Timothy A. Philpot - Mechanics of materials _ an integrated learning system-John Wiley (2017)

integration over the interval a ≤ x ≤ L
Substitute Equation (b) into Equation (10.1) to obtain
EI d 2
v Pbx
= − Px ( − a)
2 (f)
dx L
integration
Integrate Equation (f) twice to obtain the following equations:
EI dv
dx
2
Pbx P
= − x − a + C
2L
2 ( ) 2
3 (g)
3
Pbx P
EIv = − x − a + Cx + C
6L
6 ( ) 3
3 4 (h)
Equations (d), (e), (g), and (h) contain four constants of integration; therefore, four
boundary and continuity conditions are required in order to evaluate the constants.
Continuity Conditions
The beam is a single, continuous member. Consequently, the two sets of equations found
must produce the same slope and the same deflection at x = a. Consider slope equations
(d) and (g). At x = a, these two equations must produce the same slope; therefore, set the
two equations equal to each other, and substitute the value a for each variable x:
Pb( a)
2L
2
2
Pb( a)
P
+ C1
= − a − a + C ∴ C = C
2L
2 [( ) ] 2
3 1 3 (i)
Likewise, deflection equations (e) and (h) must give the same deflection v at x = a. Setting
these equations equal to each other and substituting x = a gives
Pb( a)
6L
3
3
Pb( a)
P
+ C ( a)
+ C = − a − a + C a + C ∴ C = C
6L
6 [( ) ] 3
1 2
( )
3 4 2 4 (j)
Boundary Conditions
At x = 0, the beam is supported by a pin connection; consequently, v = 0 at x = 0. Substitute
this boundary condition into Equation (e) to find
3
3
Pbx
Pb(0)
EIv = + Cx 1 + C2
⇒ EI(0)
= + C1(0) + C2 ∴ C2
= 0
6L
6L
Since C 2 = C 4 from Equation ( j),
C2 = C4 = 0
(k)
At x = L, the beam is supported by a roller connection; consequently, v = 0 at x = L.
Substitute this boundary condition into Equation (h) to find
3
3
Pbx P Pb L P
EIv = − x − a + Cx + C ⇒ EI = − L − a + C L + C
6L
6 ( ) 3
(0) ( )
6L
6 ( ) 3
3 4
( )
Noting that (L − a) = b, simplify the latter equation to obtain
3 4
2 3
2 3 2 2
PbL Pb
PbL Pb Pb( L − b )
EI(0)
= − + CL 3 ∴ C3
=− + = −
6 6 6L
6L
6L
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