Examples - Bending of statically indeterminate (continuous) beams

Examples

Example 1 represents 5-segment statically indeterminate beam with left overhang. The beam is loaded by couple, linearly distributed load and concentrated force.

Input:

Length of the beam L: 5

Number of pin supports (except left one): 3

Number of couples (external moments): 1 Number of concentrated loads: 1 Number of distributed loads: 1

l₁ - coordinate of reaction (pin support) 0<l₁≤L: 2

M₁ - value of the couple # 1 of 1: -1 a₁ - coordinate of this couple: 2

F₁ - value of the concentrated load # 1 of 1: 1 b₁ - coordinate of this concentrated load: 1
q_L1 - left value of the lineary distributed load # 1 of 1: 1 q_R1 - right value of the lineary distributed load # 1 of 1: 0

c₁ - left coordinate of this distributed load: 4 d₁ - right coordinate of this distributed load: 5

l₂ - coordinate of reaction (pin support) 0<l₂≤L: 3

l₃ - coordinate of reaction (pin support) 0<l₃≤L: 4

Output:

               Maximum shear force V is 1.2463768 at x=2.0
               Maximum bending moment M is -0.9637681 at x=2.0
               Maximum deflection EIv is -0.175729 at x=1.01

     Calculation of reactions solving system of equations.
Geometrical equations (beam deflections corresponding to pin supports are equal to 0):
EIv(l₁)=EIθ₀l₁+R₀l₁³/6-F₁(l₁-b₁)³/6=0
EIv(l₂)=EIθ₀l₂+M₁(l₂-a₁)²/2+R₀l₂³/6+R₁(l₂-l₁)³/6-F₁(l₂-b₁)³/6=0
EIv(l₃)=EIθ₀l₃+M₁(l₃-a₁)²/2+R₀l₃³/6+R₁(l₃-l₁)³/6+R₂(l₃-l₂)³/6-F₁(l₃-b₁)³/6=0
The equation of moment equilibrium about right end support:
M₀+M₁+R₀L+R₁(l₃-l₁)+R₂(l₃-l₂)-F₁(l₃-b₁)+q_L1(c₁-d₁)(l₃-(c₁+d₁)/2)+(q_R1-q_L1)(d₁-c₁)/2(l₃-(c₁+2(d₁-c₁)/3))=0
Finally we have system of 4 equations:
2.0EIθ₀+1.3333334R₀+0.0R₁+0.0R₂=0.16666667
3.0EIθ₀+4.5R₀+0.16666667R₁+0.0R₂=1.8333334
4.0EIθ₀+10.666667R₀+1.3333334R₁+0.16666667R₂=6.5
0.0EIθ₀+4.0R₀+2.0R₁+1.0R₂=3.8333333
The equation of beam equilibrium with respect to vertical axis serves for calculation of vertical reaction applied to the right support of the beam:
R₃=-R₀+F₁-R₁-R₂+q_L1(d₁-c₁)+(q_R1-q_L1)(d₁-c₁)/2
The solution is:
R₀=0.51811594; R₁=1.7282609; R₂=-1.6956521; R₃=0.9492754; EIθ₀=-0.2620773;

               The beam has 5 segments.
     Segment #1 (0.0≤x≤1.0).
Shear force:
V=R₀=
0.51811594=
0.51811594
Bending moment:
M=+R₀x=
+0.51811594x=
+0.51811594x
Deflection:
EIv=EIθ₀x+R₀x³/6=
-0.2620773x+0.51811594x³/6=
-0.2620773x+0.086352654x³
At x=0.0 V=0.51811594; M=2.5905798E-9; EIv=-1.3103865E-9;
At x=1.0 V=0.51811594; M=0.51811594; EIv=-0.17572464;

     Segment #2 (1.0≤x≤2.0).
Shear force:
V=R₀-F₁=
0.51811594-1.0=
-0.48188406
Bending moment:
M=+R₀x-F₁(x-b₁)=
+0.51811594x-1.0(x-1.0)=
+0.51811594x-1.0(x-1.0)
Deflection:
EIv=EIθ₀x+R₀x³/6-F₁(x-b₁)³/6=
-0.2620773x+0.51811594x³/6-1.0(x-1.0)³/6=
-0.2620773x+0.086352654x³-0.16666667(x-1.0)³
At x=1.0 V=-0.48188406; M=0.51811594; EIv=-0.17572464;
At x=2.0 V=-0.48188406; M=0.036231887; EIv=-1.3707736E-9;
There is mimimum deflection -0.175729 on this segment at x=1.01

