Problem 9.2. Force method – Solution Manual

Determine the bending moment diagrams of the statically indeterminate elastic structures given in the Figures a)-d) with the aid of the force method. Bending stiffness, EI of the girders are uniform, compressibility of the elements is neglected.

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Problem a)

Maximum moment, M_max[kNm]=

Problem b)

Maximum moment, M_max[kNm]=

Problem c)

Maximum moment, M_max[kNm]=

Problem d)

Maximum moment, M_max[kNm]=

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Steps
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Force method is described in Section 9.1.

Problem a)

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Step 1. Define a primary structure. Choose the redundant.

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The horizontal support of the right hinge is removed to obtain a statically determined primary structure. The redundant is a horizontal force, X acting at the right support.

Step 2. Calculate the displacements at the right support from the original load and from the redundant.

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The displacements of the frame subjected to uniformly distributed and concentrated loads are derived in Problem 6.9. The results are:

$a_{1} = e_{p} = \frac{138.7}{E I} a_{0} = e_{X} = \frac{448}{E I}$

Step 3. Determine the redundant from the compatibility condition.

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The displacement at the right support must be zero

$a_{0} + a_{1} X = 0 \to X = - \frac{a_{0}}{a_{1}} = - \frac{448}{138.7} = - 3.23 kN (\leftarrow)$

Step 4. Apply superposition to give the moment diagram of the statically indeterminate structure.

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$M = M_{0} + M_{1} X = M_{0} - 3.23 M_{1}$

The maximum moment is

$M_{\max}^{} = X h = 4.0 \times (- 3.23) = - 12.92 kNm$

Problem b)

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Step 1. Define a primary structure. Choose the redundant.

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Removing the right support the primary structure is chosen to be a cantilever. The redundant is a vertical force, X acting at the free end of the beam.

Step 2. Calculate the deflection at the beam end from the original load and from the redundant.

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See Table 3.5.

$a_{0} = v_{p} = \frac{p L^{4}}{8 E I} = \frac{4.0 \times 4 . 0^{4}}{8 E I} = \frac{128}{E I} a_{1} = v_{X} = - \frac{1 \times L^{3}}{3 E I} = - \frac{64}{3 E I}$

Step 3. Determine the redundant from the compatibility condition.

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The displacement at the beam end support must be zero

Step 4. Apply superposition to give the moment diagram of the statically indeterminate structure.

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$M = M_{0} + M_{1} X = M_{0} + \frac{3}{8} M_{1}$

The maximum moment arises at the built-in support

$M_{\max}^{} = - \frac{p L^{2}}{2} + \frac{3}{8} p L^{2} = - \frac{p L^{2}}{8} = - 8 kNm$

Problem c)

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Step 1. Define a primary structure. Choose the redundant.

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A hinge is introduced above the middle support, thus the primary structure becomes two simply supported girders. The redundant is a moment couple, X shown in the Figure.

Similarly to the solution of Problem 9.1.

Step 2. Calculate the relative rotation above the middle support from the original load and from the redundant.

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The relative rotations are calculated by Castigliano’s theorem:

See Problem 6.6.

a_{0} = \int \frac{M_{0} M_{1}}{E I} = - \frac{1}{E I} \frac{F L}{4} \frac{L}{2} \frac{1}{2} = - \frac{F L^{2}}{16 E I} a_{1} = \int \frac{M_{1} M_{1}}{E I} = 2 \frac{1 \times L}{2 E I} \frac{2}{3} = \frac{2 L}{3 E I}

Step 3. Determine the redundant from the compatibility condition.

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The relative rotation above the middle support must be zero:

$a_{0} + a_{1} X = 0 \to X = \frac{a_{0}}{a_{1}} = \frac{3}{32} F L = \frac{3}{32} 15 \times 4 = 5.625 kNm$

Step 4. Apply superposition to give the moment diagram of the statically indeterminate structure.

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$M = M_{0} + M_{1} X = M_{0} + 5.625 M_{1}$

The maximum negative moment above the support is

$M_{\max}^{-} = - X = - 5.625 kNm$

The maximum positive moment below the force is

$M_{\max}^{+} = \frac{F L}{4} - \frac{X}{2} = \frac{15.0 \times 4.0}{4} - \frac{5.625}{2} = 12.19 kNm$

Problem d)

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Step 1. Define a primary structure. Choose the redundant.

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The primary structure and the redundant is chosen similarly as in Problem a). On the cantilever a linearly distributed load, and a concentrated end force act.

Step 2. Calculate the deflection at the beam end from the original load and from the redundant.

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End deflection of a cantilever subjected to linearly distributed load is derived in Problem 6.7 a), end rotation from the redundant is equivalent to that of Problem b):
$a_{0} = v_{p} = \frac{p L^{4}}{30 E I} = \frac{4.0 \times 4 . 0^{2}}{30 E I} a_{1} = v_{X} = - \frac{1 \times L^{3}}{3 E I} = - \frac{64}{3 E I}$

Step 3. Determine the redundant from the compatibility condition.

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The deflection at the beam end must be zero

$a_{0} + a_{1} X = 0 \to X = - \frac{a_{0}}{a_{1}} = \frac{3}{30} p L = \frac{3}{30} 4.0 \times 4.0 = 1.6 kN$

Step 4. Apply superposition to give the moment diagram of the statically indeterminate structure.

