2 2

⇒ = − −

cos 1 cos cos

β α γ

( )

−

1 2 2

⇒ = − −

cos 1 cos cos .

γ α β

( )

ˆ

For a 2-D problem, ˆ ˆ

cos cos , α β

γ = = + ** e i j **and thus the cos

*z*

*A A*

term must equal zero,

i.e. 2.

γ π

= Thus we have

⎛ ⎞

( ) ( )2 2 2 cos 0 1 cos cos 1 1

⎜ ⎟ ⎝ ⎠

π β

= = − − = −

*a*

2

where for 2-D, 2 2

ˆ 1 cos cos . α β

= = +

** e **Thus the relationship between angles holds if the

problem is strictly 2-D.

1.20

Because a pin cannot support a reaction moment, the moments created by 1 ** F **and 2

*F*must balance. Therefore, if we use a standard Cartesian coordinate system, we have*r F r F*

× + × =

1 1 2 2

**0**

ˆ ˆ ˆ ˆ

*d T d T*

⇒ − × − + × − =

( ) ( ) ( )

*i j i j*

1 1 2 2

ˆ ˆ ˆ ˆ

*d T d T*

⇒ × − × =

1 1 2 2

( ) ( )

*i j i j*

**0**

ˆ ˆ ˆ ˆ

*d T d T*

⇒ × = ×

1 1 2 2

( ) ( )

*i j i j*

*T*

⇒ =

1

*d T*

2 2

*d*

1

.

**0**

1.21

(xx-Insert FBD of weight mass *m**w*, gravity force in –** j** direction, force

*T*1 in

**direction)**

*j*(xx-Insert FBD of loadcell-force *T*2 in ** i** direction)

(xx-Insert FBD of pulley, *T*2 in –** i**,

*T*1 in –

**, reaction force at angle alpha)**

*j*Assuming a weightless pulley and support, and that the pulley is ideal (frictionless) i.e.

= ** T **1 2 ,

** T **we have from the free body diagram (FBD) of the weight,

∑

∑

∑

ˆ ˆ

*F j j*

= − =

*T g*

1

*F*

*x*

=

0

*F T g*

= − =

*y*

1

**0**

0.

1 9.8m s 9.8 N, *w w **T g m *Thus 2

*m*= = = where *w **m *is the mass of the weight in kilograms and the

gravitational constant is approximated by 9.8 m/s2. Therefore, the force applied to the force

transducer is 2 1 9.8 N. *w **T T m*= =

ˆ ˆ

From the FBD of the pulley support, we have a reaction force

cos sin , *r r *α α

= + *r i j*

which gives us∑

ˆ ˆ ˆ ˆ

*F i j i j*

= + − − =

*r r T T*

cos sin

α α

1 1

**0**

∑

*F r T*

= − =

cos 0

α

*x*

1

⇒ =

*T r*

1

cos

α

∑

*F r T*

= − =

sin 0

α

*y*

1

⇒ =

*T r*

1

sin

α

ˆ ˆ

⇒ = +

*T T*

*r i j*

1 1

.

The magnitude of ** r** is thus 2

2*T *1

oriented along the axis of the support.

In addition to supporting forces, the base of the pulley support, *r*, can also sustain a

reaction moment, giving us

ˆ ˆ ˆ ˆ

∑ **0**

*M M i j + j i*

= + × − × − =

*d T d T*

*r x y*

( ) ( )

1 1

ˆ ˆ ˆ ˆ

= − + × − − + × −

*L r T L r T*

cos sin

*M i j j i*

*r*

( ) ( ) ( ) ( )

α α

1 1

ˆ ˆ ˆ ˆ

*T L r T L r*

cos sin

⇒ = + × + + ×

*M i j j i*

*r*

1 1

( ) ( ) ( ) ( )

α α

ˆ ˆ

⇒ = + + + −

*M k k*

*T L r T L r*

cos sin

*r*

1 1

( ) ( ) ( )( )

α α

ˆ

⇒ = −

*M k*

*T L*

*r*

1

( )

cos sin .

α α

Note that if α = 45o

, cos sin , α α

= and thus there is no reaction moment at the base of the pulley

support.

1.22

(xx-insert FBD of block, with coordinate system oriented on slant)

(xx-insert decomposition of gravity into *x* and *y* directions)

The block is in equilibrium just before it begins to slip. Therefore, using eq (A1.8), we

have∑

∑

*F N W*

= = −

0 sin

μ θ

*x s s*

*F N W*

= = −

0 cos

θ

*y s*

⇒ =

*N W*

cos

θ

*s*

sin

θ

*W*

*s*

⇒ = =

μ θ

tan .

*s s*

cos

θ

*W*

*s*

1.23

(xx-Insert FBD of entire truss, with joints identified as a (the top right joint), b (the bottom left

joint), c (top left joint), and d (bottom right joint))

(xx-Insert FBD of each of the four points of interest (4 panel figure))

By geometry, the angles within the equilateral triangle are all 60o, and the other triangle

is a 30o-60o-90o triangle, with sides having a length of *L/*2, 3 2 , *L *and *L*. (Useful trigonometric

quantities: ( ) ( ) sin 60 cos 30 3 2, ° = ° = ( ) ( ) sin 30 cos 60 1 2. ° = ° = ) Note that the roller can

only produce a reaction force perpendicular to the plane of the roller, and the pin connections

found throughout the truss can only support reaction forces, not moments.

