Formulas for Simple Frames | Engineering Library (2024)

The content of this page is taken from the "Stress Analysis Manual," Air Force Flight Dynamics Laboratory, October 1986. Other related chapters from the Air Force "Stress Analysis Manual" can be seen above.

Cases 1 through 9 of Table 5-3 give reaction forces and moments on rectangular frames, both of whose uprights are pinned; and such frames with both uprights fixed are treated by cases 9 through 18 of this table. Table 5-4 gives reaction forces and moments on trapezoidal frames. The first four cases treat such frames with both uprights pinned at the ends, and cases 5 and 6 treat trapezoidal frames with fixed- ended uprights. Table 5-5 gives reaction forces and moments on triangular frames.

CaseFrameForces & Moments1Formulas for Simple Frames | Engineering Library (1)
$$ V_A = { P b \over L } $$$$ V_B = P - V_A $$
$$ H_A = H_B = { 3 P L \over 8 h (2K + 3) } $$

For Special Case: \( a = b = L/2 \)

$$ V_A = V_B = {P \over 2} $$Formulas for Simple Frames | Engineering Library (2)
$$ H_A = H_B = { 3 P L \over 8 h (2K + 3) } $$2Formulas for Simple Frames | Engineering Library (3)
$$ V_A = { wb (b + 2c) \over 2L } $$$$ V_B = wb - V_A $$
$$ H_A = H_B = { wb [6ac + b(3L - 2b)] \over 4hL(2K+3) } $$

For Special Case: \( a = c = 0 \), \( b = L \)

$$ V_A = V_B = {wL \over 2} $$Formulas for Simple Frames | Engineering Library (4)
$$ H_A = H_B = { wL^2 \over 4h(2K + 3) } $$3Formulas for Simple Frames | Engineering Library (5)
$$ V_A = { wbd \over 2L } $$$$ V_B = { wb \over 2 } - V_A $$
$$ H_A = H_B = { 3wb \over 4Lh (2K + 3) } \left( dL - { b^2 \over 18 } - d^2 \right) $$

For Special Case: \( a = 0 \), \( b = L \), \( d = { L \over 3 } \)

$$ V_A = {wL \over 6} $$$$ V_B = {wL \over 2} $$Formulas for Simple Frames | Engineering Library (6)
$$ H_A = H_B = { wL^2 \over 8h(2K + 3) } $$4Formulas for Simple Frames | Engineering Library (7)

$$ V_A = -V_B = { -M \over L } $$

$$ H_A = H_B = { 3 (b - L/2) M \over Lh (2K + 3) } $$

5Formulas for Simple Frames | Engineering Library (8)

$$ V_A = -V_B = { -Pa \over L } $$

$$ H_B = { Pa \over 2h } \left[ { bK(a+h) \over h^2 (2K+3) } + 1 \right] $$

$$ H_A = H_B - P $$

6Formulas for Simple Frames | Engineering Library (9)

$$ V_A = -V_B = { w(b^2 - a^2) \over 2L } $$

$$ H_B = { w (a^2 - b^2) \over 4h } + { K [ w (a^2 - b^2) (2h^2 - a^2 - b^2) ] \over 8 h^3 (2K + 3) } $$

$$ H_A = H_B + w(b-a) $$


For Special Case: \( b = 0 \), \( a = h \)

$$ V_A = -V_B = { -wh^2 \over 2L } $$Formulas for Simple Frames | Engineering Library (10)

$$ H_B = { wh \over 4 } \left[ 1 + { K \over 2 (2K + 3) } \right] $$

$$ H_A = H_B - wh $$

7Formulas for Simple Frames | Engineering Library (11)

$$ V_A = -V_B = {w \over 6L} (2a + b) (b-a) $$

$$ H_B = { -V_A L \over 2h } + { K X_{10} \over h (2K + 3) } $$

where:

