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<!-- source-page: 221 -->
![](images/page-221_8b0de7ddc60f39c48464979e61bf4275fba5def4128c46743dafe3738e1671af.jpg)
<details>
<summary>text_image</summary>
ŷ, v̂
w
m̂(0)
L
m̂(L)
V̂(0)
V̂(L)
</details>
Figure 436 Beam element with shear forces, moments, and a distributed load
![](images/page-221_192b37a8f66d814be926c28c4705b3798d854d97215c1b808dbfb5f1cd57b50b.jpg)
<details>
<summary>text_image</summary>
V̂(L)
V̂(0)
V̂(L)
V̂(0)
</details>
Figure 437 Shear forces and moments acting on adjacent elements meeting at a node
Note that when element matrices are assembled, two shear forces and two moments from adjacent elements contribute to the concentrated force and concentrated moment at the node common to the adjacent elements as shown in Figure 437. These concentrated shear forces $\hat { V } ( 0 ) - \hat { V } ( L )$ and moments $\hat { m } ( L ) - \hat { m } ( 0 )$ are often zero; that is, $\hat { V } ( 0 ) = \hat { V } ( L )$ and $\hat { m } ( L ) = \hat { m } ( 0 )$ occur except when a concentrated nodal force or moment exists at the node. In the actual computations, we handle the expressions given by Eq. (4.8.9) by including them as concentrated nodal values making up the matrix fPg.
# References
[1] Gere, J. M., Mechanics of Materials, 5th ed., Brooks/Cole Publishers, Pacific Grove, CA, 2001.
[2] Hsieh, Y. Y., Elementary Theory of Structures, 2nd ed., Prentice-Hall, Englewood Cliffs, NJ, 1982.
[3] Fraeijes de Veubeke, B., ‘‘Upper and Lower Bounds in Matrix Structural Analysis,’’ Matrix Methods of Structural Analysis, AGAR Dograph 72, B. Fraeijes de Veubeke, ed., Macmillan, New York, 1964.
[4] Juvinall, R. C., and Marshek, K. M., Fundamentals of Machine Component Design, 4th. ed., John Wiley & Sons, New York, 2006.
[5] Przemieneicki, J. S., Theory of Matrix Structural Analysis, McGraw-Hill, New York, 1968.
[6] McGuire, W., and Gallagher, R. H., Matrix Structural Analysis, John Wiley & Sons, New York, 1979.
[7] Severn, R. T., ‘‘Inclusion of Shear Deflection in the Stiffness Matrix for a Beam Element’’, Journal of Strain Analysis, Vol. 5, No. 4, 1970, pp. 239241.
[8] Narayanaswami, R., and Adelman, H. M., ‘‘Inclusion of Transverse Shear Deformation in Finite Element Displacement Formulations’’, AIAA Journal, Vol. 12, No. 11, 1974, 16131614.
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[9] Timoshenko, S., Vibration Problems in Engineering, 3rd. ed., Van Nostrand Reinhold Company, 1955.
[10] Clark, S. K., Dynamics of Continous Elements, Prentice Hall, 1972.
[11] Algor Interactive Systems, 260 Alpha Dr., Pittsburgh, PA 15238.
# Problems
4.1 Use Eqs. (4.1.7) to plot the shape functions $N _ { 1 }$ and $N _ { 3 }$ and the derivatives $( d N _ { 2 } / d \hat { x } )$ and $( d N _ { 4 } / d \hat { x } )$ , which represent the shapes (variations) of the slopes $\hat { \phi } _ { 1 }$ and $\hat { \phi } _ { 2 }$ over the length of the beam element.
4.2 Derive the element stiffness matrix for the beam element in Figure 41 if the rotational degrees of freedom are assumed positive clockwise instead of counterclockwise. Compare the two different nodal sign conventions and discuss. Compare the resulting stiffness matrix to Eq. (4.1.14).
Solve all problems using the finite element stiffness method.
4.3 For the beam shown in Figure P43, determine the rotation at pin support A and the rotation and displacement under the load P. Determine the reactions. Draw the shear force and bending moment diagrams. Let EI be constant throughout the beam.