     Segment #3 (2.0≤x≤3.0).
Shear force:
V=R₀+R₁-F₁=
0.51811594+1.7282609-1.0=
1.2463768
Bending moment:
M=M₁+R₀x+R₁(x-l₁)-F₁(x-b₁)=
-1.0+0.51811594x+1.7282609(x-2.0)-1.0(x-1.0)=
-1.0+0.51811594x+1.7282609(x-2.0)-1.0(x-1.0)
Deflection:
EIv=EIθ₀x+M₁(x-a₁)²/2+R₀x³/6+R₁(x-l₁)³/6-F₁(x-b₁)³/6=
-0.2620773x-1.0(x-2.0)²/2+0.51811594x³/6+1.7282609(x-2.0)³/6-1.0(x-1.0)³/6=
-0.2620773x-0.5(x-2.0)²+0.086352654x³+1.7282609(x-2.0)³/6-0.16666667(x-1.0)³
At x=2.0 V=1.2463768; M=-0.9637681; EIv=1.3707725E-9;
At x=3.0 V=1.2463768; M=0.2826087; EIv=3.3212533E-10;
There is maximum deflection 0.045993216 on this segment at x=2.38

     Segment #4 (3.0≤x≤4.0).
Shear force:
V=R₀+R₁+R₂-F₁=
0.51811594+1.7282609-1.6956521-1.0=
-0.44927537
Bending moment:
M=M₁+R₀x+R₁(x-l₁)+R₂(x-l₂)-F₁(x-b₁)=
-1.0+0.51811594x+1.7282609(x-2.0)-1.6956521(x-3.0)-1.0(x-1.0)=
-1.0+0.51811594x+1.7282609(x-2.0)-1.6956521(x-3.0)-1.0(x-1.0)
Deflection:
EIv=EIθ₀x+M₁(x-a₁)²/2+R₀x³/6+R₁(x-l₁)³/6+R₂(x-l₂)³/6-F₁(x-b₁)³/6=
-0.2620773x-1.0(x-2.0)²/2+0.51811594x³/6+1.7282609(x-2.0)³/6-1.6956521(x-3.0)³/6-1.0(x-1.0)³/6=
-0.2620773x-0.5(x-2.0)²+0.086352654x³+1.7282609(x-2.0)³/6-1.6956521(x-3.0)³/6-0.16666667(x-1.0)³
At x=3.0 V=-0.44927537; M=0.2826087; EIv=-3.3212583E-10;
At x=4.0 V=-0.44927537; M=-0.16666667; EIv=4.226941E-11;
There is mimimum deflection -0.009243167 on this segment at x=3.31
There is maximum deflection 2.2373188E-4 on this segment at x=3.95

     Segment #5 (4.0≤x≤5.0).
Shear force:
V=R₀+R₁+R₂+R₃-F₁-q_L1(x-c₁)-(q_R1-q_L1)(x-c₁)²/(d₁-c₁)/2=
0.51811594+1.7282609-1.6956521+0.9492754-1.0-1.0(x-4.0)-(0.0-1.0)(x-4.0)²/(5.0-4.0)/2=
0.5-1.0(x-4.0)+0.5(x-4.0)²
Bending moment:
M=M₁+R₀x+R₁(x-l₁)+R₂(x-l₂)+R₃(x-l₃)-F₁(x-b₁)-q_L1(x-c₁)²/2-(q_R1-q_L1)(x-c₁)³/(d₁-c₁)/6=
-1.0+0.51811594x+1.7282609(x-2.0)-1.6956521(x-3.0)+0.9492754(x-4.0)-1.0(x-1.0)-1.0(x-4.0)²/2-(0.0-1.0)(x-4.0)³/(5.0-4.0)/6=
-1.0+0.51811594x+1.7282609(x-2.0)-1.6956521(x-3.0)+0.9492754(x-4.0)-1.0(x-1.0)-0.5(x-4.0)²+0.16666667(x-4.0)³
Deflection:
EIv=EIθ₀x+M₁(x-a₁)²/2+R₀x³/6+R₁(x-l₁)³/6+R₂(x-l₂)³/6+R₃(x-l₃)³/6-F₁(x-b₁)³/6-q_L1(x-c₁)⁴/24-(q_R1-q_L1)(x-c₁)⁵/(d₁-c₁)/120=
-0.2620773x-1.0(x-2.0)²/2+0.51811594x³/6+1.7282609(x-2.0)³/6-1.6956521(x-3.0)³/6+0.9492754(x-4.0)³/6-1.0(x-1.0)³/6-1.0(x-4.0)⁴/24-(0.0-1.0)(x-4.0)⁵/(5.0-4.0)/120=
-0.2620773x-0.5(x-2.0)²+0.086352654x³+1.7282609(x-2.0)³/6-1.6956521(x-3.0)³/6+0.9492754(x-4.0)³/6-0.16666667(x-1.0)³-0.041666668(x-4.0)⁴+0.008333334(x-4.0)⁵
At x=4.0 V=0.5; M=-0.16666666; EIv=-4.2272297E-11;
At x=5.0 V=0.0; M=-1.110223E-16; EIv=-0.041787438;