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$$\begin{eqnarray}X_{10} &=& { w \over 120 h^2 (a-b) } ~[-30 h^2 b (a^2 - b^2) \nonumber \\&+& 20 h^2 (a^2 - b^2) + 15 b (a^4 - b^4) - 12 (a^5 - b^5)]\end{eqnarray}$$

$$ H_A = H_B + { w (b-a) \over 2 } $$


For Special Case: \( b = 0 \), \( a = L \)

$$ V_A = -V_B = { -wh^2 \over 3L } $$Formulas for Simple Frames | Engineering Library (12)

$$ H_B = { wh \over 10 } \left({ 4K+5 \over 2K + 3 }\right) $$

$$ H_A = H_B - {wh \over 2} $$

8Formulas for Simple Frames | Engineering Library (13)

$$ V_A = -V_B = {-w \over 6L} (a^2 + ac - 2c^2) $$

$$ H_B = { -V_A L \over 2h } + { K X_7 \over (2K + 3)h } $$

where:

$$\begin{eqnarray}X_7 &=& { w \over 120 h^2 (d-c) } ~[3 (4d^5 + c^5) - 15h (3d^4 + c^4) \nonumber \\&+& 20 h^2 (2d^3 + c^3) - 15cd^2 (2h-d)^2 ]\end{eqnarray}$$


For Special Case: \( b = c = 0 \), \( a = d = h \)

$$ V_A = -V_B = { -wh^2 \over 6L } $$Formulas for Simple Frames | Engineering Library (14)

$$ H_B = { wh \over 12 } \left[ 1 + { 7K \over 10(2K + 3) } \right] $$

9Formulas for Simple Frames | Engineering Library (15)

$$ V_A = -V_B = { -M \over L } $$

$$ H_A = H_B = { 3 [ K(2ab + a^2) + h^2 ] M \over 2h^3 (2K + 3) } $$

10Formulas for Simple Frames | Engineering Library (16)
$$ V_A = { Pb \over L } \left[ 1 + { a(b-a) \over L^2 (6K+1) } \right] $$$$ V_B = P - V_A $$

$$ H_A = H_B = { 3Pab \over 2Lh(K+2) } $$

$$ M_A = {Pab \over L} \left[ {1 \over 2(K+2)} - { (b-a) \over 2L(6K+1)) } \right] $$

$$ M_B = {Pab \over L} \left[ {1 \over 2(K+2)} + { (b-a) \over 2L(6K+1)) } \right] $$


For Special Case: \( a = b = L/2 \)

$$ V_A = V_B = P/2 $$Formulas for Simple Frames | Engineering Library (17)

$$ H_A = H_B = { 3PL \over 8h(K+2) } $$

$$ M_A = M_B = { PL \over 8(K+2) } $$

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11Formulas for Simple Frames | Engineering Library (18)
$$ V_A = {wcd \over L} + { X_1 - X_2 \over L(6K + 1) } $$$$ V_B = wc - V_A $$

$$ H_A = H_B = { 3 (X_1 + X_2) \over 2h(K+2) } $$

$$ M_A = { X_1 + X_2 \over 2(K+2) } - { X_1 - X_2 \over 2(6K + 1) } $$

$$ M_B = { X_1 + X_2 \over 2(K+2) } + { X_1 - X_2 \over 2(6K + 1) } $$

where:

$$ X_1 = {-wc \over 24L} \left[ {24d^3 \over L} - {6bc^2 \over L} + {3c^2 \over L} + 4c^2 - 24d^2 \right] $$

$$ X_2 = {wc \over 24L} \left[ {24d^3 \over L} - {6bc^2 \over L} + {3c^2 \over L} + 2c^2 - 48d^2 + 24dL \right] $$


For Special Case: \( a = 0 \), \( c = b = L \), \( d = L/2 \)

$$ V_A = V_B = { wL \over 2 } $$Formulas for Simple Frames | Engineering Library (19)

$$ H_A = H_B = { wL^2 \over 4h(K+2) } $$

$$ M_A = M_B = { wL^2 \over 12(K+2) } $$

12Formulas for Simple Frames | Engineering Library (20)
$$ V_A = {wcd \over 2L} + { X_3 - X_4 \over L(6K + 1) } $$$$ V_B = {wc \over 2} - V_A $$

$$ H_A = H_B = { 3 (X_3 + X_4) \over 2h(K+2) } $$

$$ M_A = { X_3 + X_4 \over 2(K+2) } - { X_3 - X_4 \over 2(6K + 1) } $$

$$ M_B = { X_3 + X_4 \over 2(K+2) } + { X_3 - X_4 \over 2(6K + 1) } $$

where:

$$ X_3 = {-wc \over 2L} \left[ {d^3 \over L} + {c^2 \over 9} + {51c^3 \over 810L} + {c^2b \over 6L} - d^2 \right] $$

$$ X_4 = {wc \over 2L} \left[ {d^3 \over L} + {c^2 \over 18} + {51c^3 \over 810L} - {c^2b \over 6L} - 2d^2 + dL \right] $$