For the entire truss, we have

∑

ˆ ˆ ˆ ˆ

*F R R R*

*F j i + j j*

= − + + =

*ax ay by*

∑

*F R*

= =

*x ax*

0

*R*

⇒ =

*ax*

0

∑

*F F R R*

= − + + =

*y ay by*

0

⇒ + =

*R R F*

*ay by*

.

**0**

In order to fully define *ay*

*R *and , *by **R *we must introduce another equation, in this case a sum of

moments. Taking the sum of moments about point *b*, we have∑

ˆ ˆ ˆ ˆ ˆ ˆ

= = ° + ° × − + + ° ×

**0**

*L F L L R*

cos 60 sin 60 sin 60

) ( ) ( ) ( ) ( ) ( ) ( )

*M i j j i j j*

*b*

*ay*

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ

*FL FL R L R L*

cos 60 cos 60 sin 60

⇒ = − ° × − ° × + × + ° × =

∑

*M i j j j i j j j*

) ( )( ) ( ) ( ) ( ) ( ) ( )

*b*

*ay ay*

ˆ ˆ

*M k k*

*FL FL R L R L*

cos 60 cos 60 sin 60

⇒ = − ° − ° + + ° =

∑

) ( )( ) ( ) ( ) ( ) ( )( )

**0 0 0**

*b*

*ay ay*

ˆ ˆ

*M k k*

⇒ = − ° + =

∑

cos 60

*FL R L*

) ( )( ) ( )

**0**

*b*

*ay*

⇒ = ° =

*R F F*

cos 60 2

*ay*

( )

⇒

*R*

*F F F*= − =

2 2.

*b*

*y*

**0**

By making fictitious cuts between the points of interest, we can determine the forces

placed on the individual pins. The forces on the members are equal and opposite that enforced on

them by the pins. Although we consider the points in isolation, the forces felt at each pin are

*F F*=

related such that 1 1,

*a c*

*a d **F F*=

2 1,

*b c **F F*=

1 3 ,

*b d **F F*= and 2 2.

2 3 ,

*c d **F F*= Summing our forces,

we have

∑

ˆ ˆ ˆ

= − − =

2

*F F F*

*F j i j*

)

*a a*

1 2

*a*

∑

*F F*

)

= − =

1

*x a a*

0

*F*

⇒ =

*a*

1

0

∑

*F F F*

)

= − =

2 0

2

*y a a*

⇒ =

*F F*

*a*

2

2.

**0**

∑

ˆ ˆ ˆ ˆ

= + ° + ° + =

*F F F*

2 cos 60 sin 60

*F j i j i*

) ( ) ( ) ( )

*b b*

1 2

*b*

∑

*F F F*

) ( )

ˆ

= ° + =

*i*

cos 60 0

1 2

*x b b b*

⇒ = − °

*F F*

cos 60

*b b*

2 1

( )

∑

*F F F*

*y b b*

) ( )

= + ° =

2 sin 60 0

1

*F F F F*

⇒ = − = − ≈ −

0.577

*b*

1

°

2sin 60 3

( )

cos 60

°

( )

*F F F F*

⇒ = = ≈

0.289 .

*b*

2

°

2sin 60 2 3

( )

**0**∑

ˆ ˆ ˆ ˆ ˆ ˆ

*F j + i + i j i j*

) ( ) ( ) ( ) ( ) ( )

= − ° − ° − ° + ° =

*F F F F*

cos 60 sin 60 cos 60 sin 60

*c c c*

1 2 3

*c*

( )

*F F*

= =

0

*c a*

1 1

∑

*F F F*

) ( ) ( )

= ° − ° =

cos 60 cos 60 0

2 3

*x c c c*

⇒ =

*F F*

*c c*

2 3

∑

*F F F F*

) ( ) ( )

= − − ° − ° =

sin 60 sin 60 0

2 3

*y c c c*

2 sin 60

⇒ − ° =

*F F*

*c*

2

( )

−

*F F F F F*

⇒ = = = − ≈ −

0.577 .

*c c*

2 3

°

2sin 60 3

( )

**0**

where we see that 1 3 *b c **F F*= as it should.

∑

ˆ ˆ ˆ ˆ

= − ° + ° − =

*F F F*

cos 60 sin 60

) ( ) ( ) ( )

*F j + i j i*

**0**

*d d d*

1 2 3

*d*

*F F F*

= =

2

*d a*

1 2

∑

*F F F*

) ( )

= − ° − =

cos 60 0

2 3

*x d d d*

⇒ = − °

*F F*

cos 60

*d d*

3 2

( )

∑

*F F F*

*y d d*

) ( )

= + ° =

2 sin 60 0

2

− −

*F F F F*

⇒ = = ≈ −

0.577

*d*

2

°

2sin 60 3

( )

cos 60

°

( )

*F F F F*

⇒ = = ≈

0.289

*d*

3

°

2sin 60 2 3

( )

where we see that 2 3 ,

*b d **F F*= and 2 2 *c d **F F*= as expected.

Finally, the diagonal members are in compression, because the forces acting on the pins

are negative (and thus the forces on the members act into the members, or are positive), while the

vertical member and the bottom horizontal member are in tension, because the forces acting on

the pins are positive (and thus the forces on the members act out of the members, or are

negative). The top horizontal member, on which no force acts, is neither in tension nor

compression.

(xx-Insert figure showing elements in tension and compression)•

*\.1 **T*I .1 ~•

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