![](images/page-222_00f1e8191f8576fc0152276bc2280298128cb768cdb8ca6bde45aa67bace0969.jpg)
<details>
<summary>text_image</summary>
L/2
P
L/2
A
B
</details>
Figure P43
![](images/page-222_ba1733f264193a43a68bf404f0d769083444bf1e6d745ae25a29326726c60a0e.jpg)
<details>
<summary>text_image</summary>
P
A
L
B
</details>
Figure P44
4.4 For the cantilever beam subjected to the free-end load P shown in Figure P44, determine the maximum deflection and the reactions. Let EI be constant throughout the beam.
4.54.11 For the beams shown in Figures P45—P411, determine the displacements and the slopes at the nodes, the forces in each element, and the reactions. Also, draw the shear force and bending moment diagrams.
![](images/page-222_f87c4eed1cc4abad3c752890b4946db3850049b4c7e9de80dc0dcc033b65ca3d.jpg)
<details>
<summary>text_image</summary>
1 kip
1
2
3
E = 30 × 10⁶ psi
I = 100 in⁴
20 ft
20 ft
</details>
Figure P45
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![](images/page-223_f1e296edd3e98a9205213bcd4c508c3d848bc31b62f82dde23d75b1aafba28b9.jpg)
<details>
<summary>text_image</summary>
1 kip
1
3
0.5 in. gap
2
20 ft
20 ft
</details>
$$
E = 3 0 \times 1 0 ^ {6} \mathrm{psi}
$$
$$
I = 1 0 0 \mathrm{in} ^ {4}
$$
(Compare answers with P4-5.)
Figure P46
![](images/page-223_e496e869737de68945508492ee7e21c6eba8895e98fd3e943c1d5d5cbffb7a4f.jpg)
<details>
<summary>text_image</summary>
1
2
2l
5 ft
5 kip
3
5 kip
I
5 ft
</details>
$$
E = 3 0 \times 1 0 ^ {6} \mathrm{psi}
$$
$$
I = 2 0 0 \mathrm{in} ^ {4}
$$
Figure P47
![](images/page-223_7f1f06143663a31cca2c0798aeeca44278b5dd26bec9afca693a6cdf37452810.jpg)
<details>
<summary>text_image</summary>
10 kN
2
20 kN · m
3
3 m
3 m
</details>
$$
E = 2 1 0 \mathrm{GPa}
$$
$$
I = 4 \times 1 0 ^ {- 4} \mathrm{m} ^ {4}
$$
Figure P48
![](images/page-223_9b24af802e8eec64e32ecf6cfa8cafff752b7d277807e3387006afb216003ece.jpg)
<details>
<summary>text_image</summary>
8 kN
— 2 m
3
1
2
4
5
3 m
4 m
3 m
</details>
$$
E = 7 0 \mathrm{GPa}
$$
$$
I = 1 \times 1 0 ^ {- 4} \mathrm{m} ^ {4}
$$
Figure P49
![](images/page-223_b296fd68beb9b79ce8a3b469e72b3c861d9d4ba51ef76b4d894183722eecfb46.jpg)
<details>
<summary>text_image</summary>
1
20 ft
2 kip
2
k = 1000 lb/in.
3
E = 29 × 10⁶ psi
I = 200 in⁴
</details>
Figure P410
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![](images/page-224_fb8fbcb1617a206ed55ef246b6694e99021ada714dc69342f15c895a55f39594.jpg)
<details>
<summary>text_image</summary>
12 kN
4 m 2 4 m
k = 200 kN/m
1 3
4
</details>
$$
\begin{array}{r l} E & = 7 0 \mathrm{GPa} \\ I & = 2 \times 1 0 ^ {- 4} \mathrm{m} ^ {4} \end{array}
$$
Figure P411
4.12 For the fixed-fixed beam subjected to the uniform load w shown in Figure P412, determine the midspan deflection and the reactions. Draw the shear force and bending moment diagrams. The middle section of the beam has a bending stiffness of 2EI; the other sections have bending stiffnesses of EI .