For Special Case: \( a = 0 \), \( c = b = L \), \( d = L/3 \)

$$ V_A = { wL \over 6 } \left[ 1 - {1 \over 10(6K+1)} \right] $$$$ V_B = { wL \over 3 } \left[ 1 + {1 \over 20(6K+1)} \right] $$Formulas for Simple Frames | Engineering Library (21)

$$ H_A = H_B = { wL^2 \over 8h(K+2) } $$

$$ M_A = { wL^2 \over 120 } \left( {5 \over K+2} + {1 \over 6K+1} \right) $$

$$ M_B = { wL^2 \over 120 } \left( {5 \over K+2} - {1 \over 6K+1} \right) $$

13Formulas for Simple Frames | Engineering Library (22)

$$ V_A = -V_B = { -6 (ab + L^2 K) M \over L^3 (6K + 1) } $$

$$ H_A = H_B = { 3 (b-a) M \over 2Lh (K+2) } $$

$$ M_A = M \left\{ { 6ab(K+2) - L[ a(7K+3) - b(5K-1) ] \over 2L^2 (K+2) (6K+1) } \right\} $$

$$ M_B = M + M_A + V_A L $$

14Formulas for Simple Frames | Engineering Library (23)

$$ V_A = -V_B = { -3 P a^2 K \over Lh(6K+1) } $$

$$ H_B = {Pa \over 2h} \left[ {h \over b} - { h + b + K(b-a) \over h(K+2) } \right] $$

$$ H_A = H_B - P $$

$$ M_A = {Pa \over 2h} \left[ {-b (h + b + bK) \over b(K+2)} - h + {3aK \over 6(K+1)} \right] $$

$$ M_B = {Pa \over 2h} \left[ {-b (h + b + bK) \over b(K+2)} + h - {3aK \over 6(K+1)} \right] $$

15Formulas for Simple Frames | Engineering Library (24)Formulas for Simple Frames | Engineering Library (25)16Formulas for Simple Frames | Engineering Library (26)Formulas for Simple Frames | Engineering Library (27)
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Given: The trapezoidal frame shown in Figure 5-14.

Find: The reaction forces and bending moments.

$$ F = { I_2 \over I_1 } \left({s \over b}\right) = { 1.5 \over 1 } \left({5 \sqrt{5} \over 15}\right) = 1.12 $$$$ G = 3 + 2F = 3 + 2(1.12) = 5.24 $$

$$ V_A = { P(a + d) \over L } = { 1000 (5 + 7.5) \over 25 } = 500 ~\text{lb.} $$$$ V_B = P - V_A = 1000 - 500 = 500 ~\text{lb.} $$$$ H_A = H_B = { P \over 2h } \left( a + { 3 c d \over b G } \right) = { 1000 \over 2(10) } \left[ 5 + { 3(7.5)(7.5) \over 15(5.24) } \right] = 357 ~\text{lb.} $$

A free-body diagram may be constructed for a section of the frame leg, as shown in Figure 5-15. Equating the sum of the moment to zero gives

A free-body diagram may now be drawn for a section of the horizontal portion of the frame to the left of the load as in Figure 5-16. Equating the sum of the moments to zero gives

By considering symmetry, the moment diagram of the given frame may be drawn as shown in Figure 5-17.

Formulas for Simple Frames | Engineering Library (2024)

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Support Force Formula

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Definition of frame

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Area and Perimeter Formula Chart
FiguresArea FormulaVariables
Area of RectangleArea = l × wl = length w = width
Area of SquareArea = a2a = sides of the square
Area of a TriangleArea = 1/2 b×hb = base h = height
Area of a CircleArea = πr2r = radius of the circle
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SI Units for Structural Engineers
PrefixSymbolMultlplying factor
kilok103
millim10-3
microIl10-6
nanon10-9
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Types of Framing

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The setBounds() method is used in such a situation to set the position and size. To specify the position and size of the components manually, the layout manager of the frame can be null.

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If v=0.650c. Solution: γ = 1/√1−v2/c2. =1/√1−(0.650c)/c2. = 1.32.

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For example, 24fps would equal 1/24ths per second while 50fps would equal 1/50ths per second. Film is typically shot at 24 frames per second while TV can be anywhere from 25-30 depending on the country and frame rates for video games range from 20-60fps.

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