Question

the figure below. Let x be the distance measured from left Consider the simply supported beam shown end of the beam. 1. Determine the vertical reactions at A and C 2. Write the equations for shear and moment for the section of the member between B and C. 3. Draw the shear and moment diagrams for the entire beam, specifying values at changes in loading and locations where the shear is O. 48 KN 8 kN/m B с D internal pin (Moment release) -4 m 2 m

Answer 1

a)

48 KN 8 kN/m (m) 8 10

For Section BD

$\small \\\\\sum F_y=0 \\\\R_C+R_B=48 \\\\\sum M_B=0 \\\\48*6=R_B*4 \\\\R_B=72 \ KN \\\\R_C=-24 \ KN$

For section AB

$\small \\\\\sum F_y=0 \\\\R_C+R_A=-8*4=-32 \\\\R_A=-32-24\ \ \ \ \ (\because R_C \ act \ in \ \text{opp direction of beam BD } {R_C \ force}) \\\\R_A=-56 \ KN \\\\\sum M_B=0 \\\\56*4=8*4*2+M_A \\\\M_A=224-64 \\\\M_A=160 \ KN-m$

48 KN 160 kN-m 8 kN/m 1 56 KN 72 KN 1 (m) 0 8 10

b)

Shear for 0 sxs4 V 56 IN slm Slim /2 +(m) 10 0 1 Take a cut for 0 <r<4: Find the forces due to the DL from 1 = 0 to 1 = 4: The rectangular DL acts at distance of from the cut with a force of: 5 x h = 8 +8I + EF-0 -56 + [+8x) - V() = 0 V(x) = -56 +87 ..(a) = -56 +8 for 0<<<4 Shear for 4 sxs8 56 IN & Idum 32 2 + 8 10 Take a cut for 4 SIS8: + EF = 0 -56 +32-V (2)=0 V:() = -24 .. (a) = -24 for 428

Shear for 8 sxs 10 13 (0) 56 IN 72 IN & Idum ++ (m) 8 10 Take a cut for 8 <r<10: + SF, = 0 -56 +72+32 - V(+) = 0 V3 (2) - +48 ..V(x) - +48 for 85% <10 Find Bending Moments (M) and Draw BMD The 'Method of Sections' or "Method of Cuts' is used to determine the bending moment value at locations along the beam. You should aim to 'cut the beam between loads, supports and along distributed loads. After taking a cut you simply consider equilibirium about the point of the cut (i.e. let the sum of the moment about the cut equal zero and solve for the unknown bending moment M) If you follow this procedure for every cut'section then you will find the equation of bending moment for every segment of the beam and therefore you can create the bending moment diagram (BMD).

Moment for 0 sxs4 160 IdN-m M C 56 IN SA /2 (m) LO Take a cut for 0 <r<4: Find the moments due to the DL from I =0 to 1=4: The rectangular DL acts at distance of from the cut with a force of: 6 x 8 = $i = +8I So the moment force is: (+83) () + OEM, -0 +(-56) (1-0) +160+ [+4x*) - M (+) = 0 M. (a)- +160 -56% +4.r? .:M(x) = +160 -56+4.rº for OSE<4 Moment for 4 sxs8 160 Id-m M C 56 IN 2 ++ (m) 10 0 4 8 Take a cut for 4 SIS8: + OSM,=0 +(-56) (2-0)+160+[-64 +32x] - M(x) = 0 M(x) = +96 - 24.1 ...M (+) = +96 - 24 for 4SIS8

Moment for 8 sxs 10 160 IdN-m M C 56 IN 72 8 Im (m) + 0 8 T 10 Take a cut for 8 <r <10: + OEM,=0 +(-56)(x -0) + (72) (1 - 8) +160+ [-64 +32x) - M3 (7) = 0 M(I) = -480 +48 ..M:(x) = -480+48 for 857310

C)

(kN) Shear Force 48 O -24 2 - -58 1 1 4 8 10

Bending (kN-m) Moment 160 0 -96 - - - - - 4 8 10 x (m)

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