![](images/page-224_b2705ee513cd5237fc9af165eed299a44a0f018ed3e1c881795cbfe2764376f4.jpg)
<details>
<summary>text_image</summary>
w
A
L/3
L/3
L/3
B
</details>
Figure P412
4.13 Determine the midspan deflection and the reactions and draw the shear force and bending moment diagrams for the fixed-fixed beam subjected to uniformly distributed load w shown in Figure P413. Assume EI constant throughout the beam. Compare your answers with the classical solution (that is, with the appropriate equivalent joint forces given in Appendix D).
![](images/page-224_291b5daa16e3969625b1d5dd9fae542f041e407f57437644941cd9d0257490b0.jpg)
<details>
<summary>text_image</summary>
w
A
L
B
</details>
Figure P413
![](images/page-224_d46327fbc7af6a42bcb0caec9f10d388569f982ac9cc37f4a6c5a5c7052edbce.jpg)
<details>
<summary>text_image</summary>
w
A L B
O
</details>
Figure P414
4.14 Determine the midspan deflection and the reactions and draw the shear force and bending moment diagrams for the simply supported beam subjected to the uniformly distributed load w shown in Figure P414. Assume EI constant throughout the beam.
4.15 For the beam loaded as shown in Figure P415, determine the free-end deflection and the reactions and draw the shear force and bending moment diagrams. Assume EI constant throughout the beam.
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![](images/page-225_d07c7c0ca51f3ddc7e7755956e25ff9a399b9fc2230710ff3162b564356cac9a.jpg)
<details>
<summary>text_image</summary>
A
L
B
w
L
C
</details>
Figure P415
![](images/page-225_f482b9bd16ad6750c075d305fc6ac14c1c4c84081f2200620cc863f8893e5cea.jpg)
<details>
<summary>text_image</summary>
A
L
B
w
</details>
Figure P416
4.16 Using the concept of work equivalence, determine the nodal forces and moments (called equivalent nodal forces) used to replace the linearly varying distributed load shown in Figure P416.
4.17 For the beam shown in Figure 417, determine the displacement and slope at the center and the reactions. The load is symmetrical with respect to the center of the beam. Assume EI constant throughout the beam.
![](images/page-225_55e98ad15033def6f1d6d43aa80928924c89dbcfd3978b6b6e9f681420323fda.jpg)
<details>
<summary>text_image</summary>
w
L
</details>
Figure P417
4.18 For the beam subjected to the linearly varying line load w shown in Figure P418, determine the right-end rotation and the reactions. Assume EI constant throughout the beam.
![](images/page-225_9921976398ebe58f7e0145c75bfd96c678423f5bedb516308785372ca2c2cc26.jpg)
<details>
<summary>text_image</summary>
A
L
B
w
</details>
Figure P418
4.194.24 For the beams shown in Figures P419—P424, determine the nodal displacements and slopes, the forces in each element, and the reactions.
![](images/page-225_3e167203cbf577511e81a878466658c1d5fca8ea81d601f70833696ad9504b1a.jpg)
<details>
<summary>text_image</summary>
8 kN/m
1
2
3
4 m
4 m
</details>
$$
E = 7 0 \mathrm{GPa}
$$
$$
I = 3 \times 1 0 ^ {- 4} \mathrm{m} ^ {4}
$$
Figure P419
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![](images/page-226_8f932930e8e3048e7e96b80ccd3701027bae27fc27e2ff96bd3aeffb2a9db522.jpg)
<details>
<summary>text_image</summary>
10 kN/m
10 kN/m
E = 210 GPa
I = 4 × 10⁻⁴ m⁴
1
2
3
6 m
6 m
k = 10 kN/m
4
</details>
Figure P420
![](images/page-226_5288ecab8508b06d480a7c32d74715020275046c8f402a52b3dec0f6acb71d16.jpg)
<details>
<summary>text_image</summary>
2000 lb/ft
E = 29 × 10⁶ psi
I = 200 in⁴
1
2
3
15 ft
15 ft
</details>
Figure P421
![](images/page-226_2fc967150950a9378150517b93952eea84e867145896782db8daefbbbea9d92e.jpg)
<details>
<summary>text_image</summary>
4000 lb/ft
E = 29 × 10⁶ psi
I = 150 in⁴
1
2
3
10 ft
10 ft
</details>
Figure P422
![](images/page-226_b5ff79192c7ff770b7e04ea57cdf54f975e8dd3a50a941c5ef81d13178971d67.jpg)
<details>
<summary>text_image</summary>
500 lb/ft
E = 1.6 × 10⁶ psi
I = 100 in⁴
1
2
3
20 ft
20 ft
k = 4000 lb/in.