Example 2 represents 3-segment statically indeterminate beam loaded by couple, uniformly distributed load and concentrated force.

Input:

Length of the beam L: 4 Number of pin supports: 2

Number of couples (external moments): 1 Number of concentrated loads: 1 Number of distributed loads: 1

l₁ - coordinate of reaction (pin support) 0<l₁≤L: 2

M₁ - value of the couple # 1 of 1: -2 a₁ - coordinate of this couple: 4

F₁ - value of the concentrated load # 1 of 1: 1 b₁ - coordinate of this concentrated load: 1
q_L1 - left value of the lineary distributed load # 1 of 1: 1 q_R1 - right value of the lineary distributed load # 1 of 1: 1

c₁ - left coordinate of this distributed load: 2 d₁ - right coordinate of this distributed load: 4

l₂ - coordinate of reaction (pin support) 0<l₂≤L: 4

Output:

               Maximum shear force V is 2.5357144 at x=2.0
               Maximum bending moment M is 2.0 at x=4.0
               Maximum deflection EIv is -0.47082222 at x=3.22

     Calculation of reactions solving system of equations.
Geometrical equations (beam deflections corresponding to pin supports are equal to 0):
EIv(l₁)=M₀l₁²/2+R₀l₁³/6-F₁(l₁-b₁)³/6=0
EIv(l₂)=M₀l₂²/2+R₀l₂³/6+R₁(l₂-l₁)³/6-F₁(l₂-b₁)³/6-q₁(l₂-c₁)⁴/24=0
The equation of moment equilibrium about right end support:
M₀+M₁+R₀L+R₁(l₂-l₁)-F₁(l₂-b₁)+q₁(d₁-c₁)(l₂-(c₁+d₁)/2)=0
Finally we have system of 3 equations:
2.0M₀+1.3333334R₀+0.0R₁=0.33333334
8.0M₀+10.666667R₀+1.3333334R₁=9.666667
1.0M₀+4.0R₀+2.0R₁=10.0
The equation of beam equilibrium with respect to vertical axis serves for calculation of vertical reaction applied to the right support of the beam:
R₂=-R₀+F₁-R₁+q₁(d₁-c₁)
The solution is:
R₀=0.5714286; R₁=3.9642856; R₂=-0.53571427; M₀=-0.21428572;

               The beam has 3 segments.
     Segment #1 (0.0≤x≤1.0).
Shear force:
V=R₀=
0.5714286=
0.5714286
Bending moment:
M=M₀+R₀x=
-0.21428572+0.5714286x=
-0.21428572+0.5714286x
Deflection:
M₀x²/2+R₀x³/6=
-0.21428572x²/2+0.5714286x³/6=
-0.21428572x²/2+0.0952381x³
At x=0.0 V=0.5714286; M=-0.21428572; EIv=-1.7142857E-18;
At x=1.0 V=0.5714286; M=0.35714287; EIv=-0.011904762;
There is mimimum deflection -0.020089285 on this segment at x=0.75

     Segment #2 (1.0≤x≤2.0).
Shear force:
V=R₀-F₁=
0.5714286-2.0=
-1.4285715
Bending moment:
M=M₀+R₀x-F₁(x-b₁)=
-0.21428572+0.5714286x-2.0(x-1.0)=
-0.21428572+0.5714286x-2.0(x-1.0)
Deflection:
M₀x²/2+R₀x³/6-F₁(x-b₁)³/6=
-0.21428572x²/2+0.5714286x³/6-2.0(x-1.0)³/6=
-0.21428572x²/2+0.0952381x³-0.33333334(x-1.0)³
At x=1.0 V=-1.4285715; M=0.35714287; EIv=-0.011904761;
At x=2.0 V=-1.4285715; M=-1.0714285; EIv=1.142857E-9;
There is maximum deflection 0.04458333 on this segment at x=1.65

