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Q2 The beam shown in Figure Q2 is loaded with a triangular distributed load from A to B, and a uniform distributed load from B to C. The cross-section is a hollow rectangle, 60 mm x 120 mm, with constant wall thickness of 8mm. The material is Steel, with a Young's Modulus, $E =$ 200 GPa, Poisson's ratio, $\nu = 0.3$, and co-efficient of thermal expansion $\alpha = 12 \times 10^{-6}/^\circ C$. $x$ $w = 5 \text{ kN/m}$ $t = 8 \text{ mm}$ 60 mm $z$ $y$ 120 mm $w = 0 \text{ kN/m}$ A B C 2m 2m Cross-section Fig. Q2 (a) Draw a Free Body Diagram of the beam and determine the reactions at the supports. I (b) Use the Method of Sections to determine expressions for the bending moment along the beam length. (c) Determine the normal stress ($\sigma_x$) on the bottom of the beam at $x = 2.5 \text{ m}$. State whether it is tensile or compressive.

          Q2
The beam shown in Figure Q2 is loaded with a triangular distributed load from A to B, and a
uniform distributed load from B to C. The cross-section is a hollow rectangle, 60 mm x 120 mm,
with constant wall thickness of 8mm. The material is Steel, with a Young's Modulus, $E =$
200 GPa, Poisson's ratio, $\nu = 0.3$, and co-efficient of thermal expansion $\alpha = 12 \times 10^{-6}/^\circ C$.
$x$
$w = 5 \text{ kN/m}$
$t = 8 \text{ mm}$
60 mm
$z$
$y$
120 mm
$w = 0 \text{ kN/m}$
A
B
C
2m
2m
Cross-section
Fig. Q2
(a) Draw a Free Body Diagram of the beam and determine the reactions at the supports.
I
(b) Use the Method of Sections to determine expressions for the bending moment along the
beam length.
(c) Determine the normal stress ($\sigma_x$) on the bottom of the beam at $x = 2.5 \text{ m}$. State whether it is
tensile or compressive.

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Q2
The beam shown in Figure Q2 is loaded with a triangular distributed load from A to B, and a
uniform distributed load from B to C. The cross-section is a hollow rectangle, 60 mm x 120 mm,
with constant wall thickness of 8mm. The material is Steel, with a Young's Modulus, E =
200 GPa, Poisson's ratio, ν = 0.3, and co-efficient of thermal expansion α = 12 × 10^-6/^∘ C.
x
w = 5 kN/m
t = 8 mm
60 mm
z
y
120 mm
w = 0 kN/m
A
B
C
2m
2m
Cross-section
Fig. Q2
(a) Draw a Free Body Diagram of the beam and determine the reactions at the supports.
I
(b) Use the Method of Sections to determine expressions for the bending moment along the
beam length.
(c) Determine the normal stress () on the bottom of the beam at x = 2.5 m. State whether it is
tensile or compressive.

Added by Ignacio G.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Thanks for your feedback!

The answers for part (a) are Ax=0, cy=9.16 kN Ay=5.84 kN The answers for part (b) are Qab = -1.25xÂ² + 5.84 kN Mab = -0.42xÂ³ + 5.84x kNm Qbc = -5x + 10.84 kN Mbc = -2.5xÂ² + 10.84x - 3.36 Answers for part (c) = 107.7 MPa (tens) Full working out, please. Thanks Q2: The beam shown in Figure Q2 is loaded with a triangular distributed load from A to B and a uniform distributed load from B to C. The cross-section is a hollow rectangle, 60 mm x 120 mm with a constant wall thickness of 8 mm. The material is steel, with a Young's Modulus, E = 200 GPa, Poisson's ratio, v = 0.3, and coefficient of thermal expansion, a = 12 x 10^-6 /Â°C. 60 mm w = 5 kN/m 8 mm 120 mm w = 0 kN/m A B 2 m 2 m K < Cross-section Fig. Q2 (a) Draw a Free Body Diagram of the beam and determine the reactions at the supports. (b) Use the Method of Sections to determine expressions for the bending moment along the beam length. (c) Determine the normal stress (Ïƒx) on the bottom of the beam at x = 2.5 m. State whether it is tensile or compressive.

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Transcript

00:01 Hello students, it is given that the bending moment at a distance x from the fixed end a is given by mx is equal to minus w by 2 into l minus x the whole square.

00:20 So we want to determine the slope that is dy by dx and also the deflection y equations for a beam under a distributed load where l is the length of the beam and w is the load intensity.

00:46 So here in the first step we can equate the given bending moment equation with the bending moment expression from the flexural formula.

00:54 Therefore mx is equal to ei d square y by dx square.

01:03 This gives us the equations that is ei d square y by dx square is equal to minus of w by 2 into l minus x the whole square.

01:21 So let's mark this as equation number one.

01:25 Now in the step two of the equation we can integrate equation one with respect to with respect to x to get the slope dy dx in terms of x.

01:46 So for that ei into dy dx is equal to minus of integral w by 2 into l minus x the whole square dx.

02:03 So here solving the integral and adding the integration constant c1 will get ei dy by dx is equal to w by 6 l minus x the whole cube plus c1 and let's mark this as equation number two and now integrating the equation number two with respect to x to get the deflection equation y in terms of x we can write ei into y will be equal to minus of integral w by 6 into l minus x the whole cube into dx plus c1 x plus c2.

02:46 Therefore solving at the integral and adding the integration constant c2 we can write ei y will be equal to minus omega sorry minus w by 24 into l minus x the whole raised to 4 plus c1 by 2 x plus c2 and let's mark this as equation number three...

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