4
</details>
Figure P423
![](images/page-226_e321d8da30a16e388c028e9ab1258026c3c15bbf9bd589939f929abf5e0fb415.jpg)
<details>
<summary>text_image</summary>
5000 N/m
E = 210 GPa
I = 2 × 10⁻⁴ m⁴
1
2
3
5 m
4 m
</details>
Figure P424
<!-- source-page: 227 -->
4.2531 For the beams shown in Figures P425—P430, determine the maximum deflection and maximum bending stress. Let E ¼ 200 GPa or $3 0 \times 1 0 ^ { 6 }$ psi for all beams as appropriate for the rest of the units in the problem. Let c be the half-depth of each beam.
![](images/page-227_cb1cb2b8ce2b868d7feb068b7b1aaef68cbbf7e3d2ae3f8d08a832c210731a84.jpg)
<details>
<summary>text_image</summary>
w = 10 kN/m
A B C
4 m 4 m
c = 0.25m, I = 100 × 10⁻⁶ m⁴
</details>
Figure P425
![](images/page-227_dcf4ee8fa275740acacdefa43b49c1680ea68091158d7de90fef072ce577370d.jpg)
<details>
<summary>text_image</summary>
30 kN/m
A B C
10 m 20 m
I 2I
c = 0.25m, I = 500(10^-6) m^4
</details>
Figure P426
![](images/page-227_3c72d7a387da3d5846c98999f10e6bd02d3ba2cb38819cb50560389059ee8428.jpg)
<details>
<summary>text_image</summary>
75 k
2 kip/ft
A B C D
15 ft 15 ft 30 ft
3I I
c = 10 in., I = 500 in.⁴
</details>
Figure P427
![](images/page-227_6bcbcd66e48be53a465e72f11f92b05c6f15dd16b9fba57fb9ba60637aaad850.jpg)
<details>
<summary>text_image</summary>
A
25 kN/m
B
C
10 m
5 m
c = 0.30m, I = 700 × 10⁻⁶ m⁴
</details>
Figure P428
![](images/page-227_64f77304ada72fcf4aa49a7446b8797d08aca6b537b98dee75d31dda8cd1b7e0.jpg)
<details>
<summary>text_image</summary>
1.5 kip/ft
A
B
C
10 ft
10 ft
c = 10 in., I = 400 in.⁴
</details>
Figure P429
![](images/page-227_23432cd08795b471cfaafa244bac3f69ea27e336ae48d933b7199448529232d2.jpg)
<details>
<summary>text_image</summary>
10 kN/m
100 kN
A
B
C
12 m
I
6 m
2I
c = 0.30m, I = 700 × 10⁻⁶ m⁴
</details>
Figure P430
For the beam design problems shown in Figures P431 through P436, determine the size of beam to support the loads shown, based on requirements listed next to each beam.
![](images/page-227_7a020220d223614f0ba9774574315104d33f4ee17652523b549c19a49b3e72cd.jpg)
4.31 Design a beam of ASTM A36 steel with allowable bending stress of 160 MPa to support the load shown in Figure P431. Assume a standard wide flange beam from Appendix F or some other source can be used.
![](images/page-227_8e8fc519ca4c8310e26a549abeb024cffa2bc20106dfc0aeab1021256ea340ab.jpg)
<details>
<summary>text_image</summary>
w = 30 kN/m
4 m
4 m
</details>
Figure P431
![](images/page-227_7a020220d223614f0ba9774574315104d33f4ee17652523b549c19a49b3e72cd.jpg)
<!-- source-page: 228 -->
![](images/page-228_01158640be4132101738eef4b9fdfecb066575ac7293486177b23fc2a18bbe43.jpg)
4.32 Select a standard steel pipe from Appendix F to support the load shown. The allowable bending stress must not exceed 24 ksi, and the allowable deflection must not exceed $\mathrm { L } / 3 6 0$ of any span.