     Segment #3 (2.0≤x≤4.0).
Shear force:
V=R₀+R₁-F₁-q₁(x-c₁)=
0.5714286+3.9642856-2.0-1.0(x-2.0)=
2.5357144-1.0(x-2.0)
Bending moment:
M=M₀+R₀x+R₁(x-l₁)-F₁(x-b₁)-q₁(x-c₁)²/2=
-0.21428572+0.5714286x+3.9642856(x-2.0)-2.0(x-1.0)-1.0(x-2.0)²/2=
-0.21428572+0.5714286x+3.9642856(x-2.0)-2.0(x-1.0)-0.5(x-2.0)²
Deflection:
M₀x²/2+R₀x³/6+R₁(x-l₁)³/6-F₁(x-b₁)³/6-q₁(x-c₁)⁴/24=
-0.21428572x²/2+0.5714286x³/6+3.9642856(x-2.0)³/6-2.0(x-1.0)³/6-1.0(x-2.0)⁴/24=
-0.21428572x²/2+0.0952381x³+3.9642856(x-2.0)³/6-0.33333334(x-1.0)³-0.041666668(x-2.0)⁴
At x=2.0 V=2.5357144; M=-1.0714285; EIv=-1.1428573E-9;
At x=4.0 V=0.53571427; M=2.0; EIv=-5.2380953E-9;
There is mimimum deflection -0.47082222 on this segment at x=3.22

Example 3 represents 3-segment statically indeterminate beam under linearly distributed and concentrated load.

Input:

Length of the beam L: 3 Number of pin supports: 1

Number of couples (external moments): Number of concentrated loads: 1 Number of distributed loads: 1

l₁ - coordinate of reaction (pin support) 0<l₁≤L: 2

F₁ - value of the concentrated load # 1 of 1: 4 b₁ - coordinate of this concentrated load: 1
q_L1 - left value of the lineary distributed load # 1 of 1: 1 q_R1 - right value of the lineary distributed load # 1 of 1: 2

c₁ - left coordinate of this distributed load: 2 d₁ - right coordinate of this distributed load: 3

Output:

               Maximum shear force V is 2.2208333 at x=0.0
               Maximum bending moment M is -1.1472222 at x=0.0
               Maximum deflection EIv is -0.20410453 at x=1.04

     Calculation of reactions solving system of equations.
Geometrical equations (beam deflections corresponding to pin supports are equal to 0):
EIv(l₁)=M₀l₁²/2+R₀l₁³/6-F₁(l₁-b₁)³/6=0
Two conditions for the right clamped end of the beam:
EIv(L)=M₀L²/2+R₀x³/6+R₁(L-l₁)³/6-F₁(L-b₁)³/6-q_L1(L-c₁)⁴/24-(q_R1-q_L1)(L-c₁)⁵/(d₁-c₁)/120=0
EIθ(L)=M₀L+R₀x²/2+R₁(L-l₁)²/2-F₁(L-b₁)²/2-q_L1(L-c₁)³/6-(q_R1-q_L1)(L-c₁)⁴/(d₁-c₁)/24=0
Finally we have system of 3 equations:
2.0M₀+1.3333334R₀+0.0R₁=0.6666667
4.5M₀+4.5R₀+0.16666667R₁=5.383333
3.0M₀+4.5R₀+0.5R₁=8.208333
The equation of moment equilibrium about right end support serves for calculation of moment M_r:
M_r=-M₀-R₁(L-l₁)-R₀L+F₁(L-b₁)-q_L1(c₁-d₁)(L-(c₁+d₁)/2)-(q_R1-q_L1)(d₁-c₁)/2(L-(c₁+2(d₁-c₁)/3))=0
The equation of beam equilibrium with respect to vertical axis serves for calculation of vertical reaction applied to the right support of the beam:
R_r=-R₀+F₁-R₁+q_L1(d₁-c₁)+(q_R1-q_L1)(d₁-c₁)/2
The solution is:
R₀=2.2208333; R₁=3.3125; M₀=-1.1472223; M_r=-0.16111112; R_r=-0.033333335;

