Deflection of a hinged support

Problem description:

• A structure consisting of two equal steel bars, each of length \(l\) and cross-sectional area \(A\), with hinged ends is subjected to the action of a load \(F\). Determine the stress, \(\sigma\), in the bars and the deflection, \(\delta\), of point 2. Neglect the weight of the bars as a small quantity in comparison with the load \(F\).

Reference:

• S. Timoshenko, Strength of Materials, Part I, Elementary Theory and Problems, 3rd Edition, D. Van Nostrand Co., Inc., New York, NY, 1955, pg. 10, problem 2.

Analysis type(s):

• Static Analysis ANTYPE=0

Element type(s):

• 3-D Spar (or Truss) Elements (LINK180)

Material properties

• \(E = 30 \cdot 10^6 psi\)

Geometric properties:

• \(l = 15 ft\)

• \(A = 0.5 in^2\)

• \(\Theta = 30 ^\circ\)

Loading:

• \(F = 5000 lb\)

Analytical equations:

• The tensile force in the bars is \(S\) - \(S = \frac{P}{2 sin \Theta}\)

• The necessary cross-sectional area \(A\) is - \(A = \frac{S}{\sigma}\)

• The elongation of the bar \(AB\) is - \(B_1 D = \frac{\sigma l}{E}\)

• The deflection \(BB_1\) is - \(BB_1 = \frac{B_1 D}{sin \Theta}\)

Notes:

• Consistent length units are used. The dimensions \(a\) and \(b\) are calculated parametrically in the input as follows: - \(a = 2 l cos \Theta\), - \(b = l sin \Theta\).

# sphinx_gallery_thumbnail_path = '_static/vm4_setup.png'

Start MAPDL

from math import cos, pi, sin

from ansys.mapdl.core import launch_mapdl

# start mapdl and clear it
mapdl = launch_mapdl()
mapdl.clear()  # optional as MAPDL just started

# enter verification example mode and the pre-processing routine
mapdl.verify()
mapdl.prep7()

*****MAPDL VERIFICATION RUN ONLY*****
     DO NOT USE RESULTS FOR PRODUCTION

          ***** MAPDL ANALYSIS DEFINITION (PREP7) *****

Define material

Create a simple hinge geometry. We use the LINK180 element type to model this and an elastic modulus of 30e6. We store the x-coordinate of node 3 and the y-coordinate of node 2 for ease of use later on.

length_bar = 15 * 12
theta = 30
theta_rad = theta * pi / 180.0
node3_x = 2 * length_bar * cos(theta_rad)
node2_y = length_bar * sin(theta_rad)

mapdl.et(1, "LINK180")
mapdl.sectype(1, "LINK")
mapdl.secdata(0.5)
mapdl.mp("EX", 1, 30e6)

MATERIAL          1     EX   =  0.3000000E+08

Define geometry

We create three nodes in an isosceles triangle shape, with elements along the equal sides, forming a hinge.

n1 = mapdl.n(1, 0, 0, 0)
n2 = mapdl.n(2, node3_x * 0.5, -node2_y, 0)
n3 = mapdl.n(3, node3_x, 0, 0)

mapdl.e(n1, n2)
mapdl.e(n2, n3)
mapdl.eplot(show_node_numbering=True, line_width=5, cpos="xy")

Define boundary conditions

• Fix nodes 1 and 3 in place

• Apply a force of -5000 in the negative y-direction to node 2

• Then finish the prep7 section

mapdl.d(1, "ALL", "", "", 3, 2)
mapdl.f(2, "FY", -5000)
mapdl.finish()

***** ROUTINE COMPLETED *****  CP =         0.000

Solve

Enter solution mode and solve the system.

mapdl.run("/SOLU")
out = mapdl.solve()
mapdl.finish()

FINISH SOLUTION PROCESSING


 ***** ROUTINE COMPLETED *****  CP =         0.000

Post-processing

Enter post-processing, get the results and view the nodal displacement as well as the equivalent stress on the nodes.

We make the line width larger for ease of visualization as well as using two perceptually linear colormaps to enhance display of the data.

mapdl.post1()
mapdl.post_processing.plot_nodal_displacement(
    "Y",
    cmap="magma",
    line_width=5,
    cpos="xy",
    scalar_bar_args={"title": "Displacement", "vertical": False},
)

Principal nodal stress

Use the post_processing attribute to get the principal nodal stress as an array.

Note

This returns the same data as prnsol, except instead of returning text, it returns a numpy array.

seqv = mapdl.post_processing.nodal_eqv_stress()

# print out the nodes
for i, nnum in enumerate(mapdl.mesh.nnum):
    print(f"Node {nnum} : {seqv[i]} psi")

# Which is identical to:
# print(mapdl.prnsol('S', 'PRIN'))

Node 1 : 10000.0 psi
Node 2 : 10000.0 psi
Node 3 : 10000.0 psi

Check results

Now that we have the results we can compare the nodal displacement and stress experienced by node 2 to the known quantities 10000 psi and -0.12 inches. To do this we:

• Find the mid-node from the coordinates using the Query class

• Get the y-displacement from node 2

• Get the element nearest to node 2

• Get the stress on this element

• Compare

q = mapdl.queries
mid_node = q.node(node3_x * 0.5, -node2_y, 0)
displacement = mapdl.get_value("NODE", mid_node, "U", "Y")
left_element = q.enearn(mid_node)
mapdl.etable("STRS", "LS", 1)
stress = mapdl.get_value("ELEM", left_element, "ETAB", "STRS")

results = f"""
---------------------  RESULTS COMPARISON  -----------------------
|   TARGET         |  TARGET     |   Mechanical APDL   |   RATIO
------------------------------------------------------------------
Stress [psi]          10000          {stress}               {stress/10000:.2f}
Displacement [in]     -0.12          {displacement:.2f}     {abs(displacement) / 0.12:.2f}
------------------------------------------------------------------
"""

print(results)

---------------------  RESULTS COMPARISON  -----------------------
|   TARGET         |  TARGET     |   Mechanical APDL   |   RATIO
------------------------------------------------------------------
Stress [psi]          10000          10000.0               1.00
Displacement [in]     -0.12          -0.12     1.00
------------------------------------------------------------------

Stop MAPDL.

mapdl.exit()