![](images/page-228_1b401ce477e4243dd944ae11e49714d708dfe5ffb699e257d11d29a7887bbb52.jpg)
<details>
<summary>text_image</summary>
500 lb
500 lb
500 lb
6 ft
6 ft
6 ft
</details>
Figure P432
![](images/page-228_91560b5f05cc44780d70192f876980e927b2088161a86b6a8a9d9798524af160.jpg)
4.33 Select a rectangular structural tube from Appendix F to support the loads shown for the beam in Figure P433. The allowable bending stress should not exceed 24 ksi.
![](images/page-228_6d2eeee3438716002501f05ef3eb32d017c6f95d3031071c41f249855324449b.jpg)
<details>
<summary>text_image</summary>
1 kip
6 ft
6 ft
</details>
Figure P433
![](images/page-228_4d495614dbbabf798402bb1d61ff8a4de1727da72b7b0a7c50ead69446b54f35.jpg)
4.34 Select a standard W section from Appendix F or some other source to support the loads shown for the beam in Figure P434. The bending stress must not exceed 160 MPa.
![](images/page-228_5c3391fb3d072f3c2a30c02e49da24f7cfd0d2b9e4c38b2d88f113481fdd8959.jpg)
<details>
<summary>text_image</summary>
20 kN/m
60 m 60 m 60 m
</details>
Figure P434
![](images/page-228_e656152b6e2cec630ccc3da86ba9cf3f740ac0787b23f8054fbdf1e1025b0414.jpg)
4.35 For the beam shown in Figure P435, determine a suitable sized W section from Appendix F or from another suitable source such that the bending stress does not exceed 150 MPa and the maximum deflection does not exceed $\mathrm { L } / 3 6 0$ of any span.
![](images/page-228_65d5eebeb5dfb3f3ab2e941e2f23b7c6dc23a7e4ef3cb7d5743090bc3f80726c.jpg)
<details>
<summary>text_image</summary>
70 kN 70 kN 17 kN
2.5 m 2.5 m
10 m 5 m 10 m
</details>
Figure P435
<!-- source-page: 229 -->
![](images/page-229_c1cac9f5d52425c6a3bf3d6f25c7e0e1c7305a1cac03e4858b3fdccdb14ca1cb.jpg)
4.36 For the stepped shaft shown in Figure P436, determine a solid circular cross section for each section shown such that the bending stress does not exceed 160 MPa and the maximum deflection does not exceed L/360 of the span.
![](images/page-229_f15d251c11c42bd75f1c367eb7e94b543f11d71784654d230344c74ee50006c8.jpg)
<details>
<summary>text_image</summary>
200 kN
A B C D E
3 m 3 m 3 m 3 m
</details>
Figure P436
4.37 For the beam shown in Figure P437 subjected to the concentrated load P and distributed load w, determine the midspan displacement and the reactions. Let EI be constant throughout the beam.
![](images/page-229_f2fafca757ac1828b1caca0df27be12f1f964578bbbc4c9a5f76e41a07e94c27.jpg)
<details>
<summary>text_image</summary>
L/2
P
w
2
1
L
3
</details>
Figure P437
![](images/page-229_2413609cf961d554a21680320bcc7a7d5fa3d591686c0a76a8181a2010e239b1.jpg)
<details>
<summary>text_image</summary>
L/3
P
L/3
P
L/3
L
</details>
Figure P438
4.38 For the beam shown in Figure P438 subjected to the two concentrated loads P, determine the deflection at the midspan. Use the equivalent load replacement method. Let EI be constant throughout the beam.
4.39 For the beam shown in Figure P439 subjected to the concentrated load P and the linearly varying line load w, determine the free-end deflection and rotation and the reactions. Use the equivalent load replacement method. Let EI be constant throughout the beam.