               The beam has 3 segments.
     Segment #1 (0.0≤x≤1.0).
Shear force:
V=R₀=
2.2208333=
2.2208333
Bending moment:
M=M₀+R₀x=
-1.1472223+2.2208333x=
-1.1472223+2.2208333x
Deflection:
M₀x²/2+R₀x³/6=
-1.1472223x²/2+2.2208333x³/6=
-1.1472223x²/2+0.37013888x³
At x=0.0 V=2.2208333; M=-1.1472222; EIv=-5.1625E-18;
At x=1.0 V=2.2208333; M=1.0736111; EIv=-0.20347223;

     Segment #2 (1.0≤x≤2.0).
Shear force:
V=R₀-F₁=
2.2208333-4.0=
-1.7791667
Bending moment:
M=M₀+R₀x-F₁(x-b₁)=
-1.1472223+2.2208333x-4.0(x-1.0)=
-1.1472223+2.2208333x-4.0(x-1.0)
Deflection:
M₀x²/2+R₀x³/6-F₁(x-b₁)³/6=
-1.1472223x²/2+2.2208333x³/6-4.0(x-1.0)³/6=
-1.1472223x²/2+0.37013888x³-0.6666667(x-1.0)³
At x=1.0 V=-1.7791667; M=1.0736111; EIv=-0.20347223;
At x=2.0 V=-1.7791667; M=-0.70555556; EIv=-4.416667E-10;
There is mimimum deflection -0.20410453 on this segment at x=1.04

     Segment #3 (2.0≤x≤3.0).
Shear force:
V=R₀+R₁-F₁-q_L1(x-c₁)-(q_R1-q_L1)(x-c₁)²/(d₁-c₁)/2=
2.2208333+3.3125-4.0-1.0(x-2.0)-(2.0-1.0)(x-2.0)²/(3.0-2.0)/2=
1.5333333-1.0(x-2.0)-0.5(x-2.0)²
Bending moment:
M=M₀+R₀x+R₁(x-l₁)-F₁(x-b₁)-q_L1(x-c₁)²/2-(q_R1-q_L1)(x-c₁)³/(d₁-c₁)/6=
-1.1472223+2.2208333x+3.3125(x-2.0)-4.0(x-1.0)-1.0(x-2.0)²/2-(2.0-1.0)(x-2.0)³/(3.0-2.0)/6=
-1.1472223+2.2208333x+3.3125(x-2.0)-4.0(x-1.0)-0.5(x-2.0)²-0.16666667(x-2.0)³
Deflection:
M₀x²/2+R₀x³/6+R₁(x-l₁)³/6-F₁(x-b₁)³/6-q_L1(x-c₁)⁴/24-(q_R1-q_L1)(x-c₁)⁵/(d₁-c₁)/120=
-1.1472223x²/2+2.2208333x³/6+3.3125(x-2.0)³/6-4.0(x-1.0)³/6-1.0(x-2.0)⁴/24-(2.0-1.0)(x-2.0)⁵/(3.0-2.0)/120=
-1.1472223x²/2+0.37013888x³+3.3125(x-2.0)³/6-0.6666667(x-1.0)³-0.041666668(x-2.0)⁴-0.008333334(x-2.0)⁵
At x=2.0 V=1.5333333; M=-0.70555556; EIv=4.4166715E-10;
At x=3.0 V=0.03333334; M=0.16111112; EIv=-6.661338E-16;
There is maximum deflection 0.018958917 on this segment at x=2.3

M₁ - value of the couple # 1 of 1: -1	a₁ - coordinate of this couple: 2
F₁ - value of the concentrated load # 1 of 1: 1	b₁ - coordinate of this concentrated load: 1
q_L1 - left value of the lineary distributed load # 1 of 1: 1	q_R1 - right value of the lineary distributed load # 1 of 1: 0
c₁ - left coordinate of this distributed load: 4	d₁ - right coordinate of this distributed load: 5

M₁ - value of the couple # 1 of 1: -2	a₁ - coordinate of this couple: 4
F₁ - value of the concentrated load # 1 of 1: 1	b₁ - coordinate of this concentrated load: 1
q_L1 - left value of the lineary distributed load # 1 of 1: 1	q_R1 - right value of the lineary distributed load # 1 of 1: 1
c₁ - left coordinate of this distributed load: 2	d₁ - right coordinate of this distributed load: 4