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$M = M_{0} + M_{1} X = M_{0} + 1.6 M_{1}$

The maximum moment arises at the built-in support

$M_{\max}^{-} = - \frac{p L^{2}}{6} + 1.6 L = - \frac{4.0 \times 4 . 0^{2}}{6} + 1.6 \times 4.0 = - 4.27 kNm$

Results
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Force method is described in Section 9.1.

Problem a)

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The horizontal support of the right hinge is removed to obtain a statically determined primary structure. The redundant is a horizontal force, X acting at the right support.

The displacements of the frame subjected to uniformly distributed and concentrated loads are derived in Problem 6.9. The results are:

$a_{1} = e_{p} = \frac{138.7}{E I} a_{0} = e_{X} = \frac{448}{E I}$

The redundant is determined from the compatibility condition. The displacement at the right support must be zero:

$a_{0} + a_{1} X = 0 \to X = - \frac{a_{0}}{a_{1}} = - \frac{448}{138.7} = - 3.23 kN (\leftarrow)$

Superposition is applied to give the moment diagram of the statically indeterminate structure:

$M = M_{0} + M_{1} X = M_{0} - 3.23 M_{1}$

The maximum moment is

$M_{\max}^{} = X h = 4.0 \times (- 3.23) = - 12.92 kNm$

Problem b)

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Removing the right support the primary structure is chosen to be a cantilever. The redundant is a vertical force, X acting at the free end of the beam.

The deflection at the beam end from the original load and from the redundant is calculated.

See Table 3.5.

$a_{0} = v_{p} = \frac{p L^{4}}{8 E I} = \frac{4.0 \times 4 . 0^{4}}{8 E I} = \frac{128}{E I} a_{1} = v_{X} = - \frac{1 \times L^{3}}{3 E I} = - \frac{64}{3 E I}$

The redundant is determined from the compatibility condition.The displacement at the beam end support must be zero:

$a_{0} + a_{1} X = 0 \to X = - \frac{a_{0}}{a_{1}} = \frac{3}{8} p L = \frac{3}{8} 4 \times 4 = 6.0 kN$

Superposition is applied to give the moment diagram of the statically indeterminate structure:

$M = M_{0} + M_{1} X = M_{0} + \frac{3}{8} M_{1}$

The maximum moment arises at the built-in support

$M_{\max}^{} = - \frac{p L^{2}}{2} + \frac{3}{8} p L^{2} = - \frac{p L^{2}}{8} = - 8 kNm$

Problem c)

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A hinge is introduced above the middle support, thus the primary structure becomes two simply supported girders. The redundant is a moment couple, X shown in the Figure.

Similarly to the solution of Problem 9.1.

The relative rotation above the middle support from the original load and from the redundant is calculated by Castigliano’s theorem:

See Problem 6.6.

a_{0} = \int \frac{M_{0} M_{1}}{E I} = - \frac{1}{E I} \frac{F L}{4} \frac{L}{2} \frac{1}{2} = - \frac{F L^{2}}{16 E I} a_{1} = \int \frac{M_{1} M_{1}}{E I} = 2 \frac{1 \times L}{2 E I} \frac{2}{3} = \frac{2 L}{3 E I}

The redundant is determined from the compatibility condition.The relative rotation above the middle support must be zero:

$a_{0} + a_{1} X = 0 \to X = \frac{a_{0}}{a_{1}} = \frac{3}{32} F L = \frac{3}{32} 15 \times 4 = 5.625 kNm$

Superposition is applied to give the moment diagram of the statically indeterminate structure.

$M = M_{0} + M_{1} X = M_{0} + 5.625 M_{1}$

The maximum negative moment above the support is

$M_{\max}^{-} = - X = - 5.625 kNm$

The maximum positive moment below the force is

$M_{\max}^{+} = \frac{F L}{4} - \frac{X}{2} = \frac{15.0 \times 4.0}{4} - \frac{5.625}{2} = 12.19 kNm$

Problem d)

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The primary structure and the redundant is chosen similarly as in Problem b). On the cantilever a linearly distributed load, and a concentrated end force act.

The deflection at the beam end from the original load and from the redundant is calculated. End deflection of a cantilever subjected to linearly distributed load is derived in Problem 6.7 a), end rotation from the redundant is equivalent to that of Problem b):
$a_{0} = v_{p} = \frac{p L^{4}}{30 E I} = \frac{4.0 \times 4 . 0^{2}}{30 E I} a_{1} = v_{X} = - \frac{1 \times L^{3}}{3 E I} = - \frac{64}{3 E I}$

The redundant is determined from the compatibility condition.

The deflection at the beam end must be zero

$a_{0} + a_{1} X = 0 \to X = - \frac{a_{0}}{a_{1}} = \frac{3}{30} p L = \frac{3}{30} 4.0 \times 4.0 = 1.6 kN$

Superposition is applied to give the moment diagram of the statically indeterminate structure.

$M = M_{0} + M_{1} X = M_{0} + 1.6 M_{1}$

The maximum moment arises at the built-in support

$M_{\max}^{-} = - \frac{p L^{2}}{6} + 1.6 L = - \frac{4.0 \times 4 . 0^{2}}{6} + 1.6 \times 4.0 = - 4.27 kNm$