![](images/page-229_e2ab33963a0de3c6dfa28df677c9c5c63dc67940f41fe856cdab6076d15fcbd4.jpg)
<details>
<summary>text_image</summary>
L/2
P
w
L
</details>
Figure P439
![](images/page-229_d6053888e041bc3285805fada12221b43bbe4bd417481121e03a297041c2134d.jpg)
<details>
<summary>text_image</summary>
P = 5 kN
Hinge
2 m
2 m
</details>
Figure P440
4.4042 For the beams shown in Figures P440—P442, with internal hinge, determine the deflection at the hinge. Let E ¼ 210 GPa and $I = 2 \times 1 0 ^ { - 4 } ~ \mathrm { m } ^ { 4 }$ .
<!-- source-page: 230 -->
![](images/page-230_dd42e913289b76c2c1030aba8fa8972844580e6402b12ed5560cbc19905cd9f6.jpg)
<details>
<summary>text_image</summary>
P = 5 kN
Hinge
2 m
2 m
</details>
Figure P441
![](images/page-230_4dfd2107c06d84634eca9272c68c4729263e17de7d85ba3e0e9f0169dff64256.jpg)
<details>
<summary>text_image</summary>
10 kN/m
Hinge
2 m
1 m
1 m
</details>
Figure P442
4.43 Derive the stiffness matrix for a beam element with a nodal linkage—that is, the shear is 0 at node i, but the usual shear and moment resistance are present at node j (see Figure P443).
![](images/page-230_4e7aea178235775d7aa5b2fe4ffe0aff6e3cb4916ae7c70f93a3c6cc0814621c.jpg)
<details>
<summary>text_image</summary>
m̂_i
i
L
j
m̂_j
f̂_iy = 0
f̂_jy
</details>
Figure P443
4.44 Develop the stiffness matrix for a fictitious pure shear panel element (Figure P444) in terms of the shear modulus, G the shear web area, $A _ { W }$ , and the length, L. Notice the Y and v are the shear force and transverse displacement at each node, respectively.
$\mathrm { G i v e n ~ ~ \xi ~ 1 \ / } \ \tau = G _ { \gamma } , \quad \mathrm { 2 ) } \ Y = \tau _ { \mathrm { A w } } , \quad \mathrm { 3 ) } \ Y _ { \mathrm { 1 } } + Y _ { \mathrm { 2 } } = 0 , \quad \mathrm { 4 ) } \ \gamma = \frac { v _ { \mathrm { 2 } } - v _ { \mathrm { 1 } } } { \cal L }$
![](images/page-230_2152c3221b24061aee8db30e2fef305ce80e64a31b087d59f827c543ec2008dd.jpg)
<details>
<summary>text_image</summary>
L
1
2
Y₁,v₁
Y₂,v₂
</details>
Positive node force sign convention
![](images/page-230_b2f2d1e416ca92caf4aa519a8c7f16edee96b4f3f809b37ef98f5b3c9dc41432.jpg)
<details>
<summary>text_image</summary>
L
Y
Y
</details>
Element in equilibrium (neglect moments)
Figure P444
4.45 Explicitly evaluate $\pi _ { p }$ of Eq. (4.7.15); then differentiate $\pi _ { p }$ with respect to $\hat { d } _ { 1 y } , \hat { \phi } _ { 1 } , \hat { d } _ { 2 y }$ , and $\hat { \phi } _ { 2 }$ and set each of these equations to zero (that is, minimize $\pi _ { p } )$ to obtain the four element equations for the beam element. Then express these equations in matrix form.
4.46 Determine the free-end deflection for the tapered beam shown in Figure P446. Here $I ( x ) = I _ { 0 } ( 1 + n x / L )$ where $I _ { 0 }$ is the moment of inertia at $x = 0 .$ . Compare the exact beam theory solution with a two-element finite element solution for $n = 2$ .
![](images/page-230_9aa43b91b049cf87ae2833842e8f3379ea790763b69b1cb5dbfa609aa8aff1de.jpg)
<details>
<summary>text_image</summary>
M
x
L
</details>
Figure P446
![](images/page-230_4002e08a152ee8e8347e64efbe01b92bb85271b1e497c91e2145b999bea74ee9.jpg)
<details>
<summary>text_image</summary>
w(x)
L
x
k_f
</details>
Figure P447