## Strength of materials |

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Page 60

... are compatible with the deformations. Such relations are known as the

equations of compatibility. 2. By applying the conditions of equilibrium to a free-

body diagram of a portion of the body, obtain additional relations between the

stresses. These relations, resulting from considering the equilibrium between

externally applied loads and the internal resisting forces over an exploratory

section, are called the equations of equilibrium. 3. Be sure that the

equations in steps 1 ...

... are compatible with the deformations. Such relations are known as the

equations of compatibility. 2. By applying the conditions of equilibrium to a free-

body diagram of a portion of the body, obtain additional relations between the

stresses. These relations, resulting from considering the equilibrium between

externally applied loads and the internal resisting forces over an exploratory

section, are called the equations of equilibrium. 3. Be sure that the

**solution**of theequations in steps 1 ...

Page 258

Comparison with the

Eqs (d) and (e) used there are identical with Eqs. (a) and (b) here. On the other

hand, comparison with the superposition

second equation is equivalent to EUaib = 0 which we could have used here in

place of EUbia = 0. Having determined Va and Ma, we compute the shear and

moment at B in the identical manner as was used on page 251 . It is needless to

repeat it ...

Comparison with the

**solution**by double integration on page 250 discloses thatEqs (d) and (e) used there are identical with Eqs. (a) and (b) here. On the other

hand, comparison with the superposition

**solution**on page 252 discloses that itssecond equation is equivalent to EUaib = 0 which we could have used here in

place of EUbia = 0. Having determined Va and Ma, we compute the shear and

moment at B in the identical manner as was used on page 251 . It is needless to

repeat it ...

Page 308

that was described in Prob. 874 for free ends on two spans. It is inconvenient to

treat a free end as fixed, carry moment over to it, and then release it again. It is

simpler to use a modification in which the free end, initially assumed fixed, is

released only once and has no moment carried over to it for further distribution.

To understand this modification, we shall show how the moment that is

distributed to the left ...

**Solution**: There are two**solutions**. The first, in (a), involves the same procedurethat was described in Prob. 874 for free ends on two spans. It is inconvenient to

treat a free end as fixed, carry moment over to it, and then release it again. It is

simpler to use a modification in which the free end, initially assumed fixed, is

released only once and has no moment carried over to it for further distribution.

To understand this modification, we shall show how the moment that is

distributed to the left ...

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