Calculate bearing forces

can someone tell me how to calculate the bearing forces of the two bearings? I can't get any further because I don't have the angle for the second bearing. I don't have an idea anymore.

F = 20KN
q0= 2000N/m
a = 1000mm

At the bottom under Lösungen are the results. • You can find the angle using trigonometry from the right angled triangle with sides a and 2a and hypotenuse along S. Jan 22 '19 at 21:06
• Why is $M_A=0$ crossed out and replaced with $M_A=200$? The support shown in the figure is clearly a pinned support. Should that be replaced with a fixed support?
– Wasabi
Jan 22 '19 at 21:08
• Also, is $q_0$ pushing upwards or downwards? I can't see the "arrowheads" usually drawn with distributed loads. I'm therefore assuming its upwards, since the load is given with a positive sign.
– Wasabi
Jan 22 '19 at 21:12
• Take moments about any point, say point a and set the three equilibrium equations sum Fx= sum Fz = sum Ma=0. you don't need to know any angles. Jan 22 '19 at 21:13
• With the trigonometric functions it does not work, because the side lengths are not a and 2a :( Jan 22 '19 at 21:13

Solutions from RISA2D:

Case I - Both supports are pinned; V(A) = 2.86 kN (pointing up), H(A) = 15.35 kN (pointing right), V(S) = 7.14 kN (pointing up), H(S) = 4.649 kN

Case II - Left support pinned; right support on roller; V(A) = 1 kN (Pointing up), H(A) = 20 kN (pointing right), V(S) = 9 kN (pointing up), H(S) = 0 kN

Lets take the moments about point a.

$$Ma = 5a*2000(4+5/2) - 20000a = 45000*a\ Nm \\ F_{vertical S}= 45000*a/5a = 9000N$$

$$F_{vertical}A = 10000-9000= 1000N$$

$$F_{S horizontal} = 9000*-1/2 = -4500N$$

$$F_{horizonatal A}= 4500N$$

• You are assuming the diagonal member has sides $a$ and $2a$, but according to the OP in the comments, that's not the case. Also, according to the results (and confirmed in the comments), $M_A = 200\text{ Nm}$, not zero (even though the drawn support at A is pinned). Therefore, none of your results match those given at the bottom of the image.
– Wasabi
Jan 22 '19 at 21:43
• @Wasabi, I went by the diagram. I may have missed some comments. Pin support is not capable of carrying moment. I will modify my answer when I go back home. Jan 22 '19 at 21:59
• Yeah, your answer is totally understandable, it's just apparently not the right one for the particular problem OP is having (which obviously isn't what it initially seems to be from the exercise as given).
– Wasabi
Jan 22 '19 at 22:03
• @Kamran. Your equation for Ma has two errors: (4+5/2) should be (4a+5a/2), and it is missing the moment due to Fs. Jan 24 '19 at 14:16
• @JohnHoltz, I don't find any errors in my answer. Jan 24 '19 at 15:50

Case I is a statically determinate system, you can get the Vertical force on support "S" by sum moments about support "A", which shall equate to zero.

Case II is statically indeterminate to the first degree. You shall conveniently release the horizontal restrain at support "S", which then becomes identical to case I. Then you have to calculate the horizontal displacement of support point "S" (d1), and apply an unit horizontal force at "S" to get an unit displacement (d2). Now you can get a new set of reactions by equating and proportioning d1 and d2 (sum displacement = 0). Finally, the reactions at supports "A" and "S" can be obtained by superposition using the two sets results. You shall review the structural analysis textbook on "solving structurally indeterminate structures